Explanation:
Okay, so Bob and Alice each have a bag with balls of five different colors: blue, green, orange, red, and violet. Each bag starts with one ball of each color, so that's five balls in each bag.

The problem says that Alice first randomly selects one ball from her bag and puts it into Bob's bag. Then Bob randomly selects one ball from his bag (which now has six balls) and puts it into Alice's bag. After this process, we need to find the probability that the contents of the two bags are the same again.

Hmm, let me think. So initially, both bags have the same number of balls and the same colors. After Alice adds a ball to Bob's bag, Bob's bag has six balls, and Alice's bag has four. Then Bob takes a ball from his bag (which now has six) and puts it back into Alice's bag. So after this exchange, both bags should have five balls again. The question is, what's the probability that after this exchange, both bags have the same contents?

Wait, but the contents being the same would mean that each bag has the same number of each color. But since they started with one of each color, and they swapped one ball each, maybe there's a way for them to end up with the same contents?

Let me try to break it down step by step.

First, let's consider the first step: Alice takes a ball from her bag and puts it into Bob's bag. Since all balls are equally likely, the probability that she picks any particular color is 1/5. Let's denote the colors as B, G, O, R, V for blue, green, orange, red, violet.

After Alice adds a ball to Bob's bag, Bob's bag has 6 balls: the original five plus one from Alice. Then Bob takes a ball from his bag and puts it into Alice's bag. Now, depending on which ball Bob takes, the contents of both bags will change.

We need to find the probability that after these two transfers, both bags have the same contents.

So, for the contents to be the same, the net result of the transfers must be that each bag ends up with the same set of balls as before. But since they swapped one ball each, maybe the only way this can happen is if the ball that Alice gave to Bob is the same color as the ball that Bob gave back to Alice? Wait, not necessarily. Let me think.

Wait, suppose Alice gives a ball of color X to Bob. Then Bob has an extra ball of color X. Then Bob takes a ball from his bag. If he takes a ball of color X, then he gives it back to Alice, so the net result is that nothing has changed. But if he takes a different color, say Y, then Alice's bag loses a Y and gains an X, while Bob's bag loses an X and gains a Y. So in this case, the contents of the bags would be different.

But the problem says that after this process, the contents of the two bags are the same. So, for the contents to be the same, the net effect of the transfers must be that each bag ends up with the same set of balls as they started.

Wait, but if Alice gives a ball to Bob and Bob gives a ball back, then the only way for the contents to be the same is if the ball that Bob gives back is the same as the one Alice gave. Because otherwise, the contents would differ.

Wait, let me check that. Suppose Alice gives a blue ball to Bob. Then Bob's bag has an extra blue. Then Bob takes a ball from his bag. If he takes blue, then he gives blue back to Alice. So Alice's bag is back to original, and Bob's bag is back to original. If he takes a different color, say green, then Alice's bag now has one less green and one more blue, while Bob's bag has one less green and one more blue. So the contents of the bags are different.

Therefore, the only way for the contents to be the same is if Bob gives back the same color that Alice gave. Therefore, the probability that the contents are the same is equal to the probability that Alice's transferred ball is the same color as Bob's transferred ball.

But wait, is that the only possibility? Let me think again. Suppose Alice gives a blue to Bob. Then Bob has 6 balls: original 5 plus blue. Then Bob takes a ball. The chance he takes blue is 1/6. If he takes blue, then the contents are same. If he takes another color, then the contents are different.

But maybe there's another scenario where even if Bob takes a different color, the contents could still be same? Let me think. Suppose Alice gives a blue to Bob. Then Bob has 6 balls: original 5 (including blue) plus blue. Wait, no. Wait, initially, Bob's bag has one of each color. Then Alice adds one ball to Bob's bag. So if Alice adds blue, then Bob's bag has two blue balls and one each of the other colors. Then Bob takes a ball from his bag. If he takes blue, then he gives blue back to Alice. So Alice's bag now has original minus blue plus blue, so same as before. Bob's bag has original minus blue plus blue, so same as before.

If Bob takes a different color, say green. Then Alice's bag now has original minus blue (since she gave blue to Bob) plus green (since Bob gave green back). So Alice's bag has one less blue and one more green. Bob's bag has original minus green (since he took green) plus blue (since he received blue from Alice). So Bob's bag has one less green and one more blue.

So the contents of the bags are different. Therefore, the only way for the contents to be same is if Bob gives back the same color that Alice gave.

Therefore, the probability that the contents are same is equal to the probability that Bob picks the same color that Alice gave.

But how do we compute that?

First, Alice picks a color. Let's say she picks color X. The probability that she picks any color is 1/5. Then, Bob's bag has 6 balls: original 5 plus one X. So the probability that Bob picks X is 2/6 = 1/3. Therefore, the total probability would be 1/5 * 1/3 = 1/15? Wait, but that seems too low. Wait, but maybe there's another case?

Wait, no. Wait, if Alice picks a color, say X, then Bob has two Xs in his bag. So the probability that Bob picks X is 2/6. Therefore, the total probability is 1/5 * 2/6 = 1/15? But that would be the probability that Alice picks X and Bob picks X. But since there are 5 colors, each with probability 1/5, and for each color, the probability that Bob picks that color is 2/6. Therefore, total probability is 5*(1/5)*(2/6) = 2/6 = 1/3? Wait, that seems conflicting with my previous thought.

Wait, let me clarify. The total probability is the sum over all colors of the probability that Alice picks that color and Bob picks that color.

So for each color X, P(Alice picks X) = 1/5. Then, given that Alice picked X, the probability that Bob picks X is 2/6. Therefore, the total probability is sum_{X} [1/5 * 2/6] = 5*(1/5 * 2/6) = 2/6 = 1/3.

So the probability is 1/3? But let me check again.

Alternatively, maybe there's another way to think about it.

Alternatively, think of the entire process as a random exchange.

But maybe there's another way where the contents can be same even if Bob doesn't return the same color? Wait, let me think. Suppose Alice gives a blue to Bob, and Bob gives a green back. Then Alice's bag has one less blue and one more green. Bob's bag has one less green and one more blue. So the contents of the bags are different.

But suppose that Alice gives a blue to Bob, and Bob gives back a blue. Then both bags are back to original.

Alternatively, suppose that Alice gives a blue to Bob, and Bob gives back a blue. So same as original.

Alternatively, suppose that Alice gives a blue to Bob, and Bob gives back a different color. Then the contents are different.

Therefore, only when Bob gives back the same color as Alice gave, the contents are same.

Therefore, the probability is 1/3.

But let me check with an example. Let me take a smaller case. Suppose there are only two colors, say blue and green. Each has one ball. Alice gives one to Bob. Then Bob gives one back. What is the probability that the contents are same?

Original: Alice has B, G. Bob has B, G.

Alice gives one ball to Bob. Let's say she gives B. Then Bob has B, B, G. Then Bob picks a ball. Probability he picks B is 2/3, probability he picks G is 1/3. If he picks B, then he gives back B. So Alice has B, G again. Bob has B, G again. If he picks G, then Alice has G, G and Bob has B, B. So contents are different. Similarly if Alice had given G. So total probability is 1/2 (probability Alice gives B) * 2/3 (Bob gives back B) + 1/2 (Alice gives G) * 2/3 (Bob gives back G) = 1/2 * 2/3 + 1/2 * 2/3 = 2/3. Which is same as 2/ (number of colors + 1)? Wait, in this case, number of colors is 2, and the probability is 2/3. Which is same as 2/(2 + 1). Wait, but in our original problem, with 5 colors, the probability was 1/3? Wait, no. Wait in the two color case, the probability was 2/3. Wait, but according to the previous formula, if we have n colors, then the probability would be n/(n + 1)? Wait, no. Wait in the two color case, the probability was 2/3. Let me check again.

Wait, in the two color case, the probability that the contents are same is 2/3. Let me think again.

Alternatively, maybe the formula is (number of colors) * [probability Alice picks color X] * [probability Bob picks color X given Alice picked X]. Which would be n * (1/n) * (2/(n + 1)) ) = 2/(n + 1).

In the two color case, that would be 2/(2 + 1) = 2/3, which matches. In our original problem, n = 5, so 2/(5 + 1) = 2/6 = 1/3. So that seems to be the case.

Therefore, the answer would be 1/3? But let me check again with another example.

Suppose n = 1. Wait, n = 1? If there is only one color, then Alice gives the only ball to Bob. Then Bob has two balls of that color. Then Bob gives one back. The probability that he gives back the same color is 2/2 = 1. So the probability is 1. According to the formula, 2/(1 + 1) = 1. Correct.

Another example, n = 3. Then the probability would be 2/(3 + 1) = 1/2. Let me check.

Original: Alice has A, B, C. Bob has A, B, C.

Alice picks one color, say A. Then Bob has A, A, B, C. Then Bob picks a ball. Probability he picks A is 2/4 = 1/2. So for each color, the probability is 1/3 * 1/2 = 1/6. Total probability is 3 * 1/6 = 1/2. Which matches.

Therefore, the formula seems to hold. Therefore, in our original problem, with n = 5, the probability is 2/(5 + 1) = 1/3.

But wait, the answer seems too straightforward. However, let me think again.

But maybe there's another case where the contents can be same even if Bob doesn't return the same color? For example, suppose Alice gives a blue to Bob, and Bob gives back a green. Then Alice's bag has one less blue and one more green. Bob's bag has one less green and one more blue. So the contents of the bags are different. However, suppose that the original bags had multiple copies? Wait, no. Wait, in the original problem, each bag has one of each color. So after the transfer, Alice's bag has four balls, and Bob's has six. Then Bob gives back one. So after the transfer, both have five balls again.

But for the contents to be same, they need to have the same set of colors. However, since they started with one of each color, and they swapped one ball each, the only way for them to have the same contents is if the net transfer is zero. Which would require that the ball Alice gave to Bob is the same as the one Bob gave back.

Alternatively, is there a case where even if they swapped different colors, the contents are same? For example, suppose Alice gives a blue to Bob, and Bob gives a green back. Then Alice's bag has green instead of blue, and Bob's bag has blue instead of green. But since they both have one of each color, except for the swapped ones. Wait, but they both have five balls.

Wait, let me think. Suppose Alice's original bag has one of each color. She gives blue to Bob. So her bag now has green, orange, red, violet. Bob's bag has blue, blue, green, orange, red, violet. Then Bob gives green back to Alice. So Alice's bag now has green, orange, red, violet, and blue? Wait no. Wait, Alice's bag had green, orange, red, violet (four balls). Then Bob gives green back. So Alice's bag now has green, orange, red, violet, and green? Wait, no. Wait, Alice's bag had four balls: original minus blue. Then Bob gives back a green. So Alice's bag now has four balls plus one green? Wait, no. Wait, initially, Alice had five balls. She gave one to Bob, so she has four. Then Bob gives one to Alice, so she has five again.

So if Alice gave blue to Bob, her bag is missing blue. Then Bob gives back green. So Alice's bag now has green instead of blue. So her bag has green, orange, red, violet, and... Wait, original colors were blue, green, orange, red, violet. After giving blue away, she has green, orange, red, violet. Then Bob gives green back. So she has green, orange, red, violet, green? Wait, no. Wait, she had four balls: green, orange, red, violet. Then Bob gives her a green. So she has two greens, orange, red, violet.

But Bob's bag: originally had blue, green, orange, red, violet. Then Alice added blue, so he had two blues, green, orange, red, violet. Then he gave green back. So his bag now has two blues, orange, red, violet, and... Wait, he had six balls: two blues, green, orange, red, violet. Then he removes green, so he has two blues, orange, red, violet. So he has one less green and one more blue? Wait, no. Wait, he had two blues, green, orange, red, violet. Then he removes green, so he has two blues, orange, red, violet. So he has one green less, and no change in blue? Wait, he had two blues, and he didn't remove any blue. So he has two blues, orange, red, violet.

So Alice's bag has two greens, orange, red, violet. Bob's bag has two blues, orange, red, violet. So their bags are different. Therefore, the contents are different.

Therefore, only when Bob gives back the same color as Alice gave, the contents are same.

Therefore, the probability is 1/3.

But let me think again. Wait, maybe there's another case where even if they swapped different colors, the contents are same? For example, suppose Alice gives a blue to Bob, and Bob gives back a blue. Then both bags are same. If Alice gives blue to Bob, and Bob gives back a different color, then contents are different.

Alternatively, suppose that Alice gives a blue to Bob, and Bob gives back a blue. Then both bags are same.

Alternatively, suppose that Alice gives a blue to Bob, and Bob gives back a blue. So same as original.

Alternatively, suppose that Alice gives a blue to Bob, and Bob gives back a green. Then Alice's bag has green instead of blue, and Bob's bag has blue instead of green. So the contents are different.

Therefore, only when Bob gives back the same color as Alice gave, the contents are same.

Therefore, the probability is 1/3.

But let me think again. Let me compute it step by step.

First, Alice picks a color. The probability is 1/5 for each color. Let's say she picks color X. Then Bob's bag has 6 balls: original 5 plus X.

Now, Bob is going to pick a ball from his bag. The probability that he picks X is 2/6 = 1/3.

Therefore, the probability that Bob picks X is 1/3.

Therefore, the total probability is 1/5 * 1/3 = 1/15 for each color, and since there are 5 colors, total probability is 5*(1/15) = 1/3.

Therefore, the answer is 1/3.

But wait, let me check with another approach.

Alternative approach:

Let me think of the entire process as a random exchange.

After Alice and Bob exchange one ball each, what is the probability that their bags are the same?

Note that the bags will be the same if and only if the ball that Alice gave to Bob is the same as the ball that Bob gave back to Alice.

Alternatively, think of it as a permutation.

But maybe not.

Alternatively, think of the possible outcomes.

But maybe another way:

Let me think of the possible cases.

Case 1: Alice gives a ball to Bob, and Bob gives back the same ball.

Case 2: Alice gives a ball to Bob, and Bob gives back a different ball.

In Case 1, the bags are same. In Case 2, they are different.

Therefore, the probability we need is the probability of Case 1.

So, what is the probability of Case 1?

First, Alice picks a ball: 5 choices.

Then Bob picks a ball from his bag, which has 6 balls. The number of balls that are the same as Alice's is 2 (original one plus the one Alice added). Therefore, the probability that Bob picks the same color is 2/6 = 1/3.

Therefore, the total probability is 1/5 * 1/3 * 5? Wait, no. Wait, the probability is 1/5 (for Alice's choice) multiplied by 1/3 (for Bob's choice). Since Alice's choice is independent.

Wait, but since Alice's choice is any color, and for each color, the probability is 1/5, and for each color, the probability that Bob picks that color is 2/6. Therefore, total probability is sum_{colors} [P(Alice picks color) * P(Bob picks color | Alice picked color)] = sum_{colors} [1/5 * 2/6] = 5*(1/5 * 2/6) = 2/6 = 1/3.

Therefore, same result.

Alternatively, think of it as:

The probability that Bob picks the same color as Alice is equal to the expected value of the probability that Bob picks the same color given Alice's choice.

But regardless, the answer seems to be 1/3.

But let me check with actual numbers.

Suppose we have colors B, G, O, R, V.

Let me compute the total number of possible outcomes.

First, Alice has 5 choices. Then Bob has 6 choices. So total number of possible outcomes is 5 * 6 = 30.

Now, how many of these result in the bags being the same?

For each color that Alice could have given, say color X, there are 2 balls of color X in Bob's bag. So Bob can pick color X in 2 ways. Therefore, for each color X, there are 2 favorable outcomes. Since there are 5 colors, total favorable outcomes are 5 * 2 = 10.

Therefore, the probability is 10 / 30 = 1/3.

Yes! So that's another way to think about it.

Total number of possible outcomes: 5 * 6 = 30.

Number of favorable outcomes: For each color, Alice gives that color (1 way) and Bob gives back that color (2 ways). So 5 * 2 = 10.

Therefore, probability is 10/30 = 1/3.

Therefore, the answer is 1/3.

But wait, let me check with another approach.

Alternative approach:

Let me think about the final state of the bags.

After the exchange, both bags have five balls. For the bags to be the same, they must have the same multiset of colors.

Originally, each bag had one of each color.

After Alice gives a ball to Bob and Bob gives a ball to Alice, the contents of the bags will be:

Alice's bag: original minus the ball she gave plus the ball Bob gave.

Bob's bag: original plus the ball Alice gave minus the ball Bob gave.

For the bags to be the same, the following must hold:

Original minus Alice's ball + Bob's ball = Original.

Which implies that Bob's ball = Alice's ball.

Similarly, for Bob's bag: Original + Alice's ball - Bob's ball = Original. Which also implies that Alice's ball = Bob's ball.

Therefore, the only way for both bags to be same is if Alice's ball = Bob's ball.

Therefore, the probability is the probability that Bob's ball is equal to Alice's ball.

Which is the same as before.

Therefore, the answer is 1/3.

But let me think again. Suppose that Alice gives a ball to Bob, and Bob gives back a different ball. Then, the contents of the bags are different.

But suppose that Alice gives a ball to Bob, and Bob gives back a different ball. However, maybe there's a case where even though they swapped different balls, the contents are same?

Wait, for example, suppose that Alice gives a blue to Bob, and Bob gives a green back. Then Alice's bag has green instead of blue, and Bob's bag has blue instead of green. But since they both have five balls, and originally they had one of each color, then after the exchange, Alice's bag has two greens (if she had green originally and received green from Bob) or something else? Wait, no. Wait, let me think again.

Original: Alice has one of each color. She gives blue to Bob. So her bag now has green, orange, red, violet. Then Bob gives green back. So Alice's bag now has green, orange, red, violet, green. So two greens. Bob's bag originally had blue, green, orange, red, violet. Then he received blue, so he had two blues, green, orange, red, violet. Then he gave green back, so he has two blues, orange, red, violet.

So Alice's bag has two greens, and Bob's bag has two blues. Therefore, their bags are different.

Alternatively, if Alice gives blue to Bob, and Bob gives back blue. Then Alice's bag has blue, green, orange, red, violet again. Bob's bag has blue, green, orange, red, violet again.

Therefore, only when Bob gives back the same color as Alice gave, the bags are same.

Therefore, the answer is 1/3.

But let me check with another example. Suppose that Alice gives a blue to Bob, and Bob gives back a blue. Then the bags are same.

Alternatively, suppose that Alice gives a blue to Bob, and Bob gives back a blue. So same as original.

Alternatively, suppose that Alice gives a blue to Bob, and Bob gives back a green. Then Alice's bag has green instead of blue, and Bob's bag has blue instead of green.

But since they both have five balls, and the other colors are same.

But the contents are different because one has an extra green and one has an extra blue.

Therefore, the contents are different.

Therefore, only when they swap the same color, the contents are same.

Therefore, the probability is 1/3.

But let me check with another approach.

Let me think of the entire process as a permutation.

But maybe not.

Alternatively, think of the probability that the two bags are the same.

For the bags to be the same, the net transfer must be zero.

Which means that the ball that Alice gave to Bob is the same as the one Bob gave back.

Therefore, the probability is equal to the probability that Bob gives back the same color as Alice gave.

Which is 1/3.

Alternatively, think of it as follows:

The probability that Bob gives back the same color as Alice gave is equal to the probability that Bob picks the same color as Alice.

But since Alice's color is random, and Bob's choice depends on that.

Alternatively, think of it as:

What is the probability that Bob's transferred ball is the same color as Alice's transferred ball?

Which is equal to the expected value of the probability that Bob picks the same color as Alice.

Since Alice's color is uniformly random, and given Alice's color, the probability that Bob picks that color is 2/6.

Therefore, the total probability is 2/6 = 1/3.

Therefore, same answer.

Therefore, I think the answer is 1/3.

But let me check with actual numbers.

Suppose we have 5 colors.

Total number of possible ways: Alice can choose any of 5 colors, Bob can choose any of 6 colors. So 5*6=30.

Number of favorable cases: For each color, Alice chooses that color (1 way), and Bob chooses that color (2 ways). So 5*2=10.

Therefore, probability is 10/30=1/3.

Yes.

Therefore, the answer is 1/3.

But wait, I remember that in some similar problems, the answer is different. Wait, let me think again.

Wait, suppose that instead of 5 colors, we have n colors. Then the probability is 2/(n+1).

Wait, in our case, n=5, so 2/(5+1)=1/3.

But let me check with n=2.

n=2, probability is 2/3.

Yes, as we saw earlier.

Alternatively, think of the problem as follows:

After the exchange, the bags are the same if and only if the ball that Alice gave to Bob is the same as the one Bob gave back.

Therefore, the probability is equal to the probability that Bob's ball is the same as Alice's.

But how is this probability calculated?

Alternatively, think of it as follows:

The probability that Bob's ball is the same as Alice's is equal to the expected value of the probability that Bob picks the same color as Alice.

Since Alice's color is random, and given Alice's color, the probability that Bob picks that color is 2/(n+1).

Therefore, the total probability is 2/(n+1).

Therefore, for n=5, 2/6=1/3.

Therefore, answer is 1/3.

Therefore, I think the answer is 1/3.

But let me check with another approach.

Let me think of the entire process as a random variable.

Let me denote the color that Alice gives to Bob as X, and the color that Bob gives to Alice as Y.

We need to find the probability that X = Y.

Since X is uniformly random over the 5 colors.

Given X, Y is a random color from Bob's bag, which has 6 balls: original 5 colors, with X appearing twice.

Therefore, the probability that Y = X is 2/6 = 1/3.

Therefore, the total probability is E[P(Y = X | X)] = E[1/3] = 1/3.

Therefore, same answer.

Therefore, I think the answer is 1/3.

But let me check with another way.

Suppose we think of the entire process as follows:

What is the probability that after the exchange, both bags have the same contents?

This is equivalent to the probability that the two bags are identical.

Since each bag has five balls, and the only way they can be identical is if the net transfer is zero.

Which requires that the ball Alice gave to Bob is the same as the one Bob gave back.

Therefore, the probability is 1/3.

Therefore, I think the answer is 1/3.

But I want to make sure that there isn't a case where even if they swapped different colors, the bags could still be same.

For example, suppose that Alice gives a blue to Bob, and Bob gives a green back. Then Alice's bag has green instead of blue, and Bob's bag has blue instead of green. However, since they both have five balls, and the rest of the colors are same.

But the contents are different because one has an extra green and one has an extra blue.

But if there are multiple copies? Wait, no. Wait, originally, each color is present once. After the exchange, Alice's bag has one less blue and one more green. So she has two greens? No, she had one green originally. She gave away blue, so she had green, orange, red, violet. Then Bob gives her green. So she has two greens, orange, red, violet.

Bob's bag had two blues, green, orange, red, violet. Then he gives green back, so he has two blues, orange, red, violet.

Therefore, Alice's bag has two greens, and Bob's bag has two blues. Therefore, their bags are different.

Therefore, even if they swapped different colors, the bags are different.

Therefore, only when they swap the same color, the bags are same.

Therefore, the answer is 1/3.

Therefore, I think the answer is 1/3.

But let me check with another example. Suppose that Alice gives a blue to Bob, and Bob gives back a blue. Then Alice's bag is back to original, and Bob's bag is back to original.

Yes.

Therefore, the answer is 1/3.

Therefore, I think the answer is \boxed{\dfrac{1}{3}}.

**Final Answer**
\boxed{\dfrac{1}{3}}

To solve the problem, we analyze the sequence of events and the conditions under which the contents of the two bags become the same after the exchange.

---

### **Initial Setup**

- Both Alice and Bob start with **5 balls**, one of each color: **blue, green, orange, red, violet**.
- Alice randomly selects **one ball** from her bag and **adds it to Bob's bag**.
- Bob then randomly selects **one ball** from his bag (now containing **6 balls**) and **adds it back to Alice's bag**.
- After this exchange, both bags again contain **5 balls**.

We are to find the **probability that the contents of the two bags are the same** after this process.

---

### **Key Insight**

For the contents of the two bags to be the same after the exchange, the **net effect of the transfer must be zero**. That is:

- Alice gives a ball of color **X** to Bob.
- Bob gives a ball of color **Y** to Alice.
- For the bags to be the same, we must have **X = Y**.

If **X ≠ Y**, then the contents of the bags will differ: one will have an extra **X** and one will have an extra **Y**, and the rest of the colors will be the same.

Thus, the only way for the bags to be the same is if **Bob returns the same color that Alice gave him**.

---

### **Step-by-Step Probability Calculation**

1. **Alice's choice**: She picks a color uniformly at random from the 5 available.
   - Probability of choosing any specific color: $ \frac{1}{5} $

2. **Bob's choice**: After Alice adds a ball of color **X** to his bag, Bob's bag now contains:
   - The original 5 balls (including one of color **X**)
   - Plus the additional ball of color **X** from Alice
   - So, **2 balls of color X**, and 1 each of the other 4 colors

   Therefore, Bob's bag has **6 balls**, with **2 of color X**.

   - Probability that Bob picks a ball of color **X** is:
     $$
     \frac{2}{6} = \frac{1}{3}
     $$

3. **Total Probability**:
   - For each color **X**, the probability that Alice picks **X** and Bob picks **X** is:
     $$
     \frac{1}{5} \times \frac{1}{3} = \frac{1}{15}
     $$
   - Since there are 5 colors, the total probability is:
     $$
     5 \times \frac{1}{15} = \frac{1}{3}
     $$

---

### **Conclusion**

The only way for the contents of the two bags to be the same after the exchange is if the ball that Alice gives to Bob is the same color as the one Bob gives back. The probability of this happening is:

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

[Assistant]
Explanation:
Okay, so Bob and Alice each have a bag with balls of five different colors: blue, green, orange, red, and violet. Each bag starts with one ball of each color, so that's five balls in each bag.

The problem says that Alice first randomly selects one ball from her bag and puts it into Bob's bag. Then Bob randomly selects one ball from his bag (which now has six balls) and puts it into Alice's bag. After this process, we need to find the probability that the contents of the two bags are the same again.

Hmm, let me think. So initially, both bags have the same number of balls and the same colors. After Alice adds a ball to Bob's bag, Bob's bag has six balls, and Alice's bag has four. Then Bob takes a ball from his bag (which now has six) and puts it back into Alice's bag. So after this exchange, both bags should have five balls again. The question is, what's the probability that after this exchange, both bags have the same contents?

Wait, but the contents being the same would mean that each bag has the same number of each color. But since they started with one of each color, and they swapped one ball each, maybe there's a way for them to end up with the same contents?

Let me try to break it down step by step.

First, let's consider the first step: Alice takes a ball from her bag and puts it into Bob's bag. Since all balls are equally likely, the probability that she picks any particular color is 1/5. Let's denote the colors as B, G, O, R, V for blue, green, orange, red, violet.

After Alice adds a ball to Bob's bag, Bob's bag has 6 balls: the original five plus one from Alice. Then Bob takes a ball from his bag and puts it into Alice's bag. Now, depending on which ball Bob takes, the contents of both bags will change.

We need to find the probability that after these two transfers, both bags have the same contents.

So, for the contents to be the same, the net result of the transfers must be that each bag ends up with the same set of balls as before. But since they swapped one ball each, maybe the only way this can happen is if the ball that Alice gave to Bob is the same color as the ball that Bob gave back to Alice? Wait, not necessarily. Let me think.

Wait, suppose Alice gives a ball of color X to Bob. Then Bob has an extra ball of color X. Then Bob takes a ball from his bag. If he takes a ball of color X, then he gives it back to Alice, so the net result is that nothing has changed. But if he takes a different color, say Y, then Alice's bag loses a Y and gains an X, while Bob's bag loses an X and gains a Y. So in this case, the contents of the bags would be different.

But the problem says that after this process, the contents of the two bags are the same. So, for the contents to be the same, the net effect of the transfers must be that each bag ends up with the same set of balls as they started.

Wait, but if Alice gives a ball to Bob and Bob gives a ball back, then the only way for the contents to be the same is if the ball that Bob gives back is the same as the one Alice gave. Because otherwise, the contents would differ.

Wait, let me check that. Suppose Alice gives a blue ball to Bob. Then Bob's bag has an extra blue. Then Bob takes a ball from his bag. If he takes blue, then he gives blue back to Alice. So Alice's bag is back to original, and Bob's bag is back to original. If he takes a different color, say green, then Alice's bag now has one less green and one more blue, while Bob's bag has one less green and one more blue. So the contents of the bags are different.

Therefore, the only way for the contents to be the same is if Bob gives back the same color that Alice gave. Therefore, the probability that the contents are the same is equal to the probability that Alice's transferred ball is the same color as Bob's transferred ball.

But wait, is that the only possibility? Let me think again. Suppose Alice gives a blue to Bob. Then Bob has 6 balls: original 5 plus blue. Then Bob takes a ball. The chance he takes blue is 1/6. If he takes blue, then the contents are same. If he takes another color, then the contents are different.

But maybe there's another scenario where even if Bob takes a different color, the contents could still be same? Let me think. Suppose Alice gives a blue to Bob. Then Bob has 6 balls: original 5 (including blue) plus blue. Wait, no. Wait, initially, Bob's bag has one of each color. Then Alice adds one ball to Bob's bag. So if Alice adds blue, then Bob's bag has two blue balls and one each of the other colors. Then Bob takes a ball from his bag. If he takes blue, then he gives blue back to Alice. So Alice's bag now has original minus blue plus blue, so same as before. Bob's bag has original minus blue plus blue, so same as before.

If Bob takes a different color, say green. Then Alice's bag now has original minus blue (since she gave blue to Bob) plus green (since Bob gave green back). So Alice's bag has one less blue and one more green. Bob's bag has original minus green (since he took green) plus blue (since he received blue from Alice). So Bob's bag has one less green and one more blue.

So the contents of the bags are different. Therefore, the only way for the contents to be same is if Bob gives back the same color that Alice gave.

Therefore, the probability that the contents are same is equal to the probability that Bob picks the same color that Alice gave.

But how do we compute that?

First, Alice picks a color. Let's say she picks color X. The probability that she picks any color is 1/5. Then, Bob's bag has 6 balls: original 5 plus one X. So the probability that Bob picks X is 2/6 = 1/3. Therefore, the total probability would be 1/5 * 1/3 = 1/15? Wait, but that seems too low. Wait, but maybe there's another case?

Wait, no. Wait, if Alice picks a color, say X, then Bob has two Xs in his bag. So the probability that Bob picks X is 2/6. Therefore, the total probability is 1/5 * 2/6 = 1/15? But that would be the probability that Alice picks X and Bob picks X. But since there are 5 colors, each with probability 1/5, and for each color, the probability that Bob picks that color is 2/6. Therefore, total probability is 5*(1/5)*(2/6) = 2/6 = 1/3? Wait, that seems conflicting with my previous thought.

Wait, let me clarify. The total probability is the sum over all colors of the probability that Alice picks that color and Bob picks that color.

So for each color X, P(Alice picks X) = 1/5. Then, given that Alice picked X, the probability that Bob picks X is 2/6. Therefore, the total probability is sum_{X} [1/5 * 2/6] = 5*(1/5 * 2/6) = 2/6 = 1/3.

So the probability is 1/3? But let me check again.

Alternatively, maybe there's another way to think about it.

Alternatively, think of the entire process as a random exchange.

But maybe there's another way where the contents can be same even if Bob doesn't return the same color? Wait, let me think. Suppose Alice gives a blue to Bob, and Bob gives a green back. Then Alice's bag has one less blue and one more green. Bob's bag has one less green and one more blue. So the contents of the bags are different.

But suppose that Alice gives a blue to Bob, and Bob gives back a blue. Then both bags are back to original.

Alternatively, suppose that Alice gives a blue to Bob, and Bob gives back a blue. So same as original.

Alternatively, suppose that Alice gives a blue to Bob, and Bob gives back a different color. Then the contents are different.

Therefore, only when Bob gives back the same color as Alice gave, the contents are same.

Therefore, the probability is 1/3.

But let me check with an example. Let me take a smaller case. Suppose there are only two colors, say blue and green. Each has one ball. Alice gives one to Bob. Then Bob gives one back. What is the probability that the contents are same?

Original: Alice has B, G. Bob has B, G.

Alice gives one ball to Bob. Let's say she gives B. Then Bob has B, B, G. Then Bob picks a ball. Probability he picks B is 2/3, probability he picks G is 1/3. If he picks B, then he gives back B. So Alice has B, G again. Bob has B, G again. If he picks G, then Alice has G, G and Bob has B, B. So contents are different. Similarly if Alice had given G. So total probability is 1/2 (probability Alice gives B) * 2/3 (Bob gives back B) + 1/2 (Alice gives G) * 2/3 (Bob gives back G) = 1/2 * 2/3 + 1/2 * 2/3 = 2/3. Which is same as 2/ (number of colors + 1)? Wait, in this case, number of colors is 2, and the probability is 2/3. Which is same as 2/(2 + 1). Wait, but in our original problem, with 5 colors, the probability was 1/3? Wait, no. Wait in the two color case, the probability was 2/3. Wait, but according to the previous formula, if we have n colors, then the probability would be n/(n + 1)? Wait, no. Wait in the two color case, the probability was 2/3. Let me check again.

Wait, in the two color case, the probability that the contents are same is 2/3. Let me think again.

Alternatively, maybe the formula is (number of colors) * [probability Alice picks color X] * [probability Bob picks color X given Alice picked X]. Which would be n * (1/n) * (2/(n + 1)) ) = 2/(n + 1).

In the two color case, that would be 2/(2 + 1) = 2/3, which matches. In our original problem, n = 5, so 2/(5 + 1) = 2/6 = 1/3. So that seems to be the case.

Therefore, the answer would be 1/3? But let me check again with another example.

Suppose n = 1. Wait, n = 1? If there is only one color, then Alice gives the only ball to Bob. Then Bob has two balls of that color. Then Bob gives one back. The probability that he gives back the same color is 2/2 = 1. So the probability is 1. According to the formula, 2/(1 + 1) = 1. Correct.

Another example, n = 3. Then the probability would be 2/(3 + 1) = 1/2. Let me check.

Original: Alice has A, B, C. Bob has A, B, C.

Alice picks one color, say A. Then Bob has A, A, B, C. Then Bob picks a ball. Probability he picks A is 2/4 = 1/2. So for each color, the probability is 1/3 * 1/2 = 1/6. Total probability is 3 * 1/6 = 1/2. Which matches.

Therefore, the formula seems to hold. Therefore, in our original problem, with n = 5, the probability is 2/(5 + 1) = 1/3.

But wait, the answer seems too straightforward. However, let me think again.

But maybe there's another case where the contents can be same even if Bob doesn't return the same color? For example, suppose Alice gives a blue to Bob, and Bob gives back a green. Then Alice's bag has one less blue and one more green. Bob's bag has one less green and one more blue. So the contents of the bags are different. However, suppose that the original bags had multiple copies? Wait, no. Wait, in the original problem, each bag has one of each color. So after the transfer, Alice's bag has four balls, and Bob's has six. Then Bob gives back one. So after the transfer, both have five balls again.

But for the contents to be same, they need to have the same set of colors. However, since they started with one of each color, and they swapped one ball each, the only way for them to have the same contents is if the net transfer is zero. Which would require that the ball Alice gave to Bob is the same as the one Bob gave back.

Alternatively, is there a case where even if they swapped different colors, the contents are same? For example, suppose Alice gives a blue to Bob, and Bob gives a green back. Then Alice's bag has green instead of blue, and Bob's bag has blue instead of green. But since they both have one of each color, except for the swapped ones. Wait, but they both have five balls.

Wait, let me think. Suppose Alice's original bag has one of each color. She gives blue to Bob. So her bag now has green, orange, red, violet. Bob's bag has blue, blue, green, orange, red, violet. Then Bob gives green back to Alice. So Alice's bag now has green, orange, red, violet, and blue? Wait no. Wait, Alice's bag had green, orange, red, violet (four balls). Then Bob gives green back. So Alice's bag now has green, orange, red, violet, and green? Wait, no. Wait, Alice's bag had four balls: original minus blue. Then Bob gives back a green. So Alice's bag now has four balls plus one green? Wait, no. Wait, initially, Alice had five balls. She gave one to Bob, so she has four. Then Bob gives one to Alice, so she has five again.

So if Alice gave blue to Bob, her bag is missing blue. Then Bob gives back green. So Alice's bag now has green instead of blue. So her bag has green, orange, red, violet, and... Wait, original colors were blue, green, orange, red, violet. After giving blue away, she has green, orange, red, violet. Then Bob gives green back. So she has green, orange, red, violet, green? Wait, no. Wait, she had four balls: green, orange, red, violet. Then Bob gives her a green. So she has two greens, orange, red, violet.

But Bob's bag: originally had blue, green, orange, red, violet. Then Alice added blue, so he had two blues, green, orange, red, violet. Then he gave green back. So his bag now has two blues, orange, red, violet, and... Wait, he had six balls: two blues, green, orange, red, violet. Then he removes green, so he has two blues, orange, red, violet. So he has one less green and one more blue? Wait, no. Wait, he had two blues, green, orange, red, violet. Then he removes green, so he has two blues, orange, red, violet. So he has one green less, and no change in blue? Wait, he had two blues, and he didn't remove any blue. So he has two blues, orange, red, violet.

So Alice's bag has two greens, orange, red, violet. Bob's bag has two blues, orange, red, violet. So their bags are different. Therefore, the contents are different.

Therefore, only when Bob gives back the same color as Alice gave, the contents are same.

Therefore, the probability is 1/3.

But let me think again. Wait, maybe there's another case where even if they swapped different colors, the contents are same? For example, suppose Alice gives a blue to Bob, and Bob gives back a blue. Then both bags are same. If Alice gives blue to Bob, and Bob gives back a different color, then contents are different.

Alternatively, suppose that Alice gives a blue to Bob, and Bob gives back a blue. Then both bags are same.

Alternatively, suppose that Alice gives a blue to Bob, and Bob gives back a blue. So same as original.

Alternatively, suppose that Alice gives a blue to Bob, and Bob gives back a green. Then Alice's bag has green instead of blue, and Bob's bag has blue instead of green. So the contents are different.

Therefore, only when Bob gives back the same color as Alice gave, the contents are same.

Therefore, the probability is 1/3.

But let me think again. Let me compute it step by step.

First, Alice picks a color. The probability is 1/5 for each color. Let's say she picks color X. Then Bob's bag has 6 balls: original 5 plus X.

Now, Bob is going to pick a ball from his bag. The probability that he picks X is 2/6 = 1/3.

Therefore, the probability that Bob picks X is 1/3.

Therefore, the total probability is 1/5 * 1/3 = 1/15 for each color, and since there are 5 colors, total probability is 5*(1/15) = 1/3.

Therefore, the answer is 1/3.

But wait, let me check with another approach.

Alternative approach:

Let me think of the entire process as a random exchange.

After Alice and Bob exchange one ball each, what is the probability that their bags are the same?

Note that the bags will be the same if and only if the ball that Alice gave to Bob is the same as the ball that Bob gave back to Alice.

Alternatively, think of it as a permutation.

But maybe not.

Alternatively, think of the possible outcomes.

But maybe another way:

Let me think of the possible cases.

Case 1: Alice gives a ball to Bob, and Bob gives back the same ball.

Case 2: Alice gives a ball to Bob, and Bob gives back a different ball.

In Case 1, the bags are same. In Case 2, they are different.

Therefore, the probability we need is the probability of Case 1.

So, what is the probability of Case 1?

First, Alice picks a ball: 5 choices.

Then Bob picks a ball from his bag, which has 6 balls. The number of balls that are the same as Alice's is 2 (original one plus the one Alice added). Therefore, the probability that Bob picks the same color is 2/6 = 1/3.

Therefore, the total probability is 1/5 * 1/3 * 5? Wait, no. Wait, the probability is 1/5 (for Alice's choice) multiplied by 1/3 (for Bob's choice). Since Alice's choice is independent.

Wait, but since Alice's choice is any color, and for each color, the probability is 1/5, and for each color, the probability that Bob picks that color is 2/6. Therefore, total probability is sum_{colors} [P(Alice picks color) * P(Bob picks color | Alice picked color)] = sum_{colors} [1/5 * 2/6] = 5*(1/5 * 2/6) = 2/6 = 1/3.

Therefore, same result.

Alternatively, think of it as:

The probability that Bob picks the same color as Alice is equal to the expected value of the probability that Bob picks the same color given Alice's choice.

But regardless, the answer seems to be 1/3.

But let me check with actual numbers.

Suppose we have colors B, G, O, R, V.

Let me compute the total number of possible outcomes.

First, Alice has 5 choices. Then Bob has 6 choices. So total number of possible outcomes is 5 * 6 = 30.

Now, how many of these result in the bags being the same?

For each color that Alice could have given, say color X, there are 2 balls of color X in Bob's bag. So Bob can pick color X in 2 ways. Therefore, for each color X, there are 2 favorable outcomes. Since there are 5 colors, total favorable outcomes are 5 * 2 = 10.

Therefore, the probability is 10 / 30 = 1/3.

Yes! So that's another way to think about it.

Total number of possible outcomes: 5 * 6 = 30.

Number of favorable outcomes: For each color, Alice gives that color (1 way) and Bob gives back that color (2 ways). So 5 * 2 = 10.

Therefore, probability is 10/30 = 1/3.

Therefore, the answer is 1/3.

But wait, let me check with another approach.

Alternative approach:

Let me think about the final state of the bags.

After the exchange, both bags have five balls. For the bags to be the same, they must have the same multiset of colors.

Originally, each bag had one of each color.

After Alice gives a ball to Bob and Bob gives a ball to Alice, the contents of the bags will be:

Alice's bag: original minus the ball she gave plus the ball Bob gave.

Bob's bag: original plus the ball Alice gave minus the ball Bob gave.

For the bags to be the same, the following must hold:

Original minus Alice's ball + Bob's ball = Original.

Which implies that Bob's ball = Alice's ball.

Similarly, for Bob's bag: Original + Alice's ball - Bob's ball = Original. Which also implies that Alice's ball = Bob's ball.

Therefore, the only way for both bags to be same is if Alice's ball = Bob's ball.

Therefore, the probability is the probability that Bob's ball is equal to Alice's ball.

Which is the same as before.

Therefore, the answer is 1/3.

But let me think again. Suppose that Alice gives a ball to Bob, and Bob gives back a different ball. Then, the contents of the bags are different.

But suppose that Alice gives a ball to Bob, and Bob gives back a different ball. However, maybe there's a case where even though they swapped different balls, the contents are same?

Wait, for example, suppose that Alice gives a blue to Bob, and Bob gives a green back. Then Alice's bag has green instead of blue, and Bob's bag has blue instead of green. But since they both have five balls, and originally they had one of each color, then after the exchange, Alice's bag has two greens (if she had green originally and received green from Bob) or something else? Wait, no. Wait, let me think again.

Original: Alice has one of each color. She gives blue to Bob. So her bag now has green, orange, red, violet. Then Bob gives green back. So Alice's bag now has green, orange, red, violet, green. So two greens. Bob's bag originally had blue, green, orange, red, violet. Then he received blue, so he had two blues, green, orange, red, violet. Then he gave green back, so he has two blues, orange, red, violet.

So Alice's bag has two greens, and Bob's bag has two blues. Therefore, their bags are different.

Alternatively, if Alice gives blue to Bob, and Bob gives back blue. Then Alice's bag has blue, green, orange, red, violet again. Bob's bag has blue, green, orange, red, violet again.

Therefore, only when Bob gives back the same color as Alice gave, the bags are same.

Therefore, the answer is 1/3.

But let me check with another example. Suppose that Alice gives a blue to Bob, and Bob gives back a blue. Then the bags are same.

Alternatively, suppose that Alice gives a blue to Bob, and Bob gives back a blue. So same as original.

Alternatively, suppose that Alice gives a blue to Bob, and Bob gives back a green. Then Alice's bag has green instead of blue, and Bob's bag has blue instead of green.

But since they both have five balls, and the other colors are same.

But the contents are different because one has an extra green and one has an extra blue.

Therefore, the contents are different.

Therefore, only when they swap the same color, the contents are same.

Therefore, the probability is 1/3.

But let me check with another approach.

Let me think of the entire process as a permutation.

But maybe not.

Alternatively, think of the probability that the two bags are the same.

For the bags to be the same, the net transfer must be zero.

Which means that the ball that Alice gave to Bob is the same as the one Bob gave back.

Therefore, the probability is equal to the probability that Bob gives back the same color as Alice gave.

Which is 1/3.

Alternatively, think of it as follows:

The probability that Bob gives back the same color as Alice gave is equal to the probability that Bob picks the same color as Alice.

But since Alice's color is random, and Bob's choice depends on that.

Alternatively, think of it as:

What is the probability that Bob's transferred ball is the same color as Alice's transferred ball?

Which is equal to the expected value of the probability that Bob picks the same color as Alice.

Since Alice's color is uniformly random, and given Alice's color, the probability that Bob picks that color is 2/6.

Therefore, the total probability is 2/6 = 1/3.

Therefore, same answer.

Therefore, I think the answer is 1/3.

But let me check with actual numbers.

Suppose we have 5 colors.

Total number of possible ways: Alice can choose any of 5 colors, Bob can choose any of 6 colors. So 5*6=30.

Number of favorable cases: For each color, Alice chooses that color (1 way), and Bob chooses that color (2 ways). So 5*2=10.

Therefore, probability is 10/30=1/3.

Yes.

Therefore, the answer is 1/3.

But wait, I remember that in some similar problems, the answer is different. Wait, let me think again.

Wait, suppose that instead of 5 colors, we have n colors. Then the probability is 2/(n+1).

Wait, in our case, n=5, so 2/(5+1)=1/3.

But let me check with n=2.

n=2, probability is 2/3.

Yes, as we saw earlier.

Alternatively, think of the problem as follows:

After the exchange, the bags are the same if and only if the ball that Alice gave to Bob is the same as the one Bob gave back.

Therefore, the probability is equal to the probability that Bob's ball is the same as Alice's.

But how is this probability calculated?

Alternatively, think of it as follows:

The probability that Bob's ball is the same as Alice's is equal to the expected value of the probability that Bob picks the same color as Alice.

Since Alice's color is random, and given Alice's color, the probability that Bob picks that color is 2/(n+1).

Therefore, the total probability is 2/(n+1).

Therefore, for n=5, 2/6=1/3.

Therefore, answer is 1/3.

Therefore, I think the answer is 1/3.

But let me check with another approach.

Let me think of the entire process as a random variable.

Let me denote the color that Alice gives to Bob as X, and the color that Bob gives to Alice as Y.

We need to find the probability that X = Y.

Since X is uniformly random over the 5 colors.

Given X, Y is a random color from Bob's bag, which has 6 balls: original 5 colors, with X appearing twice.

Therefore, the probability that Y = X is 2/6 = 1/3.

Therefore, the total probability is E[P(Y = X | X)] = E[1/3] = 1/3.

Therefore, same answer.

Therefore, I think the answer is 1/3.

But let me check with another way.

Suppose we think of the entire process as follows:

What is the probability that after the exchange, both bags have the same contents?

This is equivalent to the probability that the two bags are identical.

Since each bag has five balls, and the only way they can be identical is if the net transfer is zero.

Which requires that the ball Alice gave to Bob is the same as the one Bob gave back.

Therefore, the probability is 1/3.

Therefore, I think the answer is 1/3.

But I want to make sure that there isn't a case where even if they swapped different colors, the bags could still be same.

For example, suppose that Alice gives a blue to Bob, and Bob gives a green back. Then Alice's bag has green instead of blue, and Bob's bag has blue instead of green. However, since they both have five balls, and the rest of the colors are same.

But the contents are different because one has an extra green and one has an extra blue.

But if there are multiple copies? Wait, no. Wait, originally, each color is present once. After the exchange, Alice's bag has one less blue and one more green. So she has two greens? No, she had one green originally. She gave away blue, so she had green, orange, red, violet. Then Bob gives her green. So she has two greens, orange, red, violet.

Bob's bag had two blues, green, orange, red, violet. Then he gives green back, so he has two blues, orange, red, violet.

Therefore, Alice's bag has two greens, and Bob's bag has two blues. Therefore, their bags are different.

Therefore, even if they swapped different colors, the bags are different.

Therefore, only when they swap the same color, the bags are same.

Therefore, the answer is 1/3.

Therefore, I think the answer is 1/3.

But let me check with another example. Suppose that Alice gives a blue to Bob, and Bob gives back a blue. Then Alice's bag is back to original, and Bob's bag is back to original.

Yes.

Therefore, the answer is 1/3.

Therefore, I think the answer is \boxed{\dfrac{1}{3}}.

**Final Answer**
\boxed{\dfrac{1}{3}}

To solve the problem, we analyze the sequence of events and the conditions under which the contents of the two bags become the same after the exchange.

---

### **Initial Setup**

- Both Alice and Bob start with **5 balls**, one of each color: **blue, green, orange, red, violet**.
- Alice randomly selects **one ball** from her bag and **adds it to Bob's bag**.
- Bob then randomly selects **one ball** from his bag (now containing **6 balls**) and **adds it back to Alice's bag**.
- After this exchange, both bags again contain **5 balls**.

We are to find the **probability that the contents of the two bags are the same** after this process.

---

### **Key Insight**

For the contents of the two bags to be the same after the exchange, the **net effect of the transfer must be zero**. That is:

- Alice gives a ball of color **X** to Bob.
- Bob gives a ball of color **Y** to Alice.
- For the bags to be the same, we must have **X = Y**.

If **X ≠ Y**, then the contents of the bags will differ: one will have an extra **X** and one will have an extra **Y**, and the rest of the colors will be the same.

Thus, the only way for the bags to be the same is if **Bob returns the same color that Alice gave him**.

---

### **Step-by-Step Probability Calculation**

1. **Alice's choice**: She picks a color uniformly at random from the 5 available.
   - Probability of choosing any specific color: $ \frac{1}{5} $

2. **Bob's choice**: After Alice adds a ball of color **X** to his bag, Bob's bag now contains:
   - The original 5 balls (including one of color **X**)
   - Plus the additional ball of color **X** from Alice
   - So, **2 balls of color X**, and 1 each of the other 4 colors

   Therefore, Bob's bag has **6 balls**, with **2 of color X**.

   - Probability that Bob picks a ball of color **X** is:
     $$
     \frac{2}{6} = \frac{1}{3}
     $$

3. **Total Probability**:
   - For each color **X**, the probability that Alice picks **X** and Bob picks **X** is:
     $$
     \frac{1}{5} \times \frac{1}{3} = \frac{1}{15}
     $$
   - Since there are 5 colors, the total probability is:
     $$
     5 \times \frac{1}{15} = \frac{1}{3}
     $$

---

### **Conclusion**

The only way for the contents of the two bags to be the same after the exchange is if the ball that Alice gives to Bob is the same color as the one Bob gives back. The probability of this happening is:

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

Solution: After Alice puts the ball into Bob's bag, his bag will contain six balls: two of one color and one of each of the other colors. After Bob selects a ball and places it into Alice's bag, the two bags will have the same contents if and only if Bob picked one of the two balls in his bag that are the same color. Because there are six balls in the bag when Bob makes his selection, the probability of selecting one of the same colored pair is $2/6=\boxed{\frac{1}{3}}$.


Explanation:
Okay, so I need to find the greatest possible value of the greatest common divisor of $ n + 7 $ and $ 2n + 1 $, where $ n $ is a positive integer. Hmm, let me think about how to approach this.

First, I remember that the greatest common divisor (gcd) of two numbers also divides their linear combinations. Maybe I can use the Euclidean algorithm here? Let me recall: the Euclidean algorithm says that $ \gcd(a, b) = \gcd(b, a - qb) $ where $ q $ is some integer. So maybe I can apply that here.

Let me denote $ a = 2n + 1 $ and $ b = n + 7 $. Then, according to the Euclidean algorithm, $ \gcd(a, b) = \gcd(n + 7, (2n + 1) - 2(n + 7)) $. Let me compute that subtraction:

$ (2n + 1) - 2(n + 7) = 2n + 1 - 2n - 14 = -13 $.

So, $ \gcd(n + 7, -13) $. But since gcd is always non-negative, this simplifies to $ \gcd(n + 7, 13) $.

Therefore, the gcd of $ n + 7 $ and $ 2n + 1 $ is the same as the gcd of $ n + 7 $ and 13. So, the problem reduces to finding the maximum possible value of $ \gcd(n + 7, 13) $, where $ n $ is a positive integer.

Now, since 13 is a prime number, the possible values of the gcd are 1 and 13. Therefore, the maximum possible value is 13, provided that $ n + 7 $ is a multiple of 13.

So, if $ n + 7 = 13k $ for some integer $ k $, then $ \gcd(n + 7, 13) = 13 $. Therefore, the maximum possible value is 13.

But wait, let me check if this is possible. Let me take $ n = 13k - 7 $. Since $ n $ has to be a positive integer, $ 13k - 7 > 0 $, so $ k \geq 1 $, because if $ k = 1 $, then $ n = 13 - 7 = 6 $, which is positive.

Let me test with $ n = 6 $. Then $ n + 7 = 13 $, and $ 2n + 1 = 13 $. Therefore, $ \gcd(13, 13) = 13 $. So that works.

But let me check another value. Let me take $ k = 2 $, so $ n = 26 - 7 = 19 $. Then $ n + 7 = 26 $, and $ 2n + 1 = 39 $. Then $ \gcd(26, 39) $. Let me compute that. The divisors of 26 are 1, 2, 13, 26. Divisors of 39 are 1, 3, 13, 39. So the gcd is 13. So that also works.

Therefore, it seems that 13 is achievable. Therefore, the maximum possible value is 13.

But let me make sure there isn't a higher value. Since 13 is prime, the gcd can't be higher than 13. So 13 is the maximum.

Wait, but let me think again. Suppose there is some other way where the gcd is higher? But since we have reduced the problem to gcd(n + 7, 13), and since 13 is prime, the gcd can only be 1 or 13. Therefore, 13 is the maximum.

Therefore, the answer should be 13.

But let me check another example. Let me take n = 6, which gives gcd(13, 13) = 13. If I take n = 1, then gcd(8, 3) = 1. If n = 2, gcd(9, 5) = 1. n = 3, gcd(10, 7) = 1. n = 4, gcd(11, 9) = 1. n = 5, gcd(12, 11) = 1. n = 6, gcd(13, 13) = 13. n = 7, gcd(14, 15) = 1. n = 8, gcd(15, 17) = 1. n = 9, gcd(16, 19) = 1. n = 10, gcd(17, 21) = 1. n = 11, gcd(18, 23) = 1. n = 12, gcd(19, 25) = 1. n = 13, gcd(20, 27) = 1. n = 14, gcd(21, 29) = 1. n = 15, gcd(22, 31) = 1. n = 16, gcd(23, 33) = 1. n = 17, gcd(24, 35) = 1. n = 18, gcd(25, 37) = 1. n = 19, gcd(26, 39) = 13. So yes, when n is 6 or 19, the gcd is 13.

Therefore, the maximum possible value is indeed 13.

But let me think again if there's a case where the gcd could be higher. Suppose someone says, what if n + 7 and 2n + 1 have a common divisor d > 13? But since we have already established that d divides 13, which is prime, so d can't be higher than 13. Therefore, 13 is the maximum.

Alternatively, suppose we didn't use the Euclidean algorithm. Let me try another approach. Let me suppose that d divides both n + 7 and 2n + 1. Then, d divides any linear combination of these two numbers. For example, d divides (2n + 1) - 2*(n + 7) = 2n + 1 - 2n - 14 = -13. Therefore, d divides -13, so d divides 13. Therefore, the possible values of d are 1 and 13. Therefore, the maximum possible value is 13.

Therefore, regardless of the approach, the answer is 13.

So, I think that's solid. Therefore, the greatest possible value is 13.

**Final Answer**
The greatest possible value of $\gcd(n + 7, 2n + 1)$ is \boxed{13}.

To determine the greatest possible value of $ \gcd(n + 7, 2n + 1) $ for a positive integer $ n $, we begin by applying the **Euclidean algorithm**, which is a powerful tool for computing the greatest common divisor (gcd) of two integers.

---

### Step 1: Apply the Euclidean Algorithm

Let:
- $ a = 2n + 1 $
- $ b = n + 7 $

We compute:
$$
\gcd(2n + 1, n + 7) = \gcd(n + 7, (2n + 1) - 2(n + 7))
$$

Simplify the expression inside the gcd:
$$
(2n + 1) - 2(n + 7) = 2n + 1 - 2n - 14 = -13
$$

So:
$$
\gcd(2n + 1, n + 7) = \gcd(n + 7, -13)
$$

Since the gcd is always non-negative, we can write:
$$
\gcd(n + 7, 13)
$$

---

### Step 2: Analyze the GCD

Now, the problem reduces to finding the maximum possible value of $ \gcd(n + 7, 13) $, where $ n $ is a positive integer.

Note that 13 is a **prime number**, so the only possible values of the gcd are:
- $ \gcd(n + 7, 13) = 1 $, if $ n + 7 $ is not divisible by 13
- $ \gcd(n + 7, 13) = 13 $, if $ n + 7 $ is divisible by 13

Thus, the **maximum possible value** of the gcd is **13**, and it is achievable if $ n + 7 $ is a multiple of 13.

---

### Step 3: Verify Achievability

Let’s find such an $ n $. Suppose:
$$
n + 7 = 13k \quad \text{for some integer } k
\Rightarrow n = 13k - 7
$$

We need $ n > 0 $, so:
$$
13k - 7 > 0 \Rightarrow k \geq 1
$$

For $ k = 1 $, we get:
$$
n = 13(1) - 7 = 6
$$

Check:
- $ n + 7 = 13 $
- $ 2n + 1 = 13 $

So:
$$
\gcd(13, 13) = 13
$$

This confirms that the maximum value of $ \gcd(n + 7, 2n + 1) $ is indeed **13**.

---

### Final Conclusion

Since the gcd can only be 1 or 13 (due to the primality of 13), and 13 is achievable, the **greatest possible value** of $ \gcd(n + 7, 2n + 1) $ is:

$$
\boxed{13}
$$

[Assistant]
Explanation:
Okay, so I need to find the greatest possible value of the greatest common divisor of $ n + 7 $ and $ 2n + 1 $, where $ n $ is a positive integer. Hmm, let me think about how to approach this.

First, I remember that the greatest common divisor (gcd) of two numbers also divides their linear combinations. Maybe I can use the Euclidean algorithm here? Let me recall: the Euclidean algorithm says that $ \gcd(a, b) = \gcd(b, a - qb) $ where $ q $ is some integer. So maybe I can apply that here.

Let me denote $ a = 2n + 1 $ and $ b = n + 7 $. Then, according to the Euclidean algorithm, $ \gcd(a, b) = \gcd(n + 7, (2n + 1) - 2(n + 7)) $. Let me compute that subtraction:

$ (2n + 1) - 2(n + 7) = 2n + 1 - 2n - 14 = -13 $.

So, $ \gcd(n + 7, -13) $. But since gcd is always non-negative, this simplifies to $ \gcd(n + 7, 13) $.

Therefore, the gcd of $ n + 7 $ and $ 2n + 1 $ is the same as the gcd of $ n + 7 $ and 13. So, the problem reduces to finding the maximum possible value of $ \gcd(n + 7, 13) $, where $ n $ is a positive integer.

Now, since 13 is a prime number, the possible values of the gcd are 1 and 13. Therefore, the maximum possible value is 13, provided that $ n + 7 $ is a multiple of 13.

So, if $ n + 7 = 13k $ for some integer $ k $, then $ \gcd(n + 7, 13) = 13 $. Therefore, the maximum possible value is 13.

But wait, let me check if this is possible. Let me take $ n = 13k - 7 $. Since $ n $ has to be a positive integer, $ 13k - 7 > 0 $, so $ k \geq 1 $, because if $ k = 1 $, then $ n = 13 - 7 = 6 $, which is positive.

Let me test with $ n = 6 $. Then $ n + 7 = 13 $, and $ 2n + 1 = 13 $. Therefore, $ \gcd(13, 13) = 13 $. So that works.

But let me check another value. Let me take $ k = 2 $, so $ n = 26 - 7 = 19 $. Then $ n + 7 = 26 $, and $ 2n + 1 = 39 $. Then $ \gcd(26, 39) $. Let me compute that. The divisors of 26 are 1, 2, 13, 26. Divisors of 39 are 1, 3, 13, 39. So the gcd is 13. So that also works.

Therefore, it seems that 13 is achievable. Therefore, the maximum possible value is 13.

But let me make sure there isn't a higher value. Since 13 is prime, the gcd can't be higher than 13. So 13 is the maximum.

Wait, but let me think again. Suppose there is some other way where the gcd is higher? But since we have reduced the problem to gcd(n + 7, 13), and since 13 is prime, the gcd can only be 1 or 13. Therefore, 13 is the maximum.

Therefore, the answer should be 13.

But let me check another example. Let me take n = 6, which gives gcd(13, 13) = 13. If I take n = 1, then gcd(8, 3) = 1. If n = 2, gcd(9, 5) = 1. n = 3, gcd(10, 7) = 1. n = 4, gcd(11, 9) = 1. n = 5, gcd(12, 11) = 1. n = 6, gcd(13, 13) = 13. n = 7, gcd(14, 15) = 1. n = 8, gcd(15, 17) = 1. n = 9, gcd(16, 19) = 1. n = 10, gcd(17, 21) = 1. n = 11, gcd(18, 23) = 1. n = 12, gcd(19, 25) = 1. n = 13, gcd(20, 27) = 1. n = 14, gcd(21, 29) = 1. n = 15, gcd(22, 31) = 1. n = 16, gcd(23, 33) = 1. n = 17, gcd(24, 35) = 1. n = 18, gcd(25, 37) = 1. n = 19, gcd(26, 39) = 13. So yes, when n is 6 or 19, the gcd is 13.

Therefore, the maximum possible value is indeed 13.

But let me think again if there's a case where the gcd could be higher. Suppose someone says, what if n + 7 and 2n + 1 have a common divisor d > 13? But since we have already established that d divides 13, which is prime, so d can't be higher than 13. Therefore, 13 is the maximum.

Alternatively, suppose we didn't use the Euclidean algorithm. Let me try another approach. Let me suppose that d divides both n + 7 and 2n + 1. Then, d divides any linear combination of these two numbers. For example, d divides (2n + 1) - 2*(n + 7) = 2n + 1 - 2n - 14 = -13. Therefore, d divides -13, so d divides 13. Therefore, the possible values of d are 1 and 13. Therefore, the maximum possible value is 13.

Therefore, regardless of the approach, the answer is 13.

So, I think that's solid. Therefore, the greatest possible value is 13.

**Final Answer**
The greatest possible value of $\gcd(n + 7, 2n + 1)$ is \boxed{13}.

To determine the greatest possible value of $ \gcd(n + 7, 2n + 1) $ for a positive integer $ n $, we begin by applying the **Euclidean algorithm**, which is a powerful tool for computing the greatest common divisor (gcd) of two integers.

---

### Step 1: Apply the Euclidean Algorithm

Let:
- $ a = 2n + 1 $
- $ b = n + 7 $

We compute:
$$
\gcd(2n + 1, n + 7) = \gcd(n + 7, (2n + 1) - 2(n + 7))
$$

Simplify the expression inside the gcd:
$$
(2n + 1) - 2(n + 7) = 2n + 1 - 2n - 14 = -13
$$

So:
$$
\gcd(2n + 1, n + 7) = \gcd(n + 7, -13)
$$

Since the gcd is always non-negative, we can write:
$$
\gcd(n + 7, 13)
$$

---

### Step 2: Analyze the GCD

Now, the problem reduces to finding the maximum possible value of $ \gcd(n + 7, 13) $, where $ n $ is a positive integer.

Note that 13 is a **prime number**, so the only possible values of the gcd are:
- $ \gcd(n + 7, 13) = 1 $, if $ n + 7 $ is not divisible by 13
- $ \gcd(n + 7, 13) = 13 $, if $ n + 7 $ is divisible by 13

Thus, the **maximum possible value** of the gcd is **13**, and it is achievable if $ n + 7 $ is a multiple of 13.

---

### Step 3: Verify Achievability

Let’s find such an $ n $. Suppose:
$$
n + 7 = 13k \quad \text{for some integer } k
\Rightarrow n = 13k - 7
$$

We need $ n > 0 $, so:
$$
13k - 7 > 0 \Rightarrow k \geq 1
$$

For $ k = 1 $, we get:
$$
n = 13(1) - 7 = 6
$$

Check:
- $ n + 7 = 13 $
- $ 2n + 1 = 13 $

So:
$$
\gcd(13, 13) = 13
$$

This confirms that the maximum value of $ \gcd(n + 7, 2n + 1) $ is indeed **13**.

---

### Final Conclusion

Since the gcd can only be 1 or 13 (due to the primality of 13), and 13 is achievable, the **greatest possible value** of $ \gcd(n + 7, 2n + 1) $ is:

$$
\boxed{13}
$$

Solution:
Let $d = \gcd(n + 7, 2n + 1)$, so $d$ divides both $n + 7$ and $2n + 1$.  Then $d$ divides $2(n + 7) - (2n + 1) = 13$, so $d$ is at most 13.

If $n = 6$, then $\gcd(n + 7, 2n + 1) = \gcd(13,13) = 13$, which shows that the value of 13 is attainable.  Therefore, the greatest possible value of $\gcd(n + 7, 2n + 1)$ is $\boxed{13}$.