Explanation:
Okay, so I need to find the probability that segments AB and CD intersect when four points A, B, C, D are chosen randomly and independently on the circumference of a circle. Hmm, interesting problem. Let me think about how to approach this.

First, since all points are chosen randomly and independently on the circumference, I can assume that their positions are uniformly distributed. Also, because the circle is symmetric, maybe I can fix one point's position without loss of generality to simplify calculations. That often helps in circular probability problems.

Let me try that. Let's fix point A at some position on the circle, say at angle 0 radians. Then, points B, C, D are randomly placed around the circle. But since the circle is continuous, fixing one point doesn't affect the generality. Now, I need to consider the positions of B, C, D relative to A.

But wait, maybe it's better to think in terms of orderings of the points. When you have four points on a circle, there are different ways they can be arranged. For any four distinct points on a circle, there are (4-1)! = 6 possible cyclic orderings, but since the circle has no starting point, actually the number of distinct arrangements is (4-1)! / 2 = 3? Wait, maybe not. Let me recall: the number of ways to arrange n points on a circle is (n-1)! because rotations are considered the same. But if we fix one point, then the number of distinct arrangements of the remaining points is (n-1)! So for four points, fixing one, the number is 3! = 6. So there are 6 possible orderings.

But maybe this is overcomplicating. Let me think again. Since the points are chosen randomly, the probability that AB and CD intersect depends on the relative positions of the points.

I remember that two chords AB and CD intersect if and only if the points are arranged alternately around the circle. That is, if you go around the circle, you encounter A, then C, then B, then D, or some cyclic permutation of that. Alternatively, if the points are interleaved such that one chord straddles the other.

Wait, let me visualize this. Suppose we have four points on a circle. If the points are in the order A, C, B, D around the circle, then chord AB would cross chord CD. Similarly, if they are in order A, D, B, C, then AB and CD would cross. But if the points are ordered A, B, C, D, then AB and CD wouldn't cross. Similarly, if they are ordered A, C, D, B, then AB and CD would not cross? Wait, let me check.

Let me draw a mental picture. Suppose the points are placed in the order A, B, C, D around the circle. Then chord AB is between A and B, chord CD is between C and D. These two chords don't intersect because they are on opposite sides of the circle. If the points are ordered A, C, B, D, then chord AB goes from A to B, passing through the arc that goes through C, and chord CD goes from C to D, passing through B. Wait, so AB would cross CD?

Yes, because if you have points A, then C, then B, then D around the circle. Then chord AB would connect A to B, which is going through the arc that goes the long way around the circle, but chord CD connects C to D, which is also going the long way around. Wait, but actually chords are straight lines, so maybe they cross. Let me think again.

Imagine a circle with points A, C, B, D placed clockwise. So starting at A, going clockwise, we have A, then C, then B, then D, and back to A. So chord AB would be from A to B, which skips over C. Chord CD would be from C to D, which skips over B. But do these chords cross? If you draw AB and CD, they would cross each other inside the circle. Yes, because the endpoints of the chords are interleaved.

So in this case, AB and CD intersect. Similarly, if the order is A, D, B, C, then similar thing. But if the order is A, B, C, D, then chords AB and CD don't cross. Similarly, if the order is A, B, D, C, then AB and CD don't cross. Wait, let me check that.

If the order is A, B, D, C, then chord AB is between A and B, chord CD is between C and D. If you go clockwise around the circle, after A is B, then D, then C. So chord CD is from C to D, which is adjacent? Wait, no. If the order is A, B, D, C, then going clockwise from A, you have A, B, D, C. So chord CD connects C to D, which are adjacent? Wait, no. Because between D and C, there's no other points. Wait, but between D and C, going clockwise, you get from D to C through the arc that goes past A and B? Wait, no. If the points are placed in the order A, B, D, C around the circle, then chord CD connects C to D, which are adjacent? Wait, if you fix A at the top, then B is next clockwise, then D, then C, then back to A. So chord CD would be from C to D, which is adjacent? Wait, no, D is between B and C. So if the order is A, B, D, C, then between D and C, there are no other points. Therefore, chord CD is just an arc from D to C, which is adjacent? Wait, no. Wait, chord CD is a straight line between C and D, which are two points on the circle. If they are adjacent in the ordering, then chord CD is just a small arc between them. But chord AB is between A and B, which are also adjacent in the ordering. Then, chords AB and CD would not intersect because they are on different parts of the circle. So in this case, they don't cross.

So the key seems to be whether the points are interleaved or not. For chords AB and CD to intersect, the points must be arranged such that when you traverse the circle, you meet A, then C, then B, then D, or some rotation of that. Therefore, for each set of four points, there are two possible ways that chords AB and CD can intersect: either the points are arranged in the order A, C, B, D or A, D, B, C, or similar cyclic permutations?

Wait, but since we are talking about AB and CD, maybe there are two distinct ways for chords to intersect. Wait, actually, if we fix AB and CD, how many different arrangements result in AB and CD crossing?

Alternatively, maybe the probability can be calculated by considering all possible orderings of the four points and finding the fraction where AB and CD intersect.

So let's suppose we fix A. Then, the positions of B, C, D can be considered as three points on the circle. However, since the circle is continuous, the exact positions can be thought of as probabilities based on their orderings.

Alternatively, since all positions are random, we can think of the problem as arranging four points on a circle, and then determining the probability that chords AB and CD cross.

Another approach I recall is that for two chords to intersect, the four points must be arranged alternately. So if we have four points, there are three ways to pair them into two chords: AB and CD, AC and BD, AD and BC. For each pairing, the probability that they intersect is 1/3? Wait, no. Wait, perhaps not. Wait, maybe for any four points, there are three ways to pair them into two chords, and only one of those pairings results in intersecting chords. Wait, is that true?

Wait, suppose you have four points on a circle. How many ways are there to connect them with two chords? Well, the number of ways to partition four points into two pairs is 3. Let me check: the first point can pair with any of the other three, then the remaining two form the other pair. But since the order of the pairs doesn't matter, it's 3. So there are three possible pairings. For example, if the points are labeled A, B, C, D, then the pairings are (AB, CD), (AC, BD), (AD, BC). Now, for each pairing, whether the chords intersect depends on the order of the points.

Now, if the four points are in a random order around the circle, then for each of these three pairings, how many of them result in intersecting chords?

If the four points are in order A, B, C, D around the circle, then the pairings (AB, CD) don't intersect, (AC, BD) intersect, and (AD, BC) intersect. Wait, let me check:

- AB and CD: endpoints are adjacent pairs, so they don't intersect.
- AC and BD: AC connects A to C, BD connects B to D. These cross each other.
- AD and BC: AD connects A to D, BC connects B to C. These also cross each other?

Wait, if points are in order A, B, C, D around the circle:

- AD is from A to D, which is a chord that spans three arcs. BC is from B to C, adjacent. Do they cross? Let me visualize. If you connect A to D, which is a long chord, and B to C, which is a short chord. Depending on the circle's geometry, they might cross. Wait, actually, they don't cross. Because if you have A, B, C, D in order, then chord AD connects A to D, and chord BC connects B to C. These two chords do not intersect. Wait, but if you draw them, chord AD would go from A through the interior of the circle to D, and chord BC is between B and C, which is adjacent. So they don't cross. Wait, but then earlier I thought (AC, BD) and (AD, BC) cross? Wait, no:

Wait, let's take points in order A, B, C, D. Then:

- AC connects A to C, BD connects B to D. These two chords cross each other inside the circle. That's correct.

- AD connects A to D, BC connects B to C. These don't cross.

Wait, but then only one pairing (AC, BD) crosses? But earlier I thought (AD, BC) might cross? Maybe I was confused. Wait, if points are ordered A, B, C, D, then chord AD and BC do not cross. Therefore, among the three pairings, only one results in intersecting chords? But that seems contradictory to my earlier thought.

Wait, but maybe if the points are ordered differently. Wait, suppose the four points are ordered A, C, B, D around the circle. Then:

- AB connects A to B, CD connects C to D. Do they cross? If the order is A, C, B, D, then chord AB goes from A to B, passing through the arc from A to B which includes C? Wait, no. Wait, chord AB is a straight line between A and B. The order around the circle is A, C, B, D. So chord AB is between A and B, which are two points with C in between. Chord CD is between C and D. Do they cross? A to B and C to D. Since the points are ordered A, C, B, D, chord AB is from A to B (which is adjacent to A and B with C in between) and chord CD is from C to D (adjacent to C and D with B in between). Wait, so chords AB and CD would cross each other? Yes, because the endpoints are interleaved: A, C, B, D. So chord AB connects A to B, which skips C, and chord CD connects C to D, which skips B. These two chords cross each other. So in this case, pairing AB and CD would cross. So if the points are ordered A, C, B, D, then AB and CD cross. Whereas if they are ordered A, B, C, D, then AB and CD don't cross. So depending on the order, different pairings cross.

So for four points on the circle, how many of the three possible pairings result in intersecting chords? It depends on the order of the points. Wait, but if we fix the order of the four points, then only one of the three pairings will result in intersecting chords. Wait, let me check again.

Take order A, B, C, D:

- Pairings (AB, CD): don't cross

- (AC, BD): cross

- (AD, BC): don't cross

So only one crossing.

Take order A, B, D, C:

- Pairings (AB, CD): AB connects A to B, CD connects C to D. Since order is A, B, D, C. So chord AB is adjacent, CD is between D and C. Do they cross? If you imagine the points in order A, B, D, C. Then AB is between A and B, CD is between C and D. Since the points are ordered A, B, D, C, chord CD is from C to D, which is adjacent? No, because between D and C, there's no points. Wait, but if you have A, B, D, C on the circle, then moving clockwise, from A to B to D to C to A. So chord CD connects C to D, which are adjacent? No, D is before C, so chord CD connects D to C, which is adjacent? Wait, if you have points ordered A, B, D, C, then D and C are adjacent? Wait, maybe not. If you fix the order as A, B, D, C clockwise, then moving from D to C is clockwise? Wait, no. Let me clarify. Suppose you fix the circle with A at the top. Then going clockwise, you have A, then B, then D, then C, and back to A. So between D and C, moving clockwise, you go from D to C, which is a short arc. Then chord CD is from C to D, which would be a chord that spans from C to D, which is adjacent? Or is it... Wait, chord CD is between C and D, which are adjacent in the order? Wait, if the order is A, B, D, C, then D and C are adjacent? Yes. Because after D comes C. So chord CD is adjacent points? Then chord AB is also adjacent points. Therefore, chords AB and CD are both adjacent, so they don't cross. Then, what about pairings (AC, BD)?

AC connects A to C, BD connects B to D. Let's see. A, B, D, C. So chord AC connects A to C, which skips B and D. Chord BD connects B to D, which skips nothing? Wait, chord BD connects B to D, which skips nothing? Wait, between B and D, moving clockwise, you have D. Wait, the order is A, B, D, C. So chord BD connects B to D, which are two points with nothing between them? Wait, no. From B to D, there's D, but also, going the other way, from B to D counterclockwise, there is A and C. Wait, chord BD is a straight line between B and D. So depending on the positions, does BD cross AC?

AC connects A to C. Let's imagine: A is at the top, B is next clockwise, then D, then C. So chord AC is from A (top) to C, which is two positions clockwise from D, which is... Wait, maybe it's better to think of specific positions. Let me assign coordinates for clarity. Let me imagine the circle as a clock.

Let’s place A at 12 o'clock. Then, moving clockwise, B is at, say, 1 o'clock, D is at 2 o'clock, and C is at 11 o'clock. Wait, no. If the order is A, B, D, C, then going clockwise, starting at A (12), then B (say, 1), D (next, say 2), then C (next, 3). Wait, but that might not be the case. Wait, if the order is A, B, D, C, then moving clockwise from A to B to D to C to A. So if A is at 12, B at 1, D at 2, C at 3. Wait, but then going clockwise from C, we get back to A? So C is at 3, D at 2, B at 1, A at 12. So chord BD connects B (1) to D (2), which is adjacent? No, because between B (1) and D (2) is just the arc from 1 to 2. So chord BD is almost like a short chord from 1 to 2. Chord AC connects A (12) to C (3). That is a long chord spanning from 12 to 3, passing through the interior. Do these cross? The chord from 12 to 3 and the chord from 1 to 2? Probably not. The chord BD is close to the top, while AC is from top to 3, which might not cross BD. Wait, but maybe they do cross?

Wait, if A is at 12, B at 1, D at 2, and C at 3. Then chord AC is from 12 to 3. Chord BD is from 1 to 2. These two chords don't cross, because chord BD is near the top and chord AC spans from top to 3, which is the right side. So they don't cross. Then pairing (AC, BD) does not cross. What about (AD, BC)? AD connects A (12) to D (2), and BC connects B (1) to C (3). So chord AD is from 12 to 2, which is a chord going from top to 2 o'clock. Chord BC is from 1 to 3, which is a chord going from 1 to 3, which is similar to AD but shifted. Do these cross? If you draw them, probably yes. Because chord AD is from 12 to 2, chord BC is from 1 to 3. These two chords would cross somewhere inside the circle. Yes, they cross. Therefore, in this ordering, pairing (AD, BC) crosses, while (AB, CD) and (AC, BD) do not. Therefore, again, only one pairing crosses.

So in each case, only one of the three pairings results in crossing chords. Therefore, if you have four random points, the probability that a particular pairing (e.g., AB and CD) intersects is equal to the probability that the four points are arranged such that this particular pairing is the crossing one. Since there are three pairings, each with equal probability? Wait, but how?

Wait, if all orderings are equally likely, then for any four points, the probability that AB and CD intersect is 1/3? Because there are three ways to pair the points, and only one of them results in intersecting chords. But wait, is that the case? Wait, no, because depending on the order of the points, different pairings are interleaved. Wait, but if all arrangements are equally likely, then regardless of the pairing, each pairing is equally likely to be the one that crosses. Wait, maybe.

But in reality, when you fix the pairing, say AB and CD, the probability that they cross is not necessarily 1/3. Wait, maybe I need to think differently. Let me recall that for four points on a circle, the probability that two randomly chosen chords intersect is 1/3. Wait, but that might be the case. Wait, if there are three possible pairings, and only one of them crosses, then the probability is 1/3. Therefore, if the problem says that we fix AB and CD as the two chords, then the probability that they cross is 1/3. But is that the case?

Wait, but maybe not. Because when you fix AB and CD, the probability that they cross depends on the relative positions of the points. Wait, maybe the answer is 1/3. But I need to verify this.

Alternatively, here's another approach. Let me fix one point, say A, and consider the positions of the other three points. Since the circle is symmetric, we can fix A at a specific position without affecting the generality. Let me fix A at angle 0. Now, the positions of B, C, D are random on the circle. Let's parameterize their positions by angles θ_B, θ_C, θ_D, each uniformly distributed between 0 and 2π.

But since the circle is rotationally symmetric, we can fix A at 0, and then consider the positions of B, C, D relative to A. However, dealing with angles might complicate things. Maybe instead, think in terms of order statistics. Since the points are randomly placed, we can assume that their order around the circle can be represented as a random permutation. Since there are four points, after fixing A, the other three points can be ordered in 3! = 6 ways around the circle. But since the circle has no fixed direction, we can fix the order starting from A and going clockwise.

Wait, but if we fix A's position, then the remaining three points can be ordered in 3! = 6 ways around the circle. However, since the circle is symmetric, the actual position of A is fixed, and the other three points can be placed in any order. So there are 6 possible orderings of B, C, D around the circle. However, the chords AB and CD will intersect only in some of these orderings.

Let me think. Let's fix A at position 0. The other three points B, C, D are randomly placed around the circle. Let me imagine the circle divided into arcs by A and the other points. But maybe it's easier to think of the positions of B, C, D as three points on the circumference, and considering their order relative to A.

Alternatively, since the circle is continuous, we can fix A at 0, and assign the positions of B, C, D as three points uniformly random on [0, 1), representing the circle as circumference 1 for simplicity. Then, we can sort the positions of B, C, D in the order they appear clockwise from A. So the positions can be denoted as x, y, z where 0 < x < y < z < 1, but since the circle wraps around, we have to be careful. Actually, since A is fixed at 0, the other points divide the circle into arcs. The positions of B, C, D can be considered as three points in [0, 1), and we can sort them in the order they appear when going clockwise from A.

However, since the circle has no endpoints, the order can be thought of as the order when moving counterclockwise from A. Let me consider the circle as a unit interval [0, 1) with 0 and 1 identified. Fix A at 0. Then, the positions of B, C, D are three independent uniform points in [0, 1). Let me denote them as B, C, D. Without loss of generality, we can assume their positions are ordered as B < C < D. But since they are random, each permutation is equally likely. Wait, actually, the probability that B < C < D is 1/6, since there are 6 permutations of three points.

But perhaps instead of considering specific permutations, we can think in terms of the arcs between the points. For AB and CD to intersect, the points must be interleaved. That is, when moving around the circle from A, we should meet B, then C, then D, or some interleaving.

Wait, but since we have fixed A at 0, let me consider the positions of B, C, D. Let me think of the circle as a line from 0 to 1 with 0 and 1 identified. Let me imagine the three points B, C, D placed somewhere on this circle. The chord AB connects A (0) to B. The chord CD connects C to D. For these chords to intersect, the points B and D must lie in different arcs determined by C and the other points?

Wait, perhaps another way. For chords AB and CD to intersect, the points must be arranged such that one of the points is in the arc AC, and the other is in the arc BD or something? Wait, maybe not. Let me recall that two chords AB and CD intersect if and only if the points are interleaved. That is, if we traverse the circle, we meet A, then C, then B, then D, or A, D, B, C, etc. So given that A is fixed at 0, the interleaving condition can be translated into the order of B, C, D.

Wait, suppose that after fixing A at 0, we have three points B, C, D. Let's think about the order in which they appear going clockwise from A. The chords AB and CD will intersect if B and D are separated by C, meaning that one of them is on one side of C and the other is on the opposite side. Wait, more precisely, if when going around the circle from A, we encounter C first, then B, then D, or C, then D, then B, or similar. Wait, this is getting confusing. Maybe I need to use a specific method.

Another approach: for two chords AB and CD to intersect, the four points must be arranged such that one of the points is in the arc between A and B, and the other is in the arc between B and C, or something? Wait, perhaps I should use the concept of cyclic order.

Suppose we fix A at 0. Let me imagine three other points B, C, D randomly placed. Now, the chord AB divides the circle into two arcs: the arc from A to B clockwise and the arc from A to B counterclockwise. Similarly, chord CD divides the circle into two arcs. The chords AB and CD intersect if one endpoint of CD is on one arc defined by AB, and the other endpoint is on the other arc.

So for AB and CD to intersect, point C and D must lie on different arcs defined by AB. Since AB is from A (fixed at 0) to B. Let me think: suppose B is at position x in [0,1). The arc from A to B clockwise is [0, x), and the arc from B to A clockwise is [x, 1). For CD to cross AB, one of C or D must be in [0, x) and the other in [x, 1). Therefore, the probability that CD crosses AB is the probability that C and D are on different sides of B.

But since the positions of B, C, D are random, we can compute this probability by considering the position of B first.

Alternatively, since all positions are random, we can fix A at 0 and consider the positions of B, C, D as three independent uniform variables in [0,1). Let me denote them as X, Y, Z. Then, the probability that AB and CD intersect is equal to the probability that one of C or D is in the arc from A to B and the other is in the arc from B to A.

But since the circle is symmetric, the position of B is random. So perhaps we can compute the expectation over B's position.

Alternatively, here's a method: given four points on a circle, the probability that AB and CD intersect is equal to the number of ways the points can be arranged so that AB and CD intersect divided by the total number of arrangements.

But since the points are chosen randomly, all arrangements are equally likely. As previously thought, for four distinct points on a circle, there are (4-1)! = 6 possible cyclic orderings, but since the circle has no fixed starting point, we can fix one point, say A, and consider the order of the other three points. There are 3! = 6 possible orderings of B, C, D around the circle.

Out of these six orderings, how many result in AB and CD intersecting?

Let me list all possible orderings of B, C, D when A is fixed at 0:

1. B, C, D
2. B, D, C
3. C, B, D
4. C, D, B
5. D, B, C
6. D, C, B

Now, for each of these orderings, determine whether AB and CD intersect.

Let me consider each case:

Case 1: B, C, D. The order around the circle is A (0), B, C, D. Then chord AB connects A to B, chord CD connects C to D. These chords are adjacent? No, chord AB is from A to B, and CD is from C to D, which is on the opposite side. These chords don't intersect.

Case 2: B, D, C. Order is A, B, D, C. Chord AB is from A to B, CD is from C to D. Do they intersect? If points are ordered A, B, D, C, then chord AB is from A to B, and CD is from C to D. Since D is between B and C, chord CD is adjacent? Wait, chord CD is from C to D, which is adjacent? No, because from D to C is the arc that goes through A and B? Wait, in the order A, B, D, C, chord CD connects C to D, which are adjacent in the order? Wait, if the order is A, B, D, C, then going clockwise from D, you reach C. Therefore, D and C are adjacent? So chord CD is adjacent points? Then chord AB is adjacent points. Therefore, chords AB and CD don't intersect.

Wait, but earlier I thought that in some orderings they do intersect. Maybe I need a better way to check. Let me think of the general condition: AB and CD intersect if and only if C and D are on different arcs defined by AB. Given that AB divides the circle into two arcs. For CD to cross AB, one of C or D must be on one arc, and the other on the other arc.

But if B is fixed, and the order is A, B, D, C, then going clockwise from A, we have B, then D, then C. Therefore, the arc from A to B is [A, B], and the arc from B to A is [B, A]. Wait, but since the circle is continuous, maybe it's better to think in terms of the arrangement.

If the order is A, B, D, C, then chord AB is between A and B. Then chord CD is between C and D. If you look at the positions, C is after D when going clockwise from A. Wait, no. If the order is A, B, D, C, then starting at A, going clockwise: A -> B -> D -> C -> A. Therefore, point D is between B and C. Therefore, the arc from A to B is from A to B, and the arc from B to A going the other way is from B through D, C back to A. Therefore, chord CD connects C to D, which are both on the arc from B back to A. Therefore, chord CD is entirely on that arc. Therefore, chord AB is on the other arc. Therefore, chords AB and CD don't intersect.

Case 3: C, B, D. Order is A, C, B, D. Then chord AB is from A to B, chord CD is from C to D. Do they cross? In this order, going from A, you have C, then B, then D. So chord AB connects A to B, which skips C. Chord CD connects C to D, which skips B. Therefore, these chords cross. Yes, they intersect. So this case results in intersection.

Case 4: C, D, B. Order is A, C, D, B. Then chord AB connects A to B, which is from A to B through D and C? Wait, chord AB is a straight line between A and B. Chord CD connects C to D. In this order, points are A, C, D, B. So chord AB goes from A to B, which is on the arc from A to B passing through D and C? Wait, chord AB is a straight line. Depending on the circle's geometry, chord AB and CD would cross? Let me check.

Imagine points A (0), C (say 1/4), D (1/2), B (3/4). Then chord AB connects 0 to 3/4, chord CD connects 1/4 to 1/2. Do they cross? The chord from 0 to 3/4 would pass near the center, and chord from 1/4 to 1/2 is near the top. They might not cross. Wait, actually, in this specific case, they don't cross. Wait, but according to the order A, C, D, B, chord AB and CD. Is there a crossing?

Wait, maybe another visualization. If the points are placed in the order A, C, D, B around the circle, then chord AB is from A to B, which skips over C and D. Chord CD is from C to D, which are adjacent. So chord AB spans from A to B through the arc that includes C and D, and chord CD is between C and D. Depending on the exact positions, these chords might or might not cross? Wait, but if the points are in order A, C, D, B, then chord AB would cross chord CD? Let me think again.

If you have four points on a circle in the order A, C, D, B. Imagine drawing chord AB, which connects A to B, and chord CD connecting C to D. If you draw the circle with A at the top, then going clockwise, C, D, B. Chord AB would go from top to B, which is clockwise from D. Chord CD connects C to D, which is adjacent points. The chord AB would pass through the interior of the circle, and chord CD is a short chord near the top. It seems they might not cross. Wait, perhaps they cross? Hmm.

Alternatively, think of the crossing condition: chords AB and CD cross if and only if the points are interleaved. In other words, the order around the circle is A, C, B, D or A, D, B, C, etc. But in the case of order A, C, D, B, the order is A, C, D, B. If we check interleaving, A, then C, then D, then B. So the chords AB and CD: A and B are separated by C and D. So chord AB goes from A to B, which skips C and D. Chord CD is between C and D. These chords do not cross, because chord AB is outside the arc CD. Therefore, in this case, chords AB and CD do not cross.

So Case 4: No intersection.

Case 5: D, B, C. Order is A, D, B, C. Chord AB connects A to B. Chord CD connects C to D. Let's see. In this order, points are A, D, B, C. So chord AB connects A to B, which is adjacent to D and C? Wait, chord AB is from A to B, which are separated by D and C? Wait, going clockwise from A, we have D, then B, then C. Therefore, chord AB connects A to B, which skips D. Chord CD connects C to D, which are adjacent? No, C is after B, so D is before B. Wait, the order is A, D, B, C. So chord CD connects C to D. Since D is before B and C is after B. Therefore, chord CD connects C to D, which are separated by points B? So chord CD would cross chord AB? Let me check with coordinates. Let A be at 0, D at 1/4, B at 1/2, C at 3/4. Then chord AB is from 0 to 1/2, chord CD is from 3/4 to 1/4. These chords do cross. Yes, they cross at some point inside the circle. So in this case, chords AB and CD intersect.

Case 6: D, C, B. Order is A, D, C, B. Then chord AB connects A to B, chord CD connects C to D. Do they cross? In this order, points are A, D, C, B. Chord AB connects A to B, which skips D and C. Chord CD connects C to D, which are adjacent. Therefore, chord AB is from A to B, which spans over D and C, while chord CD is adjacent. Do they cross? If the points are A, D, C, B, chord AB is long, chord CD is short. They don't cross. For example, if coordinates are A=0, D=1/4, C=1/2, B=3/4. Chord AB is 0 to 3/4, chord CD is 1/4 to 1/2. These don't cross. So Case 6: No intersection.

So summarizing the six cases:

Case 1: No

Case 2: No

Case 3: Yes

Case 4: No

Case 5: Yes

Case 6: No

So out of six possible orderings, two result in AB and CD intersecting. Therefore, the probability is 2/6 = 1/3.

Wait, so that suggests the probability is 1/3. However, earlier I thought that for each pairing, there's only one crossing, but here with fixing A, we have two favorable cases out of six. But since the problem is about AB and CD, not about all pairings, maybe this is the correct probability?

But according to this, fixing A and considering all orderings of B, C, D, there is a 1/3 probability that AB and CD intersect. Therefore, the answer is 1/3. But wait, let me check another way to be sure.

Alternatively, since there are three pairings (AB-CD, AC-BD, AD-BC), each equally likely? If the four points are randomly arranged, then each of the three pairings has equal probability? If that's the case, then since only one of the pairings results in intersecting chords, then the probability would be 1/3. Wait, but earlier when I considered fix A, I found that for AB-CD, the probability is 1/3. Which seems to confirm that.

But let me think again. When I fixed A, and considered different orderings, and found that two of the six orderings result in AB and CD intersecting. But since the total number of orderings is 6, and each corresponds to different configurations, this suggests that the probability is 1/3.

Alternatively, another approach is to consider that for any four points on a circle, the probability that two specific chords intersect is 1/3. But how to derive that?

Another method: Fix the four points on the circle. Label them in the order they appear around the circle as P1, P2, P3, P4. There are 4! = 24 possible orderings, but because the circle has rotational symmetry, we can fix P1 at a certain position. Then, the other three points can be arranged in 3! = 6 ways. For each such arrangement, we can check if AB and CD intersect.

But here, AB and CD are specific chords. Wait, but in this case, the labels of the points are arbitrary. Wait, no. If we fix P1 as A, then the other points are B, C, D. Wait, but the problem is about four randomly chosen points. So the labels are random. Therefore, perhaps the probability is 1/3.

Wait, but with the previous calculation, we found that when we fix A and look at the orderings of B, C, D, 2 out of 6 result in AB and CD intersecting, which is 1/3. Therefore, the probability is 1/3.

But wait, another way to think about it: for two chords AB and CD to intersect, the points must be arranged so that A and B are separated by C and D. Which is equivalent to the points being interleaved.

Alternatively, once three points are randomly placed, the probability that the fourth point falls into the correct arc... Hmm.

Wait, here's a different method. Let me imagine that after placing A, B, C, D randomly on the circle, we can fix A at a point, and then consider the positions of B, C, D. Since the circle is continuous, we can fix A at 0, and consider the circle as [0,1) with 0 identified with 1.

The probability that AB and CD intersect is equal to the probability that, after placing B, C, D randomly, the points C and D are on opposite arcs determined by AB.

But AB divides the circle into two arcs. The probability that C and D are on different arcs is equal to the probability that one is on one arc and the other is on the other arc.

But since AB is a random chord, the length of the arc from A to B is random. However, since all positions are random, the probability that C and D are on opposite arcs of AB can be calculated.

Alternatively, let's consider that once A is fixed, B is randomly placed. Then, the probability that C and D are on different sides of AB is equal to 2*(probability that C is on one side and D on the other). Since the circle is continuous, the position of B can be anywhere.

But since all positions are random, the probability that C and D are on different sides of AB is equal to the probability that, given three points B, C, D, one is in the arc from A to B and the other is in the arc from B to A. Since the positions are random, this probability is the same as the probability that, among three points, two are on one side of a randomly placed line... Hmm, not sure.

Wait, perhaps think of it as follows. Fix A, then the positions of B, C, D are three random points. For AB and CD to intersect, the chord CD must cross AB. Which happens if one of C or D is in the arc from A to B and the other is in the arc from B to A.

But since the positions of B, C, D are random, we can imagine that for any fixed B, the probability that C and D are on opposite sides of AB depends on where B is. However, due to the symmetry, we can average over all possible positions of B.

Alternatively, since all four points are random, we can think of the probability that AB and CD intersect as equal to the probability that, when choosing four points, the points are arranged such that C and D are on opposite sides of AB. But how to calculate that?

Another approach is to note that for any four distinct points on a circle, there are three ways to pair them into two chords. Each pairing is equally likely. Therefore, since only one of the three pairings results in intersecting chords, the probability that a specific pairing (AB and CD) intersects is 1/3. Therefore, the answer is 1/3.

But in my earlier analysis, fixing A and considering the orderings of B, C, D, I found that the probability is 2/6 = 1/3. So that confirms it.

But let me check with another method. Suppose we fix A and B. The circle is divided into two arcs by AB. The probability that C and D are on different arcs is equal to the probability that one is on one arc and one on the other. Since the circle is continuous, the probability that both C and D are on the same arc is 2 * [probability that both are on arc AB] or [probability that both are on arc BA]. Wait, but actually, once AB is fixed, the circle is split into two arcs. Let me denote the arc from A to B as arc AB (length θ) and arc from B to A as arc BA (length 1 - θ, assuming the circle's circumference is 1). The probability that both C and D are on arc AB is θ^2, and similarly on arc BA is (1 - θ)^2. Therefore, the probability that both are on the same arc is θ^2 + (1 - θ)^2. Therefore, the probability that they are on different arcs is 1 - [θ^2 + (1 - θ)^2] = 1 - (2θ^2 - 2θ + 1) = 2θ - 2θ^2.

But θ is the length of arc AB. However, since A and B are randomly chosen, θ is uniformly distributed between 0 and 1? Wait, no. Actually, if we fix A at 0, B is uniformly random on [0, 1). Therefore, the length of arc AB is uniform in [0, 1), but θ is the minimum arc between A and B, so actually θ is between 0 and 0.5, but since we can consider the arc in either direction, it's symmetric. Wait, this might complicate things.

Alternatively, since the positions of B, C, D are all independent, we can compute the expected value of the probability over all positions of B.

Let me denote that the probability we want is E[2θ - 2θ^2], where θ is the length of the arc from A to B. But since A is fixed and B is random, θ is uniformly distributed between 0 and 1 (assuming the circle's circumference is 1). Wait, no, θ is the arc length between A and B in one direction, but since the circle is continuous, θ is uniform in [0, 1) if we consider the arc length from A to B going clockwise. However, in reality, θ can be from 0 to 1, but due to symmetry, we can assume θ is between 0 and 1/2, but no, since the arc can be in either direction.

But perhaps integrating over all possible positions of B. Let's consider θ as the position of B on the circle, so θ ∈ [0, 1). Then, the probability that C and D are on different arcs is 2θ - 2θ²? Wait, earlier I had the expression 2θ - 2θ², but that was for a specific θ. Wait, actually, when we fix θ, the length of arc AB is θ. The probability that C and D are on different arcs is 2θ(1 - θ). Because the probability that one is on the arc AB (length θ) and the other on arc BA (length 1 - θ) is 2θ(1 - θ).

Wait, let me verify. Given θ, the probability that one of C or D is on arc AB and the other is on arc BA is equal to 2 * [θ * (1 - θ)] because for one point to be in arc AB and the other in arc BA. Since C and D are independent, the probability that C is in AB and D is in BA is θ*(1 - θ). Similarly for D in AB and C in BA. So total probability is 2θ(1 - θ). Therefore, the overall probability is E[2θ(1 - θ)] where θ is the position of B, which is uniformly distributed over [0, 1).

Since θ is uniform over [0, 1), the expected value E[2θ(1 - θ)] = 2 ∫₀¹ θ(1 - θ) dθ = 2 ∫₀¹ (θ - θ²) dθ = 2 [½θ² - (1/3)θ³] from 0 to 1 = 2 [½ - 1/3] = 2 [1/6] = 1/3.

Therefore, the probability that AB and CD intersect is 1/3. This matches our previous results!

Therefore, this confirms that the probability is 1/3. So after going through multiple approaches—fixing points, considering orderings, integrating over arc lengths—the answer seems to be 1/3.

But wait, let me check another way. Suppose I use combinatorial counting. There are four points on a circle. The total number of ways to choose two chords is three, as previously mentioned. Each pair of chords is equally likely. So the probability that a specific pair (AB and CD) intersects is 1/3, since only one of the three pairings results in intersecting chords. Therefore, the answer is 1/3.

Alternatively, another way: For any four distinct points on a circle, there are three ways to pair them into two chords. The number of pairings that result in intersecting chords is one. Therefore, the probability is 1/3. Hence, the answer is 1/3.

However, I recall that in some probability problems involving chords, the probability is 1/3 or 1/2. Wait, perhaps I need to check with an example.

Alternatively, think of choosing four points at random. The probability that AB and CD intersect is equal to the number of intersecting configurations divided by total configurations. But since the total number of configurations is infinite, we need another approach.

But our multiple methods have given us 1/3. Therefore, I think that is the correct answer.

Wait, but according to some references, the probability that two random chords intersect is 1/3. So this seems to confirm.

Wait, but here's another thought. If we have four points, the probability that AB and CD intersect is equal to the probability that the points are interleaved. How many interleaved arrangements are there?

For four points on a circle, there are 3! = 6 possible orderings (after fixing one point). The number of interleaved arrangements for AB and CD is 2 (as we saw earlier). Therefore, 2/6 = 1/3. Hence, the answer is 1/3.

Therefore, after thorough analysis, I think the probability is 1/3.

But to ensure I haven't overlooked something, let me consider a specific example with coordinates.

Imagine four points on a circle: A at (1,0), B at (0,1), C at (-1,0), D at (0,-1). These are the four cardinal directions. Now, segments AB and CD: AB connects (1,0) to (0,1), CD connects (-1,0) to (0,-1). Do these segments intersect? Let's see. AB is the line from (1,0) to (0,1), which has equation x + y = 1. CD is the line from (-1,0) to (0,-1), which has equation x + y = -1. These are parallel lines? Wait, they have the same slope but different intercepts. Therefore, they don't intersect. So in this case, AB and CD don't intersect.

Another example: if points are placed in order A, C, B, D. Let's assign coordinates. Let me place A at (1,0), then go counterclockwise to C at (0,1), then to B at (-1,0), then to D at (0,-1), and back to A. Then, chord AB connects (1,0) to (-1,0), which is the horizontal diameter. Chord CD connects (0,1) to (0,-1), which is the vertical diameter. These two diameters intersect at the center. So in this case, AB and CD intersect.

So here, we have an example where they intersect. Another example: points in order A, B, C, D. Chords AB and CD: AB is from (1,0) to (0,1), CD is from (-1,0) to (0,-1). Wait, but if points are placed in order A, B, C, D, then if A is (1,0), B is (0,1), C is (-1,0), D is (0,-1). Then AB is (1,0)-(0,1), CD is (-1,0)-(0,-1). These are the same as before, which are parallel? Wait, no. Wait, AB is from (1,0) to (0,1): slope is (1-0)/(0-1) = -1. CD is from (-1,0) to (0,-1): slope is (-1 - 0)/(0 - (-1)) = -1/1 = -1. So they are parallel, hence don't intersect. Therefore, in this case, they don't intersect. But in the previous example, when the order was A, C, B, D, they intersected.

So depending on the order, they may or may not intersect. Therefore, the probability is 1/3, as calculated earlier.

Another way, for any four points, there are three ways to pair them into two chords. Each pairing has an equal chance of being selected. Therefore, the probability that a specific pairing (AB and CD) intersects is 1/3. Therefore, the answer is 1/3.

Hence, after all these verifications and methods, I conclude that the probability is 1/3.

**Final Answer**
The probability that segments $AB$ and $CD$ intersect is \boxed{\dfrac{1}{3}}.

To determine the probability that segments $ AB $ and $ CD $ intersect when four points $ A, B, C, D $ are chosen randomly and independently on a circle, we take a geometric and combinatorial approach.

---

### Understanding the Configuration

When four distinct points are selected randomly on the circumference of a circle, we can fix the position of one point (say $ A $) without loss of generality, due to the rotational symmetry of the circle. The remaining three points $ B, C, D $ are then randomly distributed on the circle’s perimeter.

The problem becomes: *What is the probability that the chords $ AB $ and $ CD $ intersect?*

---

### Key Insight: Chord Intersection Condition

Two chords $ AB $ and $ CD $ intersect **if and only if** the points $ A, B, C, D $ are arranged **alternately** around the circle. That is, when traversing the circle, the order of points should be something like:
- $ A \to C \to B \to D \to A $, or
- $ A \to D \to B \to C \to A $,
or any cyclic permutation of this, in which the endpoints of the chords are interleaved.

This is the geometric criterion for chord intersection: the chords intersect **if and only if** the points are interleaved around the circle.

---

### Counting the Total Arrangements

Since all points are chosen independently and uniformly, the **order of the four points** around the circle is equally likely to be any of the $ 4! = 24 $ permutations. However, due to rotational symmetry and the fact that only the **cyclic order** matters, we can fix one point (say $ A $) and count the number of **cyclic orderings** of the remaining three points $ B, C, D $.

There are $ 3! = 6 $ such cyclic orderings. In each case, the **labeling of points** (which defines which chord is $ AB $ and which is $ CD $) is a specific pair of points connected by chords. Since all selections are random, the specific pairings of the points into chords are equally likely.

---

### Counting the Favorable Intersections

We can analyze the number of favorable orderings (where chords $ AB $ and $ CD $ intersect) among the 6 possible cyclic orderings. As shown in the thought process, only two of the orderings result in intersecting chords $ AB $ and $ CD $. Let’s describe one of them:

- Points in order: $ A, C, B, D $.
  - Chord $ AB $ connects $ A $ to $ B $.
  - Chord $ CD $ connects $ C $ to $ D $.
  - The endpoints of the chords are interleaved: one chord "skip" one of the other chord's endpoints, causing them to cross.

There is another similar ordering that results in intersection. Thus, in 2 out of 6 cases, the chords intersect.

---

### Generalizing the Probability

Thus, for **any four points chosen at random**, the probability that the chords $ AB $ and $ CD $ intersect is equal to the ratio of favorable arrangements to total arrangements:

$$
\text{Probability} = \frac{\text{Number of favorable arrangements}}{\text{Total number of arrangements}} = \frac{2}{6} = \frac{1}{3}
$$

Alternatively, another way to think about it is by considering that there are three possible pairings of the four points into two chords (e.g., $ AB $ with $ CD $, $ AC $ with $ BD $, and $ AD $ with $ BC $). Among these, only **one** pairing results in intersection. Hence, the probability that a specific pairing intersects is again:

$$
\text{Probability} = \frac{1}{3}
$$

---

### Final Answer

$$
\boxed{\dfrac{1}{3}}
$$

[Assistant]
Explanation:
Okay, so I need to find the probability that segments AB and CD intersect when four points A, B, C, D are chosen randomly and independently on the circumference of a circle. Hmm, interesting problem. Let me think about how to approach this.

First, since all points are chosen randomly and independently on the circumference, I can assume that their positions are uniformly distributed. Also, because the circle is symmetric, maybe I can fix one point's position without loss of generality to simplify calculations. That often helps in circular probability problems.

Let me try that. Let's fix point A at some position on the circle, say at angle 0 radians. Then, points B, C, D are randomly placed around the circle. But since the circle is continuous, fixing one point doesn't affect the generality. Now, I need to consider the positions of B, C, D relative to A.

But wait, maybe it's better to think in terms of orderings of the points. When you have four points on a circle, there are different ways they can be arranged. For any four distinct points on a circle, there are (4-1)! = 6 possible cyclic orderings, but since the circle has no starting point, actually the number of distinct arrangements is (4-1)! / 2 = 3? Wait, maybe not. Let me recall: the number of ways to arrange n points on a circle is (n-1)! because rotations are considered the same. But if we fix one point, then the number of distinct arrangements of the remaining points is (n-1)! So for four points, fixing one, the number is 3! = 6. So there are 6 possible orderings.

But maybe this is overcomplicating. Let me think again. Since the points are chosen randomly, the probability that AB and CD intersect depends on the relative positions of the points.

I remember that two chords AB and CD intersect if and only if the points are arranged alternately around the circle. That is, if you go around the circle, you encounter A, then C, then B, then D, or some cyclic permutation of that. Alternatively, if the points are interleaved such that one chord straddles the other.

Wait, let me visualize this. Suppose we have four points on a circle. If the points are in the order A, C, B, D around the circle, then chord AB would cross chord CD. Similarly, if they are in order A, D, B, C, then AB and CD would cross. But if the points are ordered A, B, C, D, then AB and CD wouldn't cross. Similarly, if they are ordered A, C, D, B, then AB and CD would not cross? Wait, let me check.

Let me draw a mental picture. Suppose the points are placed in the order A, B, C, D around the circle. Then chord AB is between A and B, chord CD is between C and D. These two chords don't intersect because they are on opposite sides of the circle. If the points are ordered A, C, B, D, then chord AB goes from A to B, passing through the arc that goes through C, and chord CD goes from C to D, passing through B. Wait, so AB would cross CD?

Yes, because if you have points A, then C, then B, then D around the circle. Then chord AB would connect A to B, which is going through the arc that goes the long way around the circle, but chord CD connects C to D, which is also going the long way around. Wait, but actually chords are straight lines, so maybe they cross. Let me think again.

Imagine a circle with points A, C, B, D placed clockwise. So starting at A, going clockwise, we have A, then C, then B, then D, and back to A. So chord AB would be from A to B, which skips over C. Chord CD would be from C to D, which skips over B. But do these chords cross? If you draw AB and CD, they would cross each other inside the circle. Yes, because the endpoints of the chords are interleaved.

So in this case, AB and CD intersect. Similarly, if the order is A, D, B, C, then similar thing. But if the order is A, B, C, D, then chords AB and CD don't cross. Similarly, if the order is A, B, D, C, then AB and CD don't cross. Wait, let me check that.

If the order is A, B, D, C, then chord AB is between A and B, chord CD is between C and D. If you go clockwise around the circle, after A is B, then D, then C. So chord CD is from C to D, which is adjacent? Wait, no. If the order is A, B, D, C, then going clockwise from A, you have A, B, D, C. So chord CD connects C to D, which are adjacent? Wait, no. Because between D and C, there's no other points. Wait, but between D and C, going clockwise, you get from D to C through the arc that goes past A and B? Wait, no. If the points are placed in the order A, B, D, C around the circle, then chord CD connects C to D, which are adjacent? Wait, if you fix A at the top, then B is next clockwise, then D, then C, then back to A. So chord CD would be from C to D, which is adjacent? Wait, no, D is between B and C. So if the order is A, B, D, C, then between D and C, there are no other points. Therefore, chord CD is just an arc from D to C, which is adjacent? Wait, no. Wait, chord CD is a straight line between C and D, which are two points on the circle. If they are adjacent in the ordering, then chord CD is just a small arc between them. But chord AB is between A and B, which are also adjacent in the ordering. Then, chords AB and CD would not intersect because they are on different parts of the circle. So in this case, they don't cross.

So the key seems to be whether the points are interleaved or not. For chords AB and CD to intersect, the points must be arranged such that when you traverse the circle, you meet A, then C, then B, then D, or some rotation of that. Therefore, for each set of four points, there are two possible ways that chords AB and CD can intersect: either the points are arranged in the order A, C, B, D or A, D, B, C, or similar cyclic permutations?

Wait, but since we are talking about AB and CD, maybe there are two distinct ways for chords to intersect. Wait, actually, if we fix AB and CD, how many different arrangements result in AB and CD crossing?

Alternatively, maybe the probability can be calculated by considering all possible orderings of the four points and finding the fraction where AB and CD intersect.

So let's suppose we fix A. Then, the positions of B, C, D can be considered as three points on the circle. However, since the circle is continuous, the exact positions can be thought of as probabilities based on their orderings.

Alternatively, since all positions are random, we can think of the problem as arranging four points on a circle, and then determining the probability that chords AB and CD cross.

Another approach I recall is that for two chords to intersect, the four points must be arranged alternately. So if we have four points, there are three ways to pair them into two chords: AB and CD, AC and BD, AD and BC. For each pairing, the probability that they intersect is 1/3? Wait, no. Wait, perhaps not. Wait, maybe for any four points, there are three ways to pair them into two chords, and only one of those pairings results in intersecting chords. Wait, is that true?

Wait, suppose you have four points on a circle. How many ways are there to connect them with two chords? Well, the number of ways to partition four points into two pairs is 3. Let me check: the first point can pair with any of the other three, then the remaining two form the other pair. But since the order of the pairs doesn't matter, it's 3. So there are three possible pairings. For example, if the points are labeled A, B, C, D, then the pairings are (AB, CD), (AC, BD), (AD, BC). Now, for each pairing, whether the chords intersect depends on the order of the points.

Now, if the four points are in a random order around the circle, then for each of these three pairings, how many of them result in intersecting chords?

If the four points are in order A, B, C, D around the circle, then the pairings (AB, CD) don't intersect, (AC, BD) intersect, and (AD, BC) intersect. Wait, let me check:

- AB and CD: endpoints are adjacent pairs, so they don't intersect.
- AC and BD: AC connects A to C, BD connects B to D. These cross each other.
- AD and BC: AD connects A to D, BC connects B to C. These also cross each other?

Wait, if points are in order A, B, C, D around the circle:

- AD is from A to D, which is a chord that spans three arcs. BC is from B to C, adjacent. Do they cross? Let me visualize. If you connect A to D, which is a long chord, and B to C, which is a short chord. Depending on the circle's geometry, they might cross. Wait, actually, they don't cross. Because if you have A, B, C, D in order, then chord AD connects A to D, and chord BC connects B to C. These two chords do not intersect. Wait, but if you draw them, chord AD would go from A through the interior of the circle to D, and chord BC is between B and C, which is adjacent. So they don't cross. Wait, but then earlier I thought (AC, BD) and (AD, BC) cross? Wait, no:

Wait, let's take points in order A, B, C, D. Then:

- AC connects A to C, BD connects B to D. These two chords cross each other inside the circle. That's correct.

- AD connects A to D, BC connects B to C. These don't cross.

Wait, but then only one pairing (AC, BD) crosses? But earlier I thought (AD, BC) might cross? Maybe I was confused. Wait, if points are ordered A, B, C, D, then chord AD and BC do not cross. Therefore, among the three pairings, only one results in intersecting chords? But that seems contradictory to my earlier thought.

Wait, but maybe if the points are ordered differently. Wait, suppose the four points are ordered A, C, B, D around the circle. Then:

- AB connects A to B, CD connects C to D. Do they cross? If the order is A, C, B, D, then chord AB goes from A to B, passing through the arc from A to B which includes C? Wait, no. Wait, chord AB is a straight line between A and B. The order around the circle is A, C, B, D. So chord AB is between A and B, which are two points with C in between. Chord CD is between C and D. Do they cross? A to B and C to D. Since the points are ordered A, C, B, D, chord AB is from A to B (which is adjacent to A and B with C in between) and chord CD is from C to D (adjacent to C and D with B in between). Wait, so chords AB and CD would cross each other? Yes, because the endpoints are interleaved: A, C, B, D. So chord AB connects A to B, which skips C, and chord CD connects C to D, which skips B. These two chords cross each other. So in this case, pairing AB and CD would cross. So if the points are ordered A, C, B, D, then AB and CD cross. Whereas if they are ordered A, B, C, D, then AB and CD don't cross. So depending on the order, different pairings cross.

So for four points on the circle, how many of the three possible pairings result in intersecting chords? It depends on the order of the points. Wait, but if we fix the order of the four points, then only one of the three pairings will result in intersecting chords. Wait, let me check again.

Take order A, B, C, D:

- Pairings (AB, CD): don't cross

- (AC, BD): cross

- (AD, BC): don't cross

So only one crossing.

Take order A, B, D, C:

- Pairings (AB, CD): AB connects A to B, CD connects C to D. Since order is A, B, D, C. So chord AB is adjacent, CD is between D and C. Do they cross? If you imagine the points in order A, B, D, C. Then AB is between A and B, CD is between C and D. Since the points are ordered A, B, D, C, chord CD is from C to D, which is adjacent? No, because between D and C, there's no points. Wait, but if you have A, B, D, C on the circle, then moving clockwise, from A to B to D to C to A. So chord CD connects C to D, which are adjacent? No, D is before C, so chord CD connects D to C, which is adjacent? Wait, if you have points ordered A, B, D, C, then D and C are adjacent? Wait, maybe not. If you fix the order as A, B, D, C clockwise, then moving from D to C is clockwise? Wait, no. Let me clarify. Suppose you fix the circle with A at the top. Then going clockwise, you have A, then B, then D, then C, and back to A. So between D and C, moving clockwise, you go from D to C, which is a short arc. Then chord CD is from C to D, which would be a chord that spans from C to D, which is adjacent? Or is it... Wait, chord CD is between C and D, which are adjacent in the order? Wait, if the order is A, B, D, C, then D and C are adjacent? Yes. Because after D comes C. So chord CD is adjacent points? Then chord AB is also adjacent points. Therefore, chords AB and CD are both adjacent, so they don't cross. Then, what about pairings (AC, BD)?

AC connects A to C, BD connects B to D. Let's see. A, B, D, C. So chord AC connects A to C, which skips B and D. Chord BD connects B to D, which skips nothing? Wait, chord BD connects B to D, which skips nothing? Wait, between B and D, moving clockwise, you have D. Wait, the order is A, B, D, C. So chord BD connects B to D, which are two points with nothing between them? Wait, no. From B to D, there's D, but also, going the other way, from B to D counterclockwise, there is A and C. Wait, chord BD is a straight line between B and D. So depending on the positions, does BD cross AC?

AC connects A to C. Let's imagine: A is at the top, B is next clockwise, then D, then C. So chord AC is from A (top) to C, which is two positions clockwise from D, which is... Wait, maybe it's better to think of specific positions. Let me assign coordinates for clarity. Let me imagine the circle as a clock.

Let’s place A at 12 o'clock. Then, moving clockwise, B is at, say, 1 o'clock, D is at 2 o'clock, and C is at 11 o'clock. Wait, no. If the order is A, B, D, C, then going clockwise, starting at A (12), then B (say, 1), D (next, say 2), then C (next, 3). Wait, but that might not be the case. Wait, if the order is A, B, D, C, then moving clockwise from A to B to D to C to A. So if A is at 12, B at 1, D at 2, C at 3. Wait, but then going clockwise from C, we get back to A? So C is at 3, D at 2, B at 1, A at 12. So chord BD connects B (1) to D (2), which is adjacent? No, because between B (1) and D (2) is just the arc from 1 to 2. So chord BD is almost like a short chord from 1 to 2. Chord AC connects A (12) to C (3). That is a long chord spanning from 12 to 3, passing through the interior. Do these cross? The chord from 12 to 3 and the chord from 1 to 2? Probably not. The chord BD is close to the top, while AC is from top to 3, which might not cross BD. Wait, but maybe they do cross?

Wait, if A is at 12, B at 1, D at 2, and C at 3. Then chord AC is from 12 to 3. Chord BD is from 1 to 2. These two chords don't cross, because chord BD is near the top and chord AC spans from top to 3, which is the right side. So they don't cross. Then pairing (AC, BD) does not cross. What about (AD, BC)? AD connects A (12) to D (2), and BC connects B (1) to C (3). So chord AD is from 12 to 2, which is a chord going from top to 2 o'clock. Chord BC is from 1 to 3, which is a chord going from 1 to 3, which is similar to AD but shifted. Do these cross? If you draw them, probably yes. Because chord AD is from 12 to 2, chord BC is from 1 to 3. These two chords would cross somewhere inside the circle. Yes, they cross. Therefore, in this ordering, pairing (AD, BC) crosses, while (AB, CD) and (AC, BD) do not. Therefore, again, only one pairing crosses.

So in each case, only one of the three pairings results in crossing chords. Therefore, if you have four random points, the probability that a particular pairing (e.g., AB and CD) intersects is equal to the probability that the four points are arranged such that this particular pairing is the crossing one. Since there are three pairings, each with equal probability? Wait, but how?

Wait, if all orderings are equally likely, then for any four points, the probability that AB and CD intersect is 1/3? Because there are three ways to pair the points, and only one of them results in intersecting chords. But wait, is that the case? Wait, no, because depending on the order of the points, different pairings are interleaved. Wait, but if all arrangements are equally likely, then regardless of the pairing, each pairing is equally likely to be the one that crosses. Wait, maybe.

But in reality, when you fix the pairing, say AB and CD, the probability that they cross is not necessarily 1/3. Wait, maybe I need to think differently. Let me recall that for four points on a circle, the probability that two randomly chosen chords intersect is 1/3. Wait, but that might be the case. Wait, if there are three possible pairings, and only one of them crosses, then the probability is 1/3. Therefore, if the problem says that we fix AB and CD as the two chords, then the probability that they cross is 1/3. But is that the case?

Wait, but maybe not. Because when you fix AB and CD, the probability that they cross depends on the relative positions of the points. Wait, maybe the answer is 1/3. But I need to verify this.

Alternatively, here's another approach. Let me fix one point, say A, and consider the positions of the other three points. Since the circle is symmetric, we can fix A at a specific position without affecting the generality. Let me fix A at angle 0. Now, the positions of B, C, D are random on the circle. Let's parameterize their positions by angles θ_B, θ_C, θ_D, each uniformly distributed between 0 and 2π.

But since the circle is rotationally symmetric, we can fix A at 0, and then consider the positions of B, C, D relative to A. However, dealing with angles might complicate things. Maybe instead, think in terms of order statistics. Since the points are randomly placed, we can assume that their order around the circle can be represented as a random permutation. Since there are four points, after fixing A, the other three points can be ordered in 3! = 6 ways around the circle. But since the circle has no fixed direction, we can fix the order starting from A and going clockwise.

Wait, but if we fix A's position, then the remaining three points can be ordered in 3! = 6 ways around the circle. However, since the circle is symmetric, the actual position of A is fixed, and the other three points can be placed in any order. So there are 6 possible orderings of B, C, D around the circle. However, the chords AB and CD will intersect only in some of these orderings.

Let me think. Let's fix A at position 0. The other three points B, C, D are randomly placed around the circle. Let me imagine the circle divided into arcs by A and the other points. But maybe it's easier to think of the positions of B, C, D as three points on the circumference, and considering their order relative to A.

Alternatively, since the circle is continuous, we can fix A at 0, and assign the positions of B, C, D as three points uniformly random on [0, 1), representing the circle as circumference 1 for simplicity. Then, we can sort the positions of B, C, D in the order they appear clockwise from A. So the positions can be denoted as x, y, z where 0 < x < y < z < 1, but since the circle wraps around, we have to be careful. Actually, since A is fixed at 0, the other points divide the circle into arcs. The positions of B, C, D can be considered as three points in [0, 1), and we can sort them in the order they appear when going clockwise from A.

However, since the circle has no endpoints, the order can be thought of as the order when moving counterclockwise from A. Let me consider the circle as a unit interval [0, 1) with 0 and 1 identified. Fix A at 0. Then, the positions of B, C, D are three independent uniform points in [0, 1). Let me denote them as B, C, D. Without loss of generality, we can assume their positions are ordered as B < C < D. But since they are random, each permutation is equally likely. Wait, actually, the probability that B < C < D is 1/6, since there are 6 permutations of three points.

But perhaps instead of considering specific permutations, we can think in terms of the arcs between the points. For AB and CD to intersect, the points must be interleaved. That is, when moving around the circle from A, we should meet B, then C, then D, or some interleaving.

Wait, but since we have fixed A at 0, let me consider the positions of B, C, D. Let me think of the circle as a line from 0 to 1 with 0 and 1 identified. Let me imagine the three points B, C, D placed somewhere on this circle. The chord AB connects A (0) to B. The chord CD connects C to D. For these chords to intersect, the points B and D must lie in different arcs determined by C and the other points?

Wait, perhaps another way. For chords AB and CD to intersect, the points must be arranged such that one of the points is in the arc AC, and the other is in the arc BD or something? Wait, maybe not. Let me recall that two chords AB and CD intersect if and only if the points are interleaved. That is, if we traverse the circle, we meet A, then C, then B, then D, or A, D, B, C, etc. So given that A is fixed at 0, the interleaving condition can be translated into the order of B, C, D.

Wait, suppose that after fixing A at 0, we have three points B, C, D. Let's think about the order in which they appear going clockwise from A. The chords AB and CD will intersect if B and D are separated by C, meaning that one of them is on one side of C and the other is on the opposite side. Wait, more precisely, if when going around the circle from A, we encounter C first, then B, then D, or C, then D, then B, or similar. Wait, this is getting confusing. Maybe I need to use a specific method.

Another approach: for two chords AB and CD to intersect, the four points must be arranged such that one of the points is in the arc between A and B, and the other is in the arc between B and C, or something? Wait, perhaps I should use the concept of cyclic order.

Suppose we fix A at 0. Let me imagine three other points B, C, D randomly placed. Now, the chord AB divides the circle into two arcs: the arc from A to B clockwise and the arc from A to B counterclockwise. Similarly, chord CD divides the circle into two arcs. The chords AB and CD intersect if one endpoint of CD is on one arc defined by AB, and the other endpoint is on the other arc.

So for AB and CD to intersect, point C and D must lie on different arcs defined by AB. Since AB is from A (fixed at 0) to B. Let me think: suppose B is at position x in [0,1). The arc from A to B clockwise is [0, x), and the arc from B to A clockwise is [x, 1). For CD to cross AB, one of C or D must be in [0, x) and the other in [x, 1). Therefore, the probability that CD crosses AB is the probability that C and D are on different sides of B.

But since the positions of B, C, D are random, we can compute this probability by considering the position of B first.

Alternatively, since all positions are random, we can fix A at 0 and consider the positions of B, C, D as three independent uniform variables in [0,1). Let me denote them as X, Y, Z. Then, the probability that AB and CD intersect is equal to the probability that one of C or D is in the arc from A to B and the other is in the arc from B to A.

But since the circle is symmetric, the position of B is random. So perhaps we can compute the expectation over B's position.

Alternatively, here's a method: given four points on a circle, the probability that AB and CD intersect is equal to the number of ways the points can be arranged so that AB and CD intersect divided by the total number of arrangements.

But since the points are chosen randomly, all arrangements are equally likely. As previously thought, for four distinct points on a circle, there are (4-1)! = 6 possible cyclic orderings, but since the circle has no fixed starting point, we can fix one point, say A, and consider the order of the other three points. There are 3! = 6 possible orderings of B, C, D around the circle.

Out of these six orderings, how many result in AB and CD intersecting?

Let me list all possible orderings of B, C, D when A is fixed at 0:

1. B, C, D
2. B, D, C
3. C, B, D
4. C, D, B
5. D, B, C
6. D, C, B

Now, for each of these orderings, determine whether AB and CD intersect.

Let me consider each case:

Case 1: B, C, D. The order around the circle is A (0), B, C, D. Then chord AB connects A to B, chord CD connects C to D. These chords are adjacent? No, chord AB is from A to B, and CD is from C to D, which is on the opposite side. These chords don't intersect.

Case 2: B, D, C. Order is A, B, D, C. Chord AB is from A to B, CD is from C to D. Do they intersect? If points are ordered A, B, D, C, then chord AB is from A to B, and CD is from C to D. Since D is between B and C, chord CD is adjacent? Wait, chord CD is from C to D, which is adjacent? No, because from D to C is the arc that goes through A and B? Wait, in the order A, B, D, C, chord CD connects C to D, which are adjacent in the order? Wait, if the order is A, B, D, C, then going clockwise from D, you reach C. Therefore, D and C are adjacent? So chord CD is adjacent points? Then chord AB is adjacent points. Therefore, chords AB and CD don't intersect.

Wait, but earlier I thought that in some orderings they do intersect. Maybe I need a better way to check. Let me think of the general condition: AB and CD intersect if and only if C and D are on different arcs defined by AB. Given that AB divides the circle into two arcs. For CD to cross AB, one of C or D must be on one arc, and the other on the other arc.

But if B is fixed, and the order is A, B, D, C, then going clockwise from A, we have B, then D, then C. Therefore, the arc from A to B is [A, B], and the arc from B to A is [B, A]. Wait, but since the circle is continuous, maybe it's better to think in terms of the arrangement.

If the order is A, B, D, C, then chord AB is between A and B. Then chord CD is between C and D. If you look at the positions, C is after D when going clockwise from A. Wait, no. If the order is A, B, D, C, then starting at A, going clockwise: A -> B -> D -> C -> A. Therefore, point D is between B and C. Therefore, the arc from A to B is from A to B, and the arc from B to A going the other way is from B through D, C back to A. Therefore, chord CD connects C to D, which are both on the arc from B back to A. Therefore, chord CD is entirely on that arc. Therefore, chord AB is on the other arc. Therefore, chords AB and CD don't intersect.

Case 3: C, B, D. Order is A, C, B, D. Then chord AB is from A to B, chord CD is from C to D. Do they cross? In this order, going from A, you have C, then B, then D. So chord AB connects A to B, which skips C. Chord CD connects C to D, which skips B. Therefore, these chords cross. Yes, they intersect. So this case results in intersection.

Case 4: C, D, B. Order is A, C, D, B. Then chord AB connects A to B, which is from A to B through D and C? Wait, chord AB is a straight line between A and B. Chord CD connects C to D. In this order, points are A, C, D, B. So chord AB goes from A to B, which is on the arc from A to B passing through D and C? Wait, chord AB is a straight line. Depending on the circle's geometry, chord AB and CD would cross? Let me check.

Imagine points A (0), C (say 1/4), D (1/2), B (3/4). Then chord AB connects 0 to 3/4, chord CD connects 1/4 to 1/2. Do they cross? The chord from 0 to 3/4 would pass near the center, and chord from 1/4 to 1/2 is near the top. They might not cross. Wait, actually, in this specific case, they don't cross. Wait, but according to the order A, C, D, B, chord AB and CD. Is there a crossing?

Wait, maybe another visualization. If the points are placed in the order A, C, D, B around the circle, then chord AB is from A to B, which skips over C and D. Chord CD is from C to D, which are adjacent. So chord AB spans from A to B through the arc that includes C and D, and chord CD is between C and D. Depending on the exact positions, these chords might or might not cross? Wait, but if the points are in order A, C, D, B, then chord AB would cross chord CD? Let me think again.

If you have four points on a circle in the order A, C, D, B. Imagine drawing chord AB, which connects A to B, and chord CD connecting C to D. If you draw the circle with A at the top, then going clockwise, C, D, B. Chord AB would go from top to B, which is clockwise from D. Chord CD connects C to D, which is adjacent points. The chord AB would pass through the interior of the circle, and chord CD is a short chord near the top. It seems they might not cross. Wait, perhaps they cross? Hmm.

Alternatively, think of the crossing condition: chords AB and CD cross if and only if the points are interleaved. In other words, the order around the circle is A, C, B, D or A, D, B, C, etc. But in the case of order A, C, D, B, the order is A, C, D, B. If we check interleaving, A, then C, then D, then B. So the chords AB and CD: A and B are separated by C and D. So chord AB goes from A to B, which skips C and D. Chord CD is between C and D. These chords do not cross, because chord AB is outside the arc CD. Therefore, in this case, chords AB and CD do not cross.

So Case 4: No intersection.

Case 5: D, B, C. Order is A, D, B, C. Chord AB connects A to B. Chord CD connects C to D. Let's see. In this order, points are A, D, B, C. So chord AB connects A to B, which is adjacent to D and C? Wait, chord AB is from A to B, which are separated by D and C? Wait, going clockwise from A, we have D, then B, then C. Therefore, chord AB connects A to B, which skips D. Chord CD connects C to D, which are adjacent? No, C is after B, so D is before B. Wait, the order is A, D, B, C. So chord CD connects C to D. Since D is before B and C is after B. Therefore, chord CD connects C to D, which are separated by points B? So chord CD would cross chord AB? Let me check with coordinates. Let A be at 0, D at 1/4, B at 1/2, C at 3/4. Then chord AB is from 0 to 1/2, chord CD is from 3/4 to 1/4. These chords do cross. Yes, they cross at some point inside the circle. So in this case, chords AB and CD intersect.

Case 6: D, C, B. Order is A, D, C, B. Then chord AB connects A to B, chord CD connects C to D. Do they cross? In this order, points are A, D, C, B. Chord AB connects A to B, which skips D and C. Chord CD connects C to D, which are adjacent. Therefore, chord AB is from A to B, which spans over D and C, while chord CD is adjacent. Do they cross? If the points are A, D, C, B, chord AB is long, chord CD is short. They don't cross. For example, if coordinates are A=0, D=1/4, C=1/2, B=3/4. Chord AB is 0 to 3/4, chord CD is 1/4 to 1/2. These don't cross. So Case 6: No intersection.

So summarizing the six cases:

Case 1: No

Case 2: No

Case 3: Yes

Case 4: No

Case 5: Yes

Case 6: No

So out of six possible orderings, two result in AB and CD intersecting. Therefore, the probability is 2/6 = 1/3.

Wait, so that suggests the probability is 1/3. However, earlier I thought that for each pairing, there's only one crossing, but here with fixing A, we have two favorable cases out of six. But since the problem is about AB and CD, not about all pairings, maybe this is the correct probability?

But according to this, fixing A and considering all orderings of B, C, D, there is a 1/3 probability that AB and CD intersect. Therefore, the answer is 1/3. But wait, let me check another way to be sure.

Alternatively, since there are three pairings (AB-CD, AC-BD, AD-BC), each equally likely? If the four points are randomly arranged, then each of the three pairings has equal probability? If that's the case, then since only one of the pairings results in intersecting chords, then the probability would be 1/3. Wait, but earlier when I considered fix A, I found that for AB-CD, the probability is 1/3. Which seems to confirm that.

But let me think again. When I fixed A, and considered different orderings, and found that two of the six orderings result in AB and CD intersecting. But since the total number of orderings is 6, and each corresponds to different configurations, this suggests that the probability is 1/3.

Alternatively, another approach is to consider that for any four points on a circle, the probability that two specific chords intersect is 1/3. But how to derive that?

Another method: Fix the four points on the circle. Label them in the order they appear around the circle as P1, P2, P3, P4. There are 4! = 24 possible orderings, but because the circle has rotational symmetry, we can fix P1 at a certain position. Then, the other three points can be arranged in 3! = 6 ways. For each such arrangement, we can check if AB and CD intersect.

But here, AB and CD are specific chords. Wait, but in this case, the labels of the points are arbitrary. Wait, no. If we fix P1 as A, then the other points are B, C, D. Wait, but the problem is about four randomly chosen points. So the labels are random. Therefore, perhaps the probability is 1/3.

Wait, but with the previous calculation, we found that when we fix A and look at the orderings of B, C, D, 2 out of 6 result in AB and CD intersecting, which is 1/3. Therefore, the probability is 1/3.

But wait, another way to think about it: for two chords AB and CD to intersect, the points must be arranged so that A and B are separated by C and D. Which is equivalent to the points being interleaved.

Alternatively, once three points are randomly placed, the probability that the fourth point falls into the correct arc... Hmm.

Wait, here's a different method. Let me imagine that after placing A, B, C, D randomly on the circle, we can fix A at a point, and then consider the positions of B, C, D. Since the circle is continuous, we can fix A at 0, and consider the circle as [0,1) with 0 identified with 1.

The probability that AB and CD intersect is equal to the probability that, after placing B, C, D randomly, the points C and D are on opposite arcs determined by AB.

But AB divides the circle into two arcs. The probability that C and D are on different arcs is equal to the probability that one is on one arc and the other is on the other arc.

But since AB is a random chord, the length of the arc from A to B is random. However, since all positions are random, the probability that C and D are on opposite arcs of AB can be calculated.

Alternatively, let's consider that once A is fixed, B is randomly placed. Then, the probability that C and D are on different sides of AB is equal to 2*(probability that C is on one side and D on the other). Since the circle is continuous, the position of B can be anywhere.

But since all positions are random, the probability that C and D are on different sides of AB is equal to the probability that, given three points B, C, D, one is in the arc from A to B and the other is in the arc from B to A. Since the positions are random, this probability is the same as the probability that, among three points, two are on one side of a randomly placed line... Hmm, not sure.

Wait, perhaps think of it as follows. Fix A, then the positions of B, C, D are three random points. For AB and CD to intersect, the chord CD must cross AB. Which happens if one of C or D is in the arc from A to B and the other is in the arc from B to A.

But since the positions of B, C, D are random, we can imagine that for any fixed B, the probability that C and D are on opposite sides of AB depends on where B is. However, due to the symmetry, we can average over all possible positions of B.

Alternatively, since all four points are random, we can think of the probability that AB and CD intersect as equal to the probability that, when choosing four points, the points are arranged such that C and D are on opposite sides of AB. But how to calculate that?

Another approach is to note that for any four distinct points on a circle, there are three ways to pair them into two chords. Each pairing is equally likely. Therefore, since only one of the three pairings results in intersecting chords, the probability that a specific pairing (AB and CD) intersects is 1/3. Therefore, the answer is 1/3.

But in my earlier analysis, fixing A and considering the orderings of B, C, D, I found that the probability is 2/6 = 1/3. So that confirms it.

But let me check with another method. Suppose we fix A and B. The circle is divided into two arcs by AB. The probability that C and D are on different arcs is equal to the probability that one is on one arc and one on the other. Since the circle is continuous, the probability that both C and D are on the same arc is 2 * [probability that both are on arc AB] or [probability that both are on arc BA]. Wait, but actually, once AB is fixed, the circle is split into two arcs. Let me denote the arc from A to B as arc AB (length θ) and arc from B to A as arc BA (length 1 - θ, assuming the circle's circumference is 1). The probability that both C and D are on arc AB is θ^2, and similarly on arc BA is (1 - θ)^2. Therefore, the probability that both are on the same arc is θ^2 + (1 - θ)^2. Therefore, the probability that they are on different arcs is 1 - [θ^2 + (1 - θ)^2] = 1 - (2θ^2 - 2θ + 1) = 2θ - 2θ^2.

But θ is the length of arc AB. However, since A and B are randomly chosen, θ is uniformly distributed between 0 and 1? Wait, no. Actually, if we fix A at 0, B is uniformly random on [0, 1). Therefore, the length of arc AB is uniform in [0, 1), but θ is the minimum arc between A and B, so actually θ is between 0 and 0.5, but since we can consider the arc in either direction, it's symmetric. Wait, this might complicate things.

Alternatively, since the positions of B, C, D are all independent, we can compute the expected value of the probability over all positions of B.

Let me denote that the probability we want is E[2θ - 2θ^2], where θ is the length of the arc from A to B. But since A is fixed and B is random, θ is uniformly distributed between 0 and 1 (assuming the circle's circumference is 1). Wait, no, θ is the arc length between A and B in one direction, but since the circle is continuous, θ is uniform in [0, 1) if we consider the arc length from A to B going clockwise. However, in reality, θ can be from 0 to 1, but due to symmetry, we can assume θ is between 0 and 1/2, but no, since the arc can be in either direction.

But perhaps integrating over all possible positions of B. Let's consider θ as the position of B on the circle, so θ ∈ [0, 1). Then, the probability that C and D are on different arcs is 2θ - 2θ²? Wait, earlier I had the expression 2θ - 2θ², but that was for a specific θ. Wait, actually, when we fix θ, the length of arc AB is θ. The probability that C and D are on different arcs is 2θ(1 - θ). Because the probability that one is on the arc AB (length θ) and the other on arc BA (length 1 - θ) is 2θ(1 - θ).

Wait, let me verify. Given θ, the probability that one of C or D is on arc AB and the other is on arc BA is equal to 2 * [θ * (1 - θ)] because for one point to be in arc AB and the other in arc BA. Since C and D are independent, the probability that C is in AB and D is in BA is θ*(1 - θ). Similarly for D in AB and C in BA. So total probability is 2θ(1 - θ). Therefore, the overall probability is E[2θ(1 - θ)] where θ is the position of B, which is uniformly distributed over [0, 1).

Since θ is uniform over [0, 1), the expected value E[2θ(1 - θ)] = 2 ∫₀¹ θ(1 - θ) dθ = 2 ∫₀¹ (θ - θ²) dθ = 2 [½θ² - (1/3)θ³] from 0 to 1 = 2 [½ - 1/3] = 2 [1/6] = 1/3.

Therefore, the probability that AB and CD intersect is 1/3. This matches our previous results!

Therefore, this confirms that the probability is 1/3. So after going through multiple approaches—fixing points, considering orderings, integrating over arc lengths—the answer seems to be 1/3.

But wait, let me check another way. Suppose I use combinatorial counting. There are four points on a circle. The total number of ways to choose two chords is three, as previously mentioned. Each pair of chords is equally likely. So the probability that a specific pair (AB and CD) intersects is 1/3, since only one of the three pairings results in intersecting chords. Therefore, the answer is 1/3.

Alternatively, another way: For any four distinct points on a circle, there are three ways to pair them into two chords. The number of pairings that result in intersecting chords is one. Therefore, the probability is 1/3. Hence, the answer is 1/3.

However, I recall that in some probability problems involving chords, the probability is 1/3 or 1/2. Wait, perhaps I need to check with an example.

Alternatively, think of choosing four points at random. The probability that AB and CD intersect is equal to the number of intersecting configurations divided by total configurations. But since the total number of configurations is infinite, we need another approach.

But our multiple methods have given us 1/3. Therefore, I think that is the correct answer.

Wait, but according to some references, the probability that two random chords intersect is 1/3. So this seems to confirm.

Wait, but here's another thought. If we have four points, the probability that AB and CD intersect is equal to the probability that the points are interleaved. How many interleaved arrangements are there?

For four points on a circle, there are 3! = 6 possible orderings (after fixing one point). The number of interleaved arrangements for AB and CD is 2 (as we saw earlier). Therefore, 2/6 = 1/3. Hence, the answer is 1/3.

Therefore, after thorough analysis, I think the probability is 1/3.

But to ensure I haven't overlooked something, let me consider a specific example with coordinates.

Imagine four points on a circle: A at (1,0), B at (0,1), C at (-1,0), D at (0,-1). These are the four cardinal directions. Now, segments AB and CD: AB connects (1,0) to (0,1), CD connects (-1,0) to (0,-1). Do these segments intersect? Let's see. AB is the line from (1,0) to (0,1), which has equation x + y = 1. CD is the line from (-1,0) to (0,-1), which has equation x + y = -1. These are parallel lines? Wait, they have the same slope but different intercepts. Therefore, they don't intersect. So in this case, AB and CD don't intersect.

Another example: if points are placed in order A, C, B, D. Let's assign coordinates. Let me place A at (1,0), then go counterclockwise to C at (0,1), then to B at (-1,0), then to D at (0,-1), and back to A. Then, chord AB connects (1,0) to (-1,0), which is the horizontal diameter. Chord CD connects (0,1) to (0,-1), which is the vertical diameter. These two diameters intersect at the center. So in this case, AB and CD intersect.

So here, we have an example where they intersect. Another example: points in order A, B, C, D. Chords AB and CD: AB is from (1,0) to (0,1), CD is from (-1,0) to (0,-1). Wait, but if points are placed in order A, B, C, D, then if A is (1,0), B is (0,1), C is (-1,0), D is (0,-1). Then AB is (1,0)-(0,1), CD is (-1,0)-(0,-1). These are the same as before, which are parallel? Wait, no. Wait, AB is from (1,0) to (0,1): slope is (1-0)/(0-1) = -1. CD is from (-1,0) to (0,-1): slope is (-1 - 0)/(0 - (-1)) = -1/1 = -1. So they are parallel, hence don't intersect. Therefore, in this case, they don't intersect. But in the previous example, when the order was A, C, B, D, they intersected.

So depending on the order, they may or may not intersect. Therefore, the probability is 1/3, as calculated earlier.

Another way, for any four points, there are three ways to pair them into two chords. Each pairing has an equal chance of being selected. Therefore, the probability that a specific pairing (AB and CD) intersects is 1/3. Therefore, the answer is 1/3.

Hence, after all these verifications and methods, I conclude that the probability is 1/3.

**Final Answer**
The probability that segments $AB$ and $CD$ intersect is \boxed{\dfrac{1}{3}}.

To determine the probability that segments $ AB $ and $ CD $ intersect when four points $ A, B, C, D $ are chosen randomly and independently on a circle, we take a geometric and combinatorial approach.

---

### Understanding the Configuration

When four distinct points are selected randomly on the circumference of a circle, we can fix the position of one point (say $ A $) without loss of generality, due to the rotational symmetry of the circle. The remaining three points $ B, C, D $ are then randomly distributed on the circle’s perimeter.

The problem becomes: *What is the probability that the chords $ AB $ and $ CD $ intersect?*

---

### Key Insight: Chord Intersection Condition

Two chords $ AB $ and $ CD $ intersect **if and only if** the points $ A, B, C, D $ are arranged **alternately** around the circle. That is, when traversing the circle, the order of points should be something like:
- $ A \to C \to B \to D \to A $, or
- $ A \to D \to B \to C \to A $,
or any cyclic permutation of this, in which the endpoints of the chords are interleaved.

This is the geometric criterion for chord intersection: the chords intersect **if and only if** the points are interleaved around the circle.

---

### Counting the Total Arrangements

Since all points are chosen independently and uniformly, the **order of the four points** around the circle is equally likely to be any of the $ 4! = 24 $ permutations. However, due to rotational symmetry and the fact that only the **cyclic order** matters, we can fix one point (say $ A $) and count the number of **cyclic orderings** of the remaining three points $ B, C, D $.

There are $ 3! = 6 $ such cyclic orderings. In each case, the **labeling of points** (which defines which chord is $ AB $ and which is $ CD $) is a specific pair of points connected by chords. Since all selections are random, the specific pairings of the points into chords are equally likely.

---

### Counting the Favorable Intersections

We can analyze the number of favorable orderings (where chords $ AB $ and $ CD $ intersect) among the 6 possible cyclic orderings. As shown in the thought process, only two of the orderings result in intersecting chords $ AB $ and $ CD $. Let’s describe one of them:

- Points in order: $ A, C, B, D $.
  - Chord $ AB $ connects $ A $ to $ B $.
  - Chord $ CD $ connects $ C $ to $ D $.
  - The endpoints of the chords are interleaved: one chord "skip" one of the other chord's endpoints, causing them to cross.

There is another similar ordering that results in intersection. Thus, in 2 out of 6 cases, the chords intersect.

---

### Generalizing the Probability

Thus, for **any four points chosen at random**, the probability that the chords $ AB $ and $ CD $ intersect is equal to the ratio of favorable arrangements to total arrangements:

$$
\text{Probability} = \frac{\text{Number of favorable arrangements}}{\text{Total number of arrangements}} = \frac{2}{6} = \frac{1}{3}
$$

Alternatively, another way to think about it is by considering that there are three possible pairings of the four points into two chords (e.g., $ AB $ with $ CD $, $ AC $ with $ BD $, and $ AD $ with $ BC $). Among these, only **one** pairing results in intersection. Hence, the probability that a specific pairing intersects is again:

$$
\text{Probability} = \frac{1}{3}
$$

---

### Final Answer

$$
\boxed{\dfrac{1}{3}}
$$

Solution:
Consider the four random points before they are labeled $A$, $B$, $C$, or $D$. In the general case, they will be distinct, forming a convex quadrilateral. Suppose $A$ is labeled. If $B$ is labeled as the vertex opposite $A$, segments $AB$ and $CD$ will intersect; otherwise, they will not. Since there are 3 points to label as $B$, the probability these segments intersect is $\boxed{\frac{1}{3}}$. [asy]
draw((0,1)..(1,0)..(0,-1)..(-1,0)..cycle);
dot((0,1)); dot((-5/13,-12/13)); dot((-1,0)); dot((4/5,3/5));
label("$A$",(0,1),N); label("$B$",(-5/13,-12/13),SSW); label("$C$",(-1,0),W); label("$D$",(4/5,3/5),NE);
draw((0,1)--(-5/13,-12/13),green); draw((-1,0)--(4/5,3/5),green);
draw((0,1)--(4/5,3/5),blue); draw((-1,0)--(-5/13,-12/13),blue);
draw((0,1)--(-1,0),red); draw((-5/13,-12/13)--(4/5,3/5),red);
[/asy] In this diagram, the green edges represent the labeling where $AB$ and $CD$ intersect, and the blue and red edges represent the equally likely labelings where $AB$ and $CD$ do not intersect.