Explanation:
Okay, so I need to find the probability that two segments AB and CD intersect when four points A, B, C, D are chosen randomly and independently on the circumference of a circle. Hmm, 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. The circle is a continuous space, so maybe I can fix one point to simplify calculations? Wait, sometimes fixing a point can help because of rotational symmetry. Let me try that.

Let me fix point A at some position on the circle, say at angle 0 radians. Since the circle is symmetric, fixing one point shouldn't affect the probability. Now, the other points B, C, D are randomly placed around the circle. But how do I model their positions?

Alternatively, maybe I can think about the order of the points around the circle. Since the circle is continuous, the probability that any two points coincide is zero, so we can consider the four points as distinct and ordered around the circle. The key idea here might be to consider all possible orderings of the four points and determine in how many of those orderings the chords AB and CD intersect.

But how many orderings are there? For four distinct points on a circle, the number of ways to arrange them around the circle is (4-1)! = 6? Wait, no. Wait, when arranging points on a circle, the number of distinct circular permutations is (n-1)! for n distinct objects. So for four points, it's 3! = 6. But since we fixed point A, maybe the number of orderings is different?

Wait, if I fix point A, then the other three points B, C, D can be arranged in 3! = 6 ways around the circle. But since the circle is rotationally symmetric, fixing A doesn't change the probability. So maybe I can fix A and consider the positions of B, C, D relative to A.

Alternatively, maybe I can think of the circle as a line segment with the endpoints identified, but that might complicate things. Let me try another approach.

Another way to think about this problem is to consider that for two chords AB and CD to intersect, the points must be arranged in an alternating order around the circle. That is, if you traverse the circle, you encounter A, then C, then B, then D, or some cyclic permutation of that. Wait, more precisely, the chords AB and CD intersect if and only if the points are interleaved. So, for example, if the points are ordered A, C, B, D around the circle, then chord AB would cross chord CD. Similarly, if they are ordered A, D, B, C, then AB and CD would also cross. But if they are ordered A, B, C, D, then AB and CD don't cross. Similarly, if they are ordered A, C, D, B, then AB and CD might not cross?

Wait, maybe I need to formalize this. Let me recall that two chords AB and CD intersect if and only if the points are arranged alternately around the circle. That is, the order around the circle is A, C, B, D or A, D, B, C or any cyclic permutation of these. So, for four points on a circle, there are two possible ways for the chords AB and CD to intersect: the points are interleaved. Whereas if they are not interleaved, the chords don't intersect.

So, how many total orderings are there? For four points on a circle, the number of distinct cyclic orderings is (4-1)! = 6. Wait, but if we fix one point, say A, then the number of orderings of the other three points is 3! = 6. So, if I fix A, then the positions of B, C, D can be arranged in 6 different ways around the circle. For each of these arrangements, I can check whether AB and CD intersect.

Alternatively, maybe I can think of the probability that AB and CD intersect as the number of favorable arrangements divided by the total number of arrangements. Since all arrangements are equally likely, the probability would be the number of favorable cases over total cases.

So, let me fix point A. Then, the positions of B, C, D can be considered as three points randomly placed on the circle. But since the circle is continuous, maybe I can fix A at 0 radians, and then parameterize the positions of B, C, D by angles between 0 and 2π. However, dealing with angles might complicate things. Alternatively, since the circle is rotationally symmetric, I can fix A at a point, and then consider the positions of B, C, D as three points randomly placed on the circumference. But how do I count the number of orderings?

Alternatively, maybe I can think of the circle as a line segment from 0 to 1, with the endpoints identified. Then, the positions of the four points can be represented as four numbers in [0,1), and we can sort them. However, since the circle is a closed loop, the orderings wrap around. But maybe by fixing one point, we can convert it into a linear problem.

Let me try fixing point A at position 0. Then, the other three points B, C, D are randomly placed on the circle. Let me imagine cutting the circle at point A, so that we can represent the circle as a line segment from 0 to 1, with 0 and 1 identified. Then, the positions of B, C, D are three points in [0,1). However, since the circle is symmetric, we can fix A at 0, and then the positions of B, C, D are three independent uniform random variables on [0,1). But since we are dealing with orderings, maybe we can consider the order statistics of these three points.

Alternatively, since the points are randomly placed, the probability that AB and CD intersect depends only on the relative positions of B, C, D with respect to A. Wait, but since A is fixed, maybe I can think of the circle as a line from 0 to 1, with A at 0, and then the other points B, C, D are placed randomly on [0,1). However, since the circle is closed, the position of a point can be on either side of A. But if I fix A at 0, then the positions of B, C, D can be anywhere on [0,1), but when considering the circle, the order around the circle is determined by their positions on the line segment [0,1) with 0 and 1 connected.

But maybe it's easier to think of the circle as a line segment with endpoints identified, and then the problem reduces to arranging the points in order. However, since we fixed A at 0, the positions of B, C, D can be considered as three points on [0,1). Let me think about the order of these points. Let me denote the positions of B, C, D as x, y, z, which are independent uniform variables on [0,1). Then, the order of these three points can be any permutation of the three variables. However, since we are dealing with a circle, the order around the circle is determined by their positions relative to each other and to A.

But since A is fixed at 0, the circle can be thought of as starting at A and going clockwise. So, the positions of B, C, D can be ordered in some way around the circle. However, since we have fixed A, the positions of B, C, D can be ordered in a linear fashion from 0 to 1, but with the caveat that if a point is near 1, it's close to A (which is at 0). So, perhaps the key is to consider the order of the points B, C, D around the circle, which can be represented as three points on a line segment from 0 to 1, with the understanding that the segment wraps around.

Alternatively, maybe I can fix A at 0, and then consider the positions of B, C, D as three points on the circle. Then, the chords AB and CD will intersect if and only if the points C and D are on opposite arcs defined by A and B. Wait, no. Wait, chord AB divides the circle into two arcs. For chord CD to intersect AB, points C and D must be on different arcs defined by AB. But since AB is a chord, the circle is split into two arcs: the arc from A to B going one way, and the arc from A to B going the other way. If points C and D are on different arcs, then chord CD will cross AB. However, since the points are randomly placed, maybe the probability that C and D are on different arcs?

But wait, but AB is a chord, but since the points are random, the position of B is random. So maybe this approach is not straightforward. Let me think again.

Alternatively, since all four points are random, maybe I can consider all possible ways to arrange the four points around the circle. For four points on a circle, there are (4-1)! = 6 possible cyclic orderings. However, each cyclic ordering can be rotated, so maybe there are 6 distinct orderings? Wait, actually, the number of distinct ways to arrange four points on a circle is (4-1)! = 6, considering rotations as the same. But if we fix one point, say A, then the number of orderings is 3! = 6. So, if I fix A, then the other three points can be arranged in 6 different ways around the circle. For each of these arrangements, I can check whether AB and CD intersect.

So, let me list all possible orderings of B, C, D around the circle, given that A is fixed. Let me fix A at position 0, and then consider the positions of B, C, D in clockwise order. The possible orderings of B, C, D are all permutations of three elements, so there are 6 permutations:

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, I need to determine whether chords AB and CD intersect.

Let me visualize each case.

Case 1: Order around the circle is A, B, C, D. So, starting at A, going clockwise, we have B, then C, then D. Then chord AB is from A to B, and chord CD is from C to D. These two chords do not intersect because they are on the same side of the circle. So, AB is between A and B, and CD is between C and D, which are both after B. So, they don't cross.

Case 2: Order is A, B, D, C. So, chord AB is from A to B, chord CD is from C to D. Now, since the order is A, B, D, C, so after B comes D, then C. So, chord CD connects D to C, which are adjacent? Wait, no. If the order is A, B, D, C, then going clockwise from A, we have B, then D, then C, then back to A. So chord CD connects C to D, which are adjacent? Wait, no. Wait, if the order is A, B, D, C, then between D and C is the next point, so chord CD is from D to C, which is adjacent? Wait, but chord CD is between D and C. However, since the points are ordered A, B, D, C, then chord CD would be from D to C, which is a chord that goes from D to C, which is adjacent? Wait, but if you have points in order A, B, D, C, then the arc from D to C is just a small arc, but chord CD is just connecting D to C. However, chord AB is from A to B, which is another chord. Do these chords intersect?

Wait, maybe I need to think about whether the chords cross. For chords AB and CD to intersect, they must cross each other inside the circle. For that to happen, the points must be arranged such that one endpoint of each chord is on one side of the other chord, and the other endpoint is on the other side.

Alternatively, another way to think is that chords AB and CD intersect if and only if the points are interleaved. That is, if you traverse the circle, you meet A, then C, then B, then D, or some cyclic permutation of that. So, in other words, the order around the circle is A, C, B, D or A, D, B, C, etc.

So, in case 1: A, B, C, D. The order is A, B, C, D. So, the chords AB and CD don't cross. In case 2: A, B, D, C. The order is A, B, D, C. So, if we check interleaving: A, B, D, C. Is there an interleaving? Let's see. If we have AB and CD, then for them to cross, the order should be A, C, B, D or A, D, B, C. In case 2, the order is A, B, D, C. So, if we look at the positions of C and D relative to AB. Since AB is between A and B, and D is after B, and C is after D. So chord CD is between D and C, which are both after B. So chord CD is entirely on the arc from B to A (since after D comes C, then back to A). Wait, but chord CD is from D to C, which is a chord that doesn't cross AB. So, chords AB and CD don't cross here.

Case 3: Order is A, C, B, D. So, points are A, then C, then B, then D. Now, chord AB connects A to B, and chord CD connects C to D. Let me see. If the order is A, C, B, D, then chord AB goes from A to B, which skips over C. Chord CD goes from C to D, which is after B. So, do these chords cross? Let me imagine the circle. Starting at A, going clockwise: A, then C, then B, then D. So chord AB connects A to B, which is two points apart (A to C to B). Chord CD connects C to D, which is adjacent? Wait, no. From C to D is one step. But chord AB is from A to B, which is two steps. Do these cross? Let me think. If you have points A, C, B, D in order around the circle, then chord AB would cross chord CD? Wait, chord AB is from A to B, which is passing through the arc that goes from A to C to B. Chord CD is from C to D, which is on the arc from C to D. Wait, but if you draw chord AB and chord CD, do they cross?

Alternatively, maybe I can use the concept that two chords AB and CD intersect if and only if the points are interleaved. So, if the order around the circle is A, C, B, D, then the order is A, C, B, D. So, if you look at the order of the points, A, then C, then B, then D. So, chord AB has endpoints A and B, and chord CD has endpoints C and D. The interleaving would be A, C, B, D. So, since C is between A and B, and D is after B, then chord CD starts at C (which is between A and B) and goes to D (which is after B). So, chord CD would cross chord AB? Wait, maybe not. Let me try to visualize.

Imagine the circle with points A, C, B, D in that order. So, if I draw chord AB, it connects A to B, which is a chord that goes from A through the arc that goes through C to B. Chord CD connects C to D, which is a chord that goes from C to D. Since D is after B, chord CD is on the arc from C to D, which is adjacent to B. Wait, but chord AB is from A to B, and chord CD is from C to D. These two chords might not cross. Wait, but if you have points A, C, B, D in order, then chord AB and chord CD would cross if you connect A to B and C to D. Wait, actually, if you have points in order A, C, B, D, then chord AB is from A to B, which is passing through the arc that goes from A to C to B. Chord CD is from C to D, which is on the arc from C to D. These two chords don't cross. Wait, but maybe I need to think of the other arc. If you connect A to B, it could also go the other way around the circle, but since we are talking about chords, they are straight lines inside the circle. So, chord AB is the straight line between A and B, and chord CD is the straight line between C and D. If the points are in order A, C, B, D, then chord AB and CD would not cross. Wait, but maybe they do?

Wait, let me think of coordinates. Let me assign coordinates to make it concrete. Let me imagine the circle as a unit circle. Let me fix A at (1, 0). Let me place C at (cos(π/2), sin(π/2)) = (0,1). Then B is somewhere between C and D. Wait, but if the order is A, C, B, D, then after C comes B, then D. Let me assign coordinates:

Let me suppose the circle is parameterized from 0 to 1, with A at 0. Let me assign positions:

A = 0

C = 1/4

B = 1/2

D = 3/4

So, the order is A (0), C (1/4), B (1/2), D (3/4). Now, chord AB connects 0 to 1/2, and chord CD connects 1/4 to 3/4. Do these chords intersect?

Drawing chord AB from 0 to 1/2, which is a straight line. Chord CD connects 1/4 to 3/4, which is a straight line. These two chords would intersect? Let me see. The chord AB is from (0,0) to (1/2, 0) if we consider the circle as a unit circle? Wait, no, if we parameterize the circle as angles, but maybe it's easier to think in terms of coordinates.

Alternatively, think of the circle as a line segment from 0 to 1 with 0 and 1 connected. If we have points at 0, 1/4, 1/2, 3/4. Chord AB is from 0 to 1/2, chord CD is from 1/4 to 3/4. These two chords would intersect? If you draw them on the circle, chord AB is from 0 to 1/2, chord CD is from 1/4 to 3/4. These chords would cross each other inside the circle. Wait, yes! Because chord AB is from 0 to 1/2, and chord CD is from 1/4 to 3/4. If you imagine the circle, chord AB is a shorter chord, and chord CD is a longer chord that spans from 1/4 to 3/4. These two chords cross each other. Wait, but in this case, they do cross. So, in this case, chords AB and CD intersect. Therefore, case 3 is a favorable case.

Wait, but earlier I thought they didn't cross, but with coordinates, they do. So, maybe my previous reasoning was wrong. So, in case 3, chords AB and CD intersect. Therefore, case 3 is favorable.

Similarly, let me check case 4: order A, C, D, B. So, points are A, C, D, B. So, chord AB connects A to B, chord CD connects C to D. Let me assign coordinates again. Let me set A at 0, C at 1/4, D at 1/2, B at 3/4. So, chord AB is from 0 to 3/4, chord CD is from 1/4 to 1/2. Do these cross? Chord AB is from 0 to 3/4, which is a long chord, and chord CD is from 1/4 to 1/2, which is a short chord. These two chords do not cross. Because chord CD is entirely within the arc from 1/4 to 1/2, which is between A and B. So, chord AB is from A to B, and chord CD is inside that arc. Therefore, they don't cross. So, case 4 is not favorable.

Case 5: Order A, D, B, C. Let me assign coordinates again. A at 0, D at 1/4, B at 1/2, C at 3/4. Chord AB is from 0 to 1/2, chord CD is from 3/4 to 1/4? Wait, chord CD connects C to D. Wait, in this case, the order is A, D, B, C. So, points are A, D, B, C. So chord CD connects C to D. Wait, C is at 3/4, D is at 1/4. So chord CD connects 3/4 to 1/4. Which is the same as connecting 1/4 to 3/4, which is the same as chord CD in case 3. So, chord AB is from 0 to 1/2, chord CD is from 1/4 to 3/4. Which we saw earlier do cross. Therefore, case 5 is favorable.

Case 6: Order A, D, C, B. Assign coordinates: A at 0, D at 1/4, C at 1/2, B at 3/4. Chord AB is from 0 to 3/4, chord CD is from 1/2 to 1/4. So chord CD is from 1/4 to 1/2, which is the same as before. Chord AB is from 0 to 3/4. These chords do not cross. So case 6 is not favorable.

So, out of the 6 possible orderings of B, C, D around the circle (with A fixed), cases 3 and 5 are favorable, where chords AB and CD intersect. Therefore, the probability is 2/6 = 1/3.

Wait, but hold on. Is that correct? So, if there are 6 possible orderings, and 2 of them result in intersecting chords, then the probability is 1/3? But I remember that the probability that two random chords intersect is 1/3? Wait, but I need to confirm.

Alternatively, maybe I made a mistake in considering the orderings. Let me check again.

Wait, when I fixed A, and considered the orderings of B, C, D, there are 6 possible orderings. But in reality, when considering the circle, some of these orderings might be equivalent? Wait, no. Since we fixed A, the orderings of B, C, D are distinct. So, with A fixed, there are 3! = 6 orderings. And in two of them, the chords intersect. Therefore, the probability is 2/6 = 1/3. Therefore, the answer is 1/3?

But I have a feeling that the answer might be 1/3, but I want to verify this with another approach.

Another approach: Let me consider all possible ways to choose four points on the circle. The total number of ways to choose four points is not important since we are dealing with continuous probability. However, the key idea is that for any four points on a circle, there are three ways to pair them into two chords: (AB and CD), (AC and BD), (AD and BC). Each pairing is equally likely? Wait, no. Wait, when choosing four points, the probability that AB and CD intersect is the same as the probability that AC and BD intersect, etc.?

But since the problem is specifically about AB and CD, maybe we can think of the probability that AB and CD intersect as 1 divided by the number of ways to pair the points? Wait, no. Wait, for four points, there are three ways to pair them into two chords. Each pairing is equally likely? If the points are randomly placed, then each pairing is equally probable? Wait, maybe not. Wait, if the points are randomly placed, then the probability that AB and CD are the chords is the same as any other pairing? Wait, but the problem is about AB and CD specifically. Wait, but since the points are randomly chosen, all pairings are equally likely? Wait, no. Wait, the way the points are chosen is that we have four points, and then we connect AB and CD. But since the points are randomly placed, the probability that AB and CD intersect is the same as the probability that any other specific pairing intersects. Therefore, since there are three possible pairings, and for each pairing, the probability that they intersect is the same, then the total probability that any two chords intersect is 3 times the probability that AB and CD intersect? Wait, no, that might not be the case.

Wait, actually, the total probability that two chords intersect is not directly related here. Wait, but if we consider that for any four points, exactly one of the three pairings results in intersecting chords? No, actually, for four points on a circle, exactly one of the three pairings will result in intersecting chords? Wait, no. For four points on a circle, if they are in general position (no three on a line, which is always true for a circle), then exactly one of the three pairings will result in intersecting chords. Wait, is that true?

Wait, suppose you have four points on a circle. If you connect them as AB and CD, they may or may not intersect. Similarly for AC and BD, and AD and BC. For any four points, exactly one of these three pairings will result in intersecting chords? Wait, no. For example, if the four points are in order A, B, C, D around the circle, then pairing AB and CD do not intersect, pairing AC and BD intersect, and pairing AD and BC also intersect? Wait, no. Let me check.

If the points are in order A, B, C, D around the circle. Then:

- AB and CD: AB is between A and B, CD is between C and D. These don't intersect.

- AC and BD: AC connects A to C, BD connects B to D. These two chords intersect.

- AD and BC: AD connects A to D, BC connects B to C. These two chords do not intersect. Wait, because AD is from A to D, which is across the circle, and BC is from B to C, which is adjacent. So, they don't intersect. Wait, so only one pairing (AC and BD) intersect. So, for this configuration, only one pairing intersects.

Another example: suppose the points are in order A, B, D, C. Then:

- AB and CD: AB is from A to B, CD is from C to D. Do they intersect? If the order is A, B, D, C, then chord AB is from A to B, chord CD is from C to D. If you imagine the circle, chord AB is between A and B, chord CD is between C and D, which are on the other side. Do they cross? If the order is A, B, D, C, then chord CD is from D to C, which is adjacent? Wait, no. If the order is A, B, D, C, then going clockwise from A, you have B, then D, then C, then back to A. So chord CD connects D to C, which is adjacent? Wait, no. If you have points A, B, D, C in order, then chord CD connects C to D, which are adjacent? Wait, no. If you have points in order A, B, D, C, then between D and C is the next point, so chord CD is from D to C, which is adjacent? Wait, but chord CD is between D and C, which are two points apart? Wait, no. If the order is A, B, D, C, then between D and C is the arc that goes from D to C, which is one step. So chord CD is just connecting D to C, which is adjacent? Wait, but if you have four points on a circle, then any two points are connected by a chord. If the points are in order A, B, D, C, then chord CD is between D and C, which are adjacent? No, because between D and C is the arc that goes from D to C, which is one step. But chord CD is just connecting D to C, which is a chord that doesn't cross anything else. However, chord AB is from A to B, which is adjacent. So, chords AB and CD don't cross. However, pairing AC and BD: chord AC connects A to C, chord BD connects B to D. Do they cross? If the order is A, B, D, C, then chord AC goes from A to C, which skips over B and D? Wait, chord AC would cross BD? Let me think. If the order is A, B, D, C, then chord AC connects A to C, which is passing through the arc that goes from A to B to D to C. Chord BD connects B to D. These two chords would cross? Yes, because chord AC goes from A to C, which is passing through the arc that includes B and D, and chord BD is from B to D. So, they cross. Similarly, pairing AD and BC: chord AD connects A to D, chord BC connects B to C. Do they cross? Chord AD is from A to D, which is a long chord, and chord BC is from B to C, which is adjacent. These don't cross. Therefore, only one pairing (AC and BD) intersect. Therefore, for this configuration, only one pairing intersects.

Therefore, in general, for any four points on a circle, exactly one of the three pairings results in intersecting chords. Therefore, if we fix a specific pairing, say AB and CD, then the probability that they intersect is 1/3. Because for any four points, there are three pairings, and only one of them intersects. Therefore, the probability that AB and CD intersect is 1/3. Therefore, the answer is 1/3.

But wait, this seems to contradict my previous calculation where I fixed A and found that 2 out of 6 orderings resulted in AB and CD intersecting. But according to this, it's 1/3. Which is consistent, because 2/6 is 1/3. So, both approaches agree. Therefore, the probability is 1/3.

But let me check another way. Let me think about choosing four points on the circle. The probability that AB and CD intersect is equal to the probability that the points are arranged such that one of the chords crosses the other. Since there are three ways to pair the points, and only one of them results in crossing, then the probability is 1/3. Therefore, the answer is 1/3.

Alternatively, another approach: Fix the four points on the circle. The probability that AB and CD intersect is equal to the probability that C and D are in different arcs defined by A and B. Wait, but since A and B are random points, how do we compute that?

Wait, suppose we fix points A and B. The circle is divided into two arcs by A and B. For chords AB and CD to intersect, points C and D must be on different arcs. The probability that C and D are on different arcs is 2*(1/2)*(1/2) = 1/2? Wait, no. Wait, if we fix A and B, then the circle is split into two arcs. The probability that C is on one arc and D is on the other arc is 2*(probability that C is on arc AB and D is on arc BA) or something? Wait, actually, if we fix A and B, the circle is split into two arcs. Let me assume that the arc length between A and B is some value, but since the points are random, the arc length is random. However, since the circle is continuous, we can fix A and B such that the arc from A to B is some length, but since the circle is symmetric, we can fix A at 0 and B at some position θ, but integrating over θ might complicate things.

Alternatively, since all positions are random, the probability that C and D are on different arcs defined by A and B is equal to the probability that one is on one arc and the other is on the other arc. Since the circle is continuous, the probability that C is on one arc and D is on the other arc is equal to 2 * (length of arc AB / circumference) * (length of arc BA / circumference). Wait, but since the arc AB and arc BA are the same as the two arcs between A and B. Wait, but since the circle is continuous, the probability that C is on one arc and D is on the other arc is equal to 2 * (probability that C is on arc AB) * (probability that D is on arc BA). But since the positions of C and D are independent, and the arc AB is some length, but since A and B are random, the length of arc AB is random. Wait, this seems complicated.

Alternatively, since the positions of A, B, C, D are all random, we can fix A and B, and then consider the positions of C and D. However, since the circle is symmetric, we can fix A at 0 and B at some position, say, θ, but θ is random. However, integrating over θ might be complex. Alternatively, since the positions are random, the probability that C and D are on different arcs defined by A and B is equal to the probability that C and D are on opposite sides of the chord AB. But since the circle is symmetric, the probability that C is on one side of AB is 1/2, and similarly for D. However, since the positions are independent, the probability that C and D are on different sides is 2*(1/2)*(1/2) = 1/2. Wait, but this would suggest that the probability is 1/2, which contradicts the previous result of 1/3. So, which one is correct?

Wait, there must be a mistake here. Because if we fix A and B, then the probability that C and D are on different arcs is 1/2, but then the probability that AB and CD intersect is 1/2? But earlier reasoning suggested 1/3. So, which is correct?

Wait, let me think again. If we fix A and B, then the circle is split into two arcs. The probability that C is on one arc and D is on the other arc is 1/2. However, even if C and D are on different arcs, the chords AB and CD might not intersect. Wait, no. If C is on one arc and D is on the other arc, then chord CD must cross chord AB. Wait, is that true?

Wait, suppose chord AB divides the circle into two arcs. If C is on one arc and D is on the other arc, then chord CD must cross AB. Because chord CD connects a point on one arc to a point on the other arc, so it must cross AB. Therefore, if C and D are on different arcs, then CD crosses AB. Therefore, the probability that AB and CD cross is equal to the probability that C and D are on different arcs. But if that is the case, then the probability is 1/2. But this contradicts the previous result of 1/3. Therefore, there must be a mistake in one of the reasonings.

Wait, let me check with an example. Suppose we fix A and B. Let me assume that the arc AB is 1/2 the circumference. Then, the probability that C is on one arc and D is on the other arc is 1/2 * 1/2 + 1/2 * 1/2 = 1/2. Wait, no. If the arc AB is 1/2 the circumference, then the arc AB is 1/2, and the other arc is also 1/2. Then, the probability that C is on arc AB is 1/2, and D is on the other arc is 1/2. So, the probability that C is on arc AB and D is on the other arc is 1/2 * 1/2 = 1/4. Similarly, the probability that C is on the other arc and D is on arc AB is 1/4. So total probability is 1/2. Therefore, if arc AB is 1/2, then the probability is 1/2. However, if arc AB is something else, say, length L, then the probability that C and D are on different arcs is 2 * (L / circumference) * ((1 - L)/circumference). Wait, but since circumference is 1, it's 2L(1 - L). However, since L is random, we need to average over all possible positions of A and B.

Wait, but since A and B are random, the distribution of L is uniform? Wait, no. The length between A and B is uniformly distributed between 0 and 1/2 (since beyond 1/2, it's the same as the other arc). Wait, actually, the arc length between A and B can be considered as a random variable uniformly distributed between 0 and 1, but since the circle is symmetric, we can consider it between 0 and 1/2, and double it? Wait, no. Actually, the arc length between A and B is uniformly distributed between 0 and 1, but since the circle is symmetric, we can fix A at 0 and let B be uniformly distributed between 0 and 1. Then, the arc length between A and B is the minimum of B and 1 - B. Wait, no. Actually, the arc length between A and B can be considered as the length from A to B in the clockwise direction, which is uniformly distributed between 0 and 1. However, since the circle is symmetric, we can consider the arc length between A and B as a random variable L, which is uniformly distributed between 0 and 1. However, for the purpose of calculating the probability that C and D are on different arcs, we need to consider that the arc length is L, and the other arc is 1 - L.

Therefore, the probability that C is on one arc and D is on the other arc is 2 * (L * (1 - L)). Wait, no. Wait, the probability that C is on the arc of length L is L, and the probability that D is on the other arc is (1 - L). Therefore, the probability that C is on one arc and D is on the other is L*(1 - L) + (1 - L)*L = 2L(1 - L). However, since L is uniformly distributed between 0 and 1, we need to compute the expected value of 2L(1 - L).

Wait, but L is not uniformly distributed. Wait, if we fix A at 0 and let B be uniformly random on [0,1), then the arc length between A and B is uniformly distributed between 0 and 1? No, actually, the arc length between A and B is uniformly distributed between 0 and 1, but since we can consider the arc in either direction, but for the purpose of calculating the probability that C and D are on different arcs, we can consider the arc length as L, which is between 0 and 1, and the other arc is 1 - L. However, since the circle is symmetric, we can assume that L is between 0 and 1/2, and double the probability? Wait, no. Actually, the distribution of L is not uniform. If B is uniformly random on [0,1), then the arc length from A to B clockwise is uniform on [0,1). However, the arc length between A and B is the minimum of the clockwise and counter-clockwise arcs. Wait, no. If we fix A at 0, and B is uniformly random on [0,1), then the arc length from A to B clockwise is just B, and the arc length counter-clockwise is 1 - B. However, for the purpose of considering the two arcs defined by A and B, we can take the arc length as L = B, and the other arc is 1 - L. However, since the circle is symmetric, we can consider L to be in [0, 1/2], because beyond 1/2, it's the same as the other arc. Wait, but actually, the distribution of L is not uniform. If B is uniformly distributed on [0,1), then the distribution of L (the arc length from A to B clockwise) is uniform on [0,1). However, when considering the two arcs, the length of the arc between A and B is L, and the other arc is 1 - L. Therefore, the probability that C is on one arc and D is on the other is 2 * L * (1 - L). Wait, no. Wait, the probability that C is on the arc of length L is L, and the probability that D is on the arc of length 1 - L is (1 - L). Therefore, the probability that C and D are on different arcs is L*(1 - L) + (1 - L)*L = 2L(1 - L). However, since L is uniformly distributed between 0 and 1, we need to compute the expected value of 2L(1 - L).

Wait, the expected value of 2L(1 - L) where L is uniform on [0,1] is 2 * ∫₀¹ L(1 - L) dL = 2 * [∫₀¹ L dL - ∫₀¹ L² dL] = 2 * [ (1/2) - (1/3) ] = 2 * (1/6) = 1/3. Therefore, the expected probability that C and D are on different arcs is 1/3. Therefore, the probability that AB and CD intersect is 1/3. Which matches our previous result. Therefore, this confirms that the probability is 1/3.

Therefore, both approaches give the same answer, 1/3. Therefore, the probability that segments AB and CD intersect is 1/3.

But let me check with yet another approach. Let me think about choosing four points on the circle. The total number of ways to connect them into two chords is 3, as mentioned before. For each set of four points, exactly one of these three pairings results in intersecting chords. Therefore, if we randomly choose a pairing, the probability that it is the intersecting one is 1/3. Since the problem is about a specific pairing (AB and CD), the probability is 1/3. Therefore, this also gives 1/3.

Therefore, all approaches lead to the same answer, 1/3. Therefore, I think the probability is 1/3.

But just to be thorough, let me think of another way. Suppose we fix the four points on the circle. Let me label them in the order they appear around the circle as P1, P2, P3, P4. The probability that AB and CD intersect depends on how we label the points. However, since the points are randomly chosen, all labelings are equally likely. There are 4! = 24 ways to label the points, but considering rotational symmetry, it's equivalent to fixing one point and permuting the others. However, since we are dealing with probability, we can consider all possible orderings.

Alternatively, for four points on a circle, there are 3 ways to pair them into two chords. As previously established, only one of these pairings results in intersecting chords. Therefore, if we randomly select a pairing, the probability is 1/3. Since the problem is about a specific pairing (AB and CD), the probability is 1/3.

Therefore, after multiple approaches and verifications, I can be confident 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 the circumference of a circle, we can approach the problem using geometric and combinatorial reasoning.

---

### **Key Insight: Intersecting Chords on a Circle**

Two chords $ AB $ and $ CD $ intersect **if and only if** the points are arranged **alternately** around the circle. That is, the order of the points around the circle must be such that one endpoint of each chord lies on opposite arcs defined by the other chord.

This is equivalent to saying that the points are **interleaved** in the circular order. For example, if the points are arranged in the order $ A, C, B, D $, then chord $ AB $ and chord $ CD $ intersect.

---

### **Counting the Total Number of Arrangements**

Since the points are chosen randomly and independently, we can assume that all orderings of the four points around the circle are equally likely. Fixing one point (say, $ A $) due to the symmetry of the circle, the remaining three points $ B, C, D $ can be arranged in $ 3! = 6 $ different ways around the circle.

Each of these 6 orderings corresponds to a unique way the points are placed around the circle, and we can check whether the chords $ AB $ and $ CD $ intersect in each case.

---

### **Checking Which Arrangements Result in Intersections**

Let’s consider the 6 possible orderings of $ B, C, D $ around the circle (with $ A $ fixed at a specific position):

1. $ A, B, C, D $ → Chords $ AB $ and $ CD $ do **not** intersect.
2. $ A, B, D, C $ → Chords $ AB $ and $ CD $ do **not** intersect.
3. $ A, C, B, D $ → Chords $ AB $ and $ CD $ **do** intersect.
4. $ A, C, D, B $ → Chords $ AB $ and $ CD $ do **not** intersect.
5. $ A, D, B, C $ → Chords $ AB $ and $ CD $ **do** intersect.
6. $ A, D, C, B $ → Chords $ AB $ and $ CD $ do **not** intersect.

So, out of the 6 possible orderings, **2** result in intersecting chords.

---

### **Probability Calculation**

Since all orderings are equally likely, the probability that $ AB $ and $ CD $ intersect is:

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

---

### **Alternative Confirmation via Pairings**

Another way to think about this is to consider that for any four points on a circle, there are exactly **three** ways to pair them into two chords:

- $ AB $ and $ CD $
- $ AC $ and $ BD $
- $ AD $ and $ BC $

For any set of four points, **exactly one** of these pairings results in intersecting chords. Therefore, the probability that a **specific** pairing (like $ AB $ and $ CD $) results in intersecting chords is:

$$
\frac{1}{3}
$$

---

### **Conclusion**

All approaches—whether through direct enumeration of orderings, geometric reasoning, or combinatorial pairing—lead to the same conclusion.

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

[Assistant]
Explanation:
Okay, so I need to find the probability that two segments AB and CD intersect when four points A, B, C, D are chosen randomly and independently on the circumference of a circle. Hmm, 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. The circle is a continuous space, so maybe I can fix one point to simplify calculations? Wait, sometimes fixing a point can help because of rotational symmetry. Let me try that.

Let me fix point A at some position on the circle, say at angle 0 radians. Since the circle is symmetric, fixing one point shouldn't affect the probability. Now, the other points B, C, D are randomly placed around the circle. But how do I model their positions?

Alternatively, maybe I can think about the order of the points around the circle. Since the circle is continuous, the probability that any two points coincide is zero, so we can consider the four points as distinct and ordered around the circle. The key idea here might be to consider all possible orderings of the four points and determine in how many of those orderings the chords AB and CD intersect.

But how many orderings are there? For four distinct points on a circle, the number of ways to arrange them around the circle is (4-1)! = 6? Wait, no. Wait, when arranging points on a circle, the number of distinct circular permutations is (n-1)! for n distinct objects. So for four points, it's 3! = 6. But since we fixed point A, maybe the number of orderings is different?

Wait, if I fix point A, then the other three points B, C, D can be arranged in 3! = 6 ways around the circle. But since the circle is rotationally symmetric, fixing A doesn't change the probability. So maybe I can fix A and consider the positions of B, C, D relative to A.

Alternatively, maybe I can think of the circle as a line segment with the endpoints identified, but that might complicate things. Let me try another approach.

Another way to think about this problem is to consider that for two chords AB and CD to intersect, the points must be arranged in an alternating order around the circle. That is, if you traverse the circle, you encounter A, then C, then B, then D, or some cyclic permutation of that. Wait, more precisely, the chords AB and CD intersect if and only if the points are interleaved. So, for example, if the points are ordered A, C, B, D around the circle, then chord AB would cross chord CD. Similarly, if they are ordered A, D, B, C, then AB and CD would also cross. But if they are ordered A, B, C, D, then AB and CD don't cross. Similarly, if they are ordered A, C, D, B, then AB and CD might not cross?

Wait, maybe I need to formalize this. Let me recall that two chords AB and CD intersect if and only if the points are arranged alternately around the circle. That is, the order around the circle is A, C, B, D or A, D, B, C or any cyclic permutation of these. So, for four points on a circle, there are two possible ways for the chords AB and CD to intersect: the points are interleaved. Whereas if they are not interleaved, the chords don't intersect.

So, how many total orderings are there? For four points on a circle, the number of distinct cyclic orderings is (4-1)! = 6. Wait, but if we fix one point, say A, then the number of orderings of the other three points is 3! = 6. So, if I fix A, then the positions of B, C, D can be arranged in 6 different ways around the circle. For each of these arrangements, I can check whether AB and CD intersect.

Alternatively, maybe I can think of the probability that AB and CD intersect as the number of favorable arrangements divided by the total number of arrangements. Since all arrangements are equally likely, the probability would be the number of favorable cases over total cases.

So, let me fix point A. Then, the positions of B, C, D can be considered as three points randomly placed on the circle. But since the circle is continuous, maybe I can fix A at 0 radians, and then parameterize the positions of B, C, D by angles between 0 and 2π. However, dealing with angles might complicate things. Alternatively, since the circle is rotationally symmetric, I can fix A at a point, and then consider the positions of B, C, D as three points randomly placed on the circumference. But how do I count the number of orderings?

Alternatively, maybe I can think of the circle as a line segment from 0 to 1, with the endpoints identified. Then, the positions of the four points can be represented as four numbers in [0,1), and we can sort them. However, since the circle is a closed loop, the orderings wrap around. But maybe by fixing one point, we can convert it into a linear problem.

Let me try fixing point A at position 0. Then, the other three points B, C, D are randomly placed on the circle. Let me imagine cutting the circle at point A, so that we can represent the circle as a line segment from 0 to 1, with 0 and 1 identified. Then, the positions of B, C, D are three points in [0,1). However, since the circle is symmetric, we can fix A at 0, and then the positions of B, C, D are three independent uniform random variables on [0,1). But since we are dealing with orderings, maybe we can consider the order statistics of these three points.

Alternatively, since the points are randomly placed, the probability that AB and CD intersect depends only on the relative positions of B, C, D with respect to A. Wait, but since A is fixed, maybe I can think of the circle as a line from 0 to 1, with A at 0, and then the other points B, C, D are placed randomly on [0,1). However, since the circle is closed, the position of a point can be on either side of A. But if I fix A at 0, then the positions of B, C, D can be anywhere on [0,1), but when considering the circle, the order around the circle is determined by their positions on the line segment [0,1) with 0 and 1 connected.

But maybe it's easier to think of the circle as a line segment with endpoints identified, and then the problem reduces to arranging the points in order. However, since we fixed A at 0, the positions of B, C, D can be considered as three points on [0,1). Let me think about the order of these points. Let me denote the positions of B, C, D as x, y, z, which are independent uniform variables on [0,1). Then, the order of these three points can be any permutation of the three variables. However, since we are dealing with a circle, the order around the circle is determined by their positions relative to each other and to A.

But since A is fixed at 0, the circle can be thought of as starting at A and going clockwise. So, the positions of B, C, D can be ordered in some way around the circle. However, since we have fixed A, the positions of B, C, D can be ordered in a linear fashion from 0 to 1, but with the caveat that if a point is near 1, it's close to A (which is at 0). So, perhaps the key is to consider the order of the points B, C, D around the circle, which can be represented as three points on a line segment from 0 to 1, with the understanding that the segment wraps around.

Alternatively, maybe I can fix A at 0, and then consider the positions of B, C, D as three points on the circle. Then, the chords AB and CD will intersect if and only if the points C and D are on opposite arcs defined by A and B. Wait, no. Wait, chord AB divides the circle into two arcs. For chord CD to intersect AB, points C and D must be on different arcs defined by AB. But since AB is a chord, the circle is split into two arcs: the arc from A to B going one way, and the arc from A to B going the other way. If points C and D are on different arcs, then chord CD will cross AB. However, since the points are randomly placed, maybe the probability that C and D are on different arcs?

But wait, but AB is a chord, but since the points are random, the position of B is random. So maybe this approach is not straightforward. Let me think again.

Alternatively, since all four points are random, maybe I can consider all possible ways to arrange the four points around the circle. For four points on a circle, there are (4-1)! = 6 possible cyclic orderings. However, each cyclic ordering can be rotated, so maybe there are 6 distinct orderings? Wait, actually, the number of distinct ways to arrange four points on a circle is (4-1)! = 6, considering rotations as the same. But if we fix one point, say A, then the number of orderings is 3! = 6. So, if I fix A, then the other three points can be arranged in 6 different ways around the circle. For each of these arrangements, I can check whether AB and CD intersect.

So, let me list all possible orderings of B, C, D around the circle, given that A is fixed. Let me fix A at position 0, and then consider the positions of B, C, D in clockwise order. The possible orderings of B, C, D are all permutations of three elements, so there are 6 permutations:

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, I need to determine whether chords AB and CD intersect.

Let me visualize each case.

Case 1: Order around the circle is A, B, C, D. So, starting at A, going clockwise, we have B, then C, then D. Then chord AB is from A to B, and chord CD is from C to D. These two chords do not intersect because they are on the same side of the circle. So, AB is between A and B, and CD is between C and D, which are both after B. So, they don't cross.

Case 2: Order is A, B, D, C. So, chord AB is from A to B, chord CD is from C to D. Now, since the order is A, B, D, C, so after B comes D, then C. So, chord CD connects D to C, which are adjacent? Wait, no. If the order is A, B, D, C, then going clockwise from A, we have B, then D, then C, then back to A. So chord CD connects C to D, which are adjacent? Wait, no. Wait, if the order is A, B, D, C, then between D and C is the next point, so chord CD is from D to C, which is adjacent? Wait, but chord CD is between D and C. However, since the points are ordered A, B, D, C, then chord CD would be from D to C, which is a chord that goes from D to C, which is adjacent? Wait, but if you have points in order A, B, D, C, then the arc from D to C is just a small arc, but chord CD is just connecting D to C. However, chord AB is from A to B, which is another chord. Do these chords intersect?

Wait, maybe I need to think about whether the chords cross. For chords AB and CD to intersect, they must cross each other inside the circle. For that to happen, the points must be arranged such that one endpoint of each chord is on one side of the other chord, and the other endpoint is on the other side.

Alternatively, another way to think is that chords AB and CD intersect if and only if the points are interleaved. That is, if you traverse the circle, you meet A, then C, then B, then D, or some cyclic permutation of that. So, in other words, the order around the circle is A, C, B, D or A, D, B, C, etc.

So, in case 1: A, B, C, D. The order is A, B, C, D. So, the chords AB and CD don't cross. In case 2: A, B, D, C. The order is A, B, D, C. So, if we check interleaving: A, B, D, C. Is there an interleaving? Let's see. If we have AB and CD, then for them to cross, the order should be A, C, B, D or A, D, B, C. In case 2, the order is A, B, D, C. So, if we look at the positions of C and D relative to AB. Since AB is between A and B, and D is after B, and C is after D. So chord CD is between D and C, which are both after B. So chord CD is entirely on the arc from B to A (since after D comes C, then back to A). Wait, but chord CD is from D to C, which is a chord that doesn't cross AB. So, chords AB and CD don't cross here.

Case 3: Order is A, C, B, D. So, points are A, then C, then B, then D. Now, chord AB connects A to B, and chord CD connects C to D. Let me see. If the order is A, C, B, D, then chord AB goes from A to B, which skips over C. Chord CD goes from C to D, which is after B. So, do these chords cross? Let me imagine the circle. Starting at A, going clockwise: A, then C, then B, then D. So chord AB connects A to B, which is two points apart (A to C to B). Chord CD connects C to D, which is adjacent? Wait, no. From C to D is one step. But chord AB is from A to B, which is two steps. Do these cross? Let me think. If you have points A, C, B, D in order around the circle, then chord AB would cross chord CD? Wait, chord AB is from A to B, which is passing through the arc that goes from A to C to B. Chord CD is from C to D, which is on the arc from C to D. Wait, but if you draw chord AB and chord CD, do they cross?

Alternatively, maybe I can use the concept that two chords AB and CD intersect if and only if the points are interleaved. So, if the order around the circle is A, C, B, D, then the order is A, C, B, D. So, if you look at the order of the points, A, then C, then B, then D. So, chord AB has endpoints A and B, and chord CD has endpoints C and D. The interleaving would be A, C, B, D. So, since C is between A and B, and D is after B, then chord CD starts at C (which is between A and B) and goes to D (which is after B). So, chord CD would cross chord AB? Wait, maybe not. Let me try to visualize.

Imagine the circle with points A, C, B, D in that order. So, if I draw chord AB, it connects A to B, which is a chord that goes from A through the arc that goes through C to B. Chord CD connects C to D, which is a chord that goes from C to D. Since D is after B, chord CD is on the arc from C to D, which is adjacent to B. Wait, but chord AB is from A to B, and chord CD is from C to D. These two chords might not cross. Wait, but if you have points A, C, B, D in order, then chord AB and chord CD would cross if you connect A to B and C to D. Wait, actually, if you have points in order A, C, B, D, then chord AB is from A to B, which is passing through the arc that goes from A to C to B. Chord CD is from C to D, which is on the arc from C to D. These two chords don't cross. Wait, but maybe I need to think of the other arc. If you connect A to B, it could also go the other way around the circle, but since we are talking about chords, they are straight lines inside the circle. So, chord AB is the straight line between A and B, and chord CD is the straight line between C and D. If the points are in order A, C, B, D, then chord AB and CD would not cross. Wait, but maybe they do?

Wait, let me think of coordinates. Let me assign coordinates to make it concrete. Let me imagine the circle as a unit circle. Let me fix A at (1, 0). Let me place C at (cos(π/2), sin(π/2)) = (0,1). Then B is somewhere between C and D. Wait, but if the order is A, C, B, D, then after C comes B, then D. Let me assign coordinates:

Let me suppose the circle is parameterized from 0 to 1, with A at 0. Let me assign positions:

A = 0

C = 1/4

B = 1/2

D = 3/4

So, the order is A (0), C (1/4), B (1/2), D (3/4). Now, chord AB connects 0 to 1/2, and chord CD connects 1/4 to 3/4. Do these chords intersect?

Drawing chord AB from 0 to 1/2, which is a straight line. Chord CD connects 1/4 to 3/4, which is a straight line. These two chords would intersect? Let me see. The chord AB is from (0,0) to (1/2, 0) if we consider the circle as a unit circle? Wait, no, if we parameterize the circle as angles, but maybe it's easier to think in terms of coordinates.

Alternatively, think of the circle as a line segment from 0 to 1 with 0 and 1 connected. If we have points at 0, 1/4, 1/2, 3/4. Chord AB is from 0 to 1/2, chord CD is from 1/4 to 3/4. These two chords would intersect? If you draw them on the circle, chord AB is from 0 to 1/2, chord CD is from 1/4 to 3/4. These chords would cross each other inside the circle. Wait, yes! Because chord AB is from 0 to 1/2, and chord CD is from 1/4 to 3/4. If you imagine the circle, chord AB is a shorter chord, and chord CD is a longer chord that spans from 1/4 to 3/4. These two chords cross each other. Wait, but in this case, they do cross. So, in this case, chords AB and CD intersect. Therefore, case 3 is a favorable case.

Wait, but earlier I thought they didn't cross, but with coordinates, they do. So, maybe my previous reasoning was wrong. So, in case 3, chords AB and CD intersect. Therefore, case 3 is favorable.

Similarly, let me check case 4: order A, C, D, B. So, points are A, C, D, B. So, chord AB connects A to B, chord CD connects C to D. Let me assign coordinates again. Let me set A at 0, C at 1/4, D at 1/2, B at 3/4. So, chord AB is from 0 to 3/4, chord CD is from 1/4 to 1/2. Do these cross? Chord AB is from 0 to 3/4, which is a long chord, and chord CD is from 1/4 to 1/2, which is a short chord. These two chords do not cross. Because chord CD is entirely within the arc from 1/4 to 1/2, which is between A and B. So, chord AB is from A to B, and chord CD is inside that arc. Therefore, they don't cross. So, case 4 is not favorable.

Case 5: Order A, D, B, C. Let me assign coordinates again. A at 0, D at 1/4, B at 1/2, C at 3/4. Chord AB is from 0 to 1/2, chord CD is from 3/4 to 1/4? Wait, chord CD connects C to D. Wait, in this case, the order is A, D, B, C. So, points are A, D, B, C. So chord CD connects C to D. Wait, C is at 3/4, D is at 1/4. So chord CD connects 3/4 to 1/4. Which is the same as connecting 1/4 to 3/4, which is the same as chord CD in case 3. So, chord AB is from 0 to 1/2, chord CD is from 1/4 to 3/4. Which we saw earlier do cross. Therefore, case 5 is favorable.

Case 6: Order A, D, C, B. Assign coordinates: A at 0, D at 1/4, C at 1/2, B at 3/4. Chord AB is from 0 to 3/4, chord CD is from 1/2 to 1/4. So chord CD is from 1/4 to 1/2, which is the same as before. Chord AB is from 0 to 3/4. These chords do not cross. So case 6 is not favorable.

So, out of the 6 possible orderings of B, C, D around the circle (with A fixed), cases 3 and 5 are favorable, where chords AB and CD intersect. Therefore, the probability is 2/6 = 1/3.

Wait, but hold on. Is that correct? So, if there are 6 possible orderings, and 2 of them result in intersecting chords, then the probability is 1/3? But I remember that the probability that two random chords intersect is 1/3? Wait, but I need to confirm.

Alternatively, maybe I made a mistake in considering the orderings. Let me check again.

Wait, when I fixed A, and considered the orderings of B, C, D, there are 6 possible orderings. But in reality, when considering the circle, some of these orderings might be equivalent? Wait, no. Since we fixed A, the orderings of B, C, D are distinct. So, with A fixed, there are 3! = 6 orderings. And in two of them, the chords intersect. Therefore, the probability is 2/6 = 1/3. Therefore, the answer is 1/3?

But I have a feeling that the answer might be 1/3, but I want to verify this with another approach.

Another approach: Let me consider all possible ways to choose four points on the circle. The total number of ways to choose four points is not important since we are dealing with continuous probability. However, the key idea is that for any four points on a circle, there are three ways to pair them into two chords: (AB and CD), (AC and BD), (AD and BC). Each pairing is equally likely? Wait, no. Wait, when choosing four points, the probability that AB and CD intersect is the same as the probability that AC and BD intersect, etc.?

But since the problem is specifically about AB and CD, maybe we can think of the probability that AB and CD intersect as 1 divided by the number of ways to pair the points? Wait, no. Wait, for four points, there are three ways to pair them into two chords. Each pairing is equally likely? If the points are randomly placed, then each pairing is equally probable? Wait, maybe not. Wait, if the points are randomly placed, then the probability that AB and CD are the chords is the same as any other pairing? Wait, but the problem is about AB and CD specifically. Wait, but since the points are randomly chosen, all pairings are equally likely? Wait, no. Wait, the way the points are chosen is that we have four points, and then we connect AB and CD. But since the points are randomly placed, the probability that AB and CD intersect is the same as the probability that any other specific pairing intersects. Therefore, since there are three possible pairings, and for each pairing, the probability that they intersect is the same, then the total probability that any two chords intersect is 3 times the probability that AB and CD intersect? Wait, no, that might not be the case.

Wait, actually, the total probability that two chords intersect is not directly related here. Wait, but if we consider that for any four points, exactly one of the three pairings results in intersecting chords? No, actually, for four points on a circle, exactly one of the three pairings will result in intersecting chords? Wait, no. For four points on a circle, if they are in general position (no three on a line, which is always true for a circle), then exactly one of the three pairings will result in intersecting chords. Wait, is that true?

Wait, suppose you have four points on a circle. If you connect them as AB and CD, they may or may not intersect. Similarly for AC and BD, and AD and BC. For any four points, exactly one of these three pairings will result in intersecting chords? Wait, no. For example, if the four points are in order A, B, C, D around the circle, then pairing AB and CD do not intersect, pairing AC and BD intersect, and pairing AD and BC also intersect? Wait, no. Let me check.

If the points are in order A, B, C, D around the circle. Then:

- AB and CD: AB is between A and B, CD is between C and D. These don't intersect.

- AC and BD: AC connects A to C, BD connects B to D. These two chords intersect.

- AD and BC: AD connects A to D, BC connects B to C. These two chords do not intersect. Wait, because AD is from A to D, which is across the circle, and BC is from B to C, which is adjacent. So, they don't intersect. Wait, so only one pairing (AC and BD) intersect. So, for this configuration, only one pairing intersects.

Another example: suppose the points are in order A, B, D, C. Then:

- AB and CD: AB is from A to B, CD is from C to D. Do they intersect? If the order is A, B, D, C, then chord AB is from A to B, chord CD is from C to D. If you imagine the circle, chord AB is between A and B, chord CD is between C and D, which are on the other side. Do they cross? If the order is A, B, D, C, then chord CD is from D to C, which is adjacent? Wait, no. If the order is A, B, D, C, then going clockwise from A, you have B, then D, then C, then back to A. So chord CD connects D to C, which is adjacent? Wait, no. If you have points A, B, D, C in order, then chord CD connects C to D, which are adjacent? Wait, no. If you have points in order A, B, D, C, then between D and C is the next point, so chord CD is from D to C, which is adjacent? Wait, but chord CD is between D and C, which are two points apart? Wait, no. If the order is A, B, D, C, then between D and C is the arc that goes from D to C, which is one step. So chord CD is just connecting D to C, which is adjacent? Wait, but if you have four points on a circle, then any two points are connected by a chord. If the points are in order A, B, D, C, then chord CD is between D and C, which are adjacent? No, because between D and C is the arc that goes from D to C, which is one step. But chord CD is just connecting D to C, which is a chord that doesn't cross anything else. However, chord AB is from A to B, which is adjacent. So, chords AB and CD don't cross. However, pairing AC and BD: chord AC connects A to C, chord BD connects B to D. Do they cross? If the order is A, B, D, C, then chord AC goes from A to C, which skips over B and D? Wait, chord AC would cross BD? Let me think. If the order is A, B, D, C, then chord AC connects A to C, which is passing through the arc that goes from A to B to D to C. Chord BD connects B to D. These two chords would cross? Yes, because chord AC goes from A to C, which is passing through the arc that includes B and D, and chord BD is from B to D. So, they cross. Similarly, pairing AD and BC: chord AD connects A to D, chord BC connects B to C. Do they cross? Chord AD is from A to D, which is a long chord, and chord BC is from B to C, which is adjacent. These don't cross. Therefore, only one pairing (AC and BD) intersect. Therefore, for this configuration, only one pairing intersects.

Therefore, in general, for any four points on a circle, exactly one of the three pairings results in intersecting chords. Therefore, if we fix a specific pairing, say AB and CD, then the probability that they intersect is 1/3. Because for any four points, there are three pairings, and only one of them intersects. Therefore, the probability that AB and CD intersect is 1/3. Therefore, the answer is 1/3.

But wait, this seems to contradict my previous calculation where I fixed A and found that 2 out of 6 orderings resulted in AB and CD intersecting. But according to this, it's 1/3. Which is consistent, because 2/6 is 1/3. So, both approaches agree. Therefore, the probability is 1/3.

But let me check another way. Let me think about choosing four points on the circle. The probability that AB and CD intersect is equal to the probability that the points are arranged such that one of the chords crosses the other. Since there are three ways to pair the points, and only one of them results in crossing, then the probability is 1/3. Therefore, the answer is 1/3.

Alternatively, another approach: Fix the four points on the circle. The probability that AB and CD intersect is equal to the probability that C and D are in different arcs defined by A and B. Wait, but since A and B are random points, how do we compute that?

Wait, suppose we fix points A and B. The circle is divided into two arcs by A and B. For chords AB and CD to intersect, points C and D must be on different arcs. The probability that C and D are on different arcs is 2*(1/2)*(1/2) = 1/2? Wait, no. Wait, if we fix A and B, then the circle is split into two arcs. The probability that C is on one arc and D is on the other arc is 2*(probability that C is on arc AB and D is on arc BA) or something? Wait, actually, if we fix A and B, the circle is split into two arcs. Let me assume that the arc length between A and B is some value, but since the points are random, the arc length is random. However, since the circle is continuous, we can fix A and B such that the arc from A to B is some length, but since the circle is symmetric, we can fix A at 0 and B at some position θ, but integrating over θ might complicate things.

Alternatively, since all positions are random, the probability that C and D are on different arcs defined by A and B is equal to the probability that one is on one arc and the other is on the other arc. Since the circle is continuous, the probability that C is on one arc and D is on the other arc is equal to 2 * (length of arc AB / circumference) * (length of arc BA / circumference). Wait, but since the arc AB and arc BA are the same as the two arcs between A and B. Wait, but since the circle is continuous, the probability that C is on one arc and D is on the other arc is equal to 2 * (probability that C is on arc AB) * (probability that D is on arc BA). But since the positions of C and D are independent, and the arc AB is some length, but since A and B are random, the length of arc AB is random. Wait, this seems complicated.

Alternatively, since the positions of A, B, C, D are all random, we can fix A and B, and then consider the positions of C and D. However, since the circle is symmetric, we can fix A at 0 and B at some position, say, θ, but θ is random. However, integrating over θ might be complex. Alternatively, since the positions are random, the probability that C and D are on different arcs defined by A and B is equal to the probability that C and D are on opposite sides of the chord AB. But since the circle is symmetric, the probability that C is on one side of AB is 1/2, and similarly for D. However, since the positions are independent, the probability that C and D are on different sides is 2*(1/2)*(1/2) = 1/2. Wait, but this would suggest that the probability is 1/2, which contradicts the previous result of 1/3. So, which one is correct?

Wait, there must be a mistake here. Because if we fix A and B, then the probability that C and D are on different arcs is 1/2, but then the probability that AB and CD intersect is 1/2? But earlier reasoning suggested 1/3. So, which is correct?

Wait, let me think again. If we fix A and B, then the circle is split into two arcs. The probability that C is on one arc and D is on the other arc is 1/2. However, even if C and D are on different arcs, the chords AB and CD might not intersect. Wait, no. If C is on one arc and D is on the other arc, then chord CD must cross chord AB. Wait, is that true?

Wait, suppose chord AB divides the circle into two arcs. If C is on one arc and D is on the other arc, then chord CD must cross AB. Because chord CD connects a point on one arc to a point on the other arc, so it must cross AB. Therefore, if C and D are on different arcs, then CD crosses AB. Therefore, the probability that AB and CD cross is equal to the probability that C and D are on different arcs. But if that is the case, then the probability is 1/2. But this contradicts the previous result of 1/3. Therefore, there must be a mistake in one of the reasonings.

Wait, let me check with an example. Suppose we fix A and B. Let me assume that the arc AB is 1/2 the circumference. Then, the probability that C is on one arc and D is on the other arc is 1/2 * 1/2 + 1/2 * 1/2 = 1/2. Wait, no. If the arc AB is 1/2 the circumference, then the arc AB is 1/2, and the other arc is also 1/2. Then, the probability that C is on arc AB is 1/2, and D is on the other arc is 1/2. So, the probability that C is on arc AB and D is on the other arc is 1/2 * 1/2 = 1/4. Similarly, the probability that C is on the other arc and D is on arc AB is 1/4. So total probability is 1/2. Therefore, if arc AB is 1/2, then the probability is 1/2. However, if arc AB is something else, say, length L, then the probability that C and D are on different arcs is 2 * (L / circumference) * ((1 - L)/circumference). Wait, but since circumference is 1, it's 2L(1 - L). However, since L is random, we need to average over all possible positions of A and B.

Wait, but since A and B are random, the distribution of L is uniform? Wait, no. The length between A and B is uniformly distributed between 0 and 1/2 (since beyond 1/2, it's the same as the other arc). Wait, actually, the arc length between A and B can be considered as a random variable uniformly distributed between 0 and 1, but since the circle is symmetric, we can consider it between 0 and 1/2, and double it? Wait, no. Actually, the arc length between A and B is uniformly distributed between 0 and 1, but since the circle is symmetric, we can fix A at 0 and let B be uniformly distributed between 0 and 1. Then, the arc length between A and B is the minimum of B and 1 - B. Wait, no. Actually, the arc length between A and B can be considered as the length from A to B in the clockwise direction, which is uniformly distributed between 0 and 1. However, since the circle is symmetric, we can consider the arc length between A and B as a random variable L, which is uniformly distributed between 0 and 1. However, for the purpose of calculating the probability that C and D are on different arcs, we need to consider that the arc length is L, and the other arc is 1 - L.

Therefore, the probability that C is on one arc and D is on the other arc is 2 * (L * (1 - L)). Wait, no. Wait, the probability that C is on the arc of length L is L, and the probability that D is on the other arc is (1 - L). Therefore, the probability that C is on one arc and D is on the other is L*(1 - L) + (1 - L)*L = 2L(1 - L). However, since L is uniformly distributed between 0 and 1, we need to compute the expected value of 2L(1 - L).

Wait, but L is not uniformly distributed. Wait, if we fix A at 0 and let B be uniformly random on [0,1), then the arc length between A and B is uniformly distributed between 0 and 1? No, actually, the arc length between A and B is uniformly distributed between 0 and 1, but since we can consider the arc in either direction, but for the purpose of calculating the probability that C and D are on different arcs, we can consider the arc length as L, which is between 0 and 1, and the other arc is 1 - L. However, since the circle is symmetric, we can assume that L is between 0 and 1/2, and double the probability? Wait, no. Actually, the distribution of L is not uniform. If B is uniformly random on [0,1), then the arc length from A to B clockwise is uniform on [0,1). However, the arc length between A and B is the minimum of the clockwise and counter-clockwise arcs. Wait, no. If we fix A at 0, and B is uniformly random on [0,1), then the arc length from A to B clockwise is just B, and the arc length counter-clockwise is 1 - B. However, for the purpose of considering the two arcs defined by A and B, we can take the arc length as L = B, and the other arc is 1 - L. However, since the circle is symmetric, we can consider L to be in [0, 1/2], because beyond 1/2, it's the same as the other arc. Wait, but actually, the distribution of L is not uniform. If B is uniformly distributed on [0,1), then the distribution of L (the arc length from A to B clockwise) is uniform on [0,1). However, when considering the two arcs, the length of the arc between A and B is L, and the other arc is 1 - L. Therefore, the probability that C is on one arc and D is on the other is 2 * L * (1 - L). Wait, no. Wait, the probability that C is on the arc of length L is L, and the probability that D is on the arc of length 1 - L is (1 - L). Therefore, the probability that C and D are on different arcs is L*(1 - L) + (1 - L)*L = 2L(1 - L). However, since L is uniformly distributed between 0 and 1, we need to compute the expected value of 2L(1 - L).

Wait, the expected value of 2L(1 - L) where L is uniform on [0,1] is 2 * ∫₀¹ L(1 - L) dL = 2 * [∫₀¹ L dL - ∫₀¹ L² dL] = 2 * [ (1/2) - (1/3) ] = 2 * (1/6) = 1/3. Therefore, the expected probability that C and D are on different arcs is 1/3. Therefore, the probability that AB and CD intersect is 1/3. Which matches our previous result. Therefore, this confirms that the probability is 1/3.

Therefore, both approaches give the same answer, 1/3. Therefore, the probability that segments AB and CD intersect is 1/3.

But let me check with yet another approach. Let me think about choosing four points on the circle. The total number of ways to connect them into two chords is 3, as mentioned before. For each set of four points, exactly one of these three pairings results in intersecting chords. Therefore, if we randomly choose a pairing, the probability that it is the intersecting one is 1/3. Since the problem is about a specific pairing (AB and CD), the probability is 1/3. Therefore, this also gives 1/3.

Therefore, all approaches lead to the same answer, 1/3. Therefore, I think the probability is 1/3.

But just to be thorough, let me think of another way. Suppose we fix the four points on the circle. Let me label them in the order they appear around the circle as P1, P2, P3, P4. The probability that AB and CD intersect depends on how we label the points. However, since the points are randomly chosen, all labelings are equally likely. There are 4! = 24 ways to label the points, but considering rotational symmetry, it's equivalent to fixing one point and permuting the others. However, since we are dealing with probability, we can consider all possible orderings.

Alternatively, for four points on a circle, there are 3 ways to pair them into two chords. As previously established, only one of these pairings results in intersecting chords. Therefore, if we randomly select a pairing, the probability is 1/3. Since the problem is about a specific pairing (AB and CD), the probability is 1/3.

Therefore, after multiple approaches and verifications, I can be confident 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 the circumference of a circle, we can approach the problem using geometric and combinatorial reasoning.

---

### **Key Insight: Intersecting Chords on a Circle**

Two chords $ AB $ and $ CD $ intersect **if and only if** the points are arranged **alternately** around the circle. That is, the order of the points around the circle must be such that one endpoint of each chord lies on opposite arcs defined by the other chord.

This is equivalent to saying that the points are **interleaved** in the circular order. For example, if the points are arranged in the order $ A, C, B, D $, then chord $ AB $ and chord $ CD $ intersect.

---

### **Counting the Total Number of Arrangements**

Since the points are chosen randomly and independently, we can assume that all orderings of the four points around the circle are equally likely. Fixing one point (say, $ A $) due to the symmetry of the circle, the remaining three points $ B, C, D $ can be arranged in $ 3! = 6 $ different ways around the circle.

Each of these 6 orderings corresponds to a unique way the points are placed around the circle, and we can check whether the chords $ AB $ and $ CD $ intersect in each case.

---

### **Checking Which Arrangements Result in Intersections**

Let’s consider the 6 possible orderings of $ B, C, D $ around the circle (with $ A $ fixed at a specific position):

1. $ A, B, C, D $ → Chords $ AB $ and $ CD $ do **not** intersect.
2. $ A, B, D, C $ → Chords $ AB $ and $ CD $ do **not** intersect.
3. $ A, C, B, D $ → Chords $ AB $ and $ CD $ **do** intersect.
4. $ A, C, D, B $ → Chords $ AB $ and $ CD $ do **not** intersect.
5. $ A, D, B, C $ → Chords $ AB $ and $ CD $ **do** intersect.
6. $ A, D, C, B $ → Chords $ AB $ and $ CD $ do **not** intersect.

So, out of the 6 possible orderings, **2** result in intersecting chords.

---

### **Probability Calculation**

Since all orderings are equally likely, the probability that $ AB $ and $ CD $ intersect is:

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

---

### **Alternative Confirmation via Pairings**

Another way to think about this is to consider that for any four points on a circle, there are exactly **three** ways to pair them into two chords:

- $ AB $ and $ CD $
- $ AC $ and $ BD $
- $ AD $ and $ BC $

For any set of four points, **exactly one** of these pairings results in intersecting chords. Therefore, the probability that a **specific** pairing (like $ AB $ and $ CD $) results in intersecting chords is:

$$
\frac{1}{3}
$$

---

### **Conclusion**

All approaches—whether through direct enumeration of orderings, geometric reasoning, or combinatorial pairing—lead to the same conclusion.

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

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.