Okay, let's try to figure out this probability problem. So, Bob and Alice each start with a bag containing five balls: blue, green, orange, red, and violet. Alice randomly takes one ball from her bag and puts it into Bob's bag. Then Bob randomly takes one ball from his now-increased bag (which now has six balls) and puts it back into Alice's original bag. We need to find the probability that after this exchange, both bags have the same set of balls. Hmm.

First, let me make sure I understand the problem correctly. Initially, both bags have the same five colors. Alice moves one color from her bag to Bob's. Then Bob moves one color back. After these two moves, we want the probability that both bags have exactly the same five colors again. So, essentially, the net effect of the two moves should result in the original configuration.

Wait, but the problem says "the contents of the two bags are the same." So, does that mean each bag has the same set of balls? Since Alice's bag was originally five, Bob's was five. After Alice moves one to Bob, Bob has six, Alice has four. Then Bob moves one back, so Bob has five again, and Alice has five. So, after both moves, both have five balls. For them to be the same, each must have the same five colors as the other. But since they started with the same five, the only way this can happen is if the ball Alice took from her original bag and put into Bob's bag is the same ball that Bob later takes back into Alice's original bag.

Wait, is that the case? Let me think. Suppose Alice takes a red ball and puts it into Bob's bag. Then Bob has six balls: his original five plus the red. Bob then randomly selects one to put back into Alice's bag. If Bob picks red, then Alice's bag will have her original four plus red, which is five, and Bob's bag will have the original five minus red plus red, which is five. So, both bags are the same as they started. But if Bob picks a different color, say blue, then Alice's bag would have her original four plus blue, and Bob's bag would have the original five plus red minus blue. So, the contents would differ.

Therefore, the only way the contents of the two bags are the same is if the ball Alice moved to Bob's bag is the same one that Bob moves back. Therefore, the probability we need is the probability that Bob selects the same color that Alice initially moved.

So, let's model this step by step.

First, Alice selects a ball from her bag. Since all balls are equally likely, each color has a probability of 1/5. Let's denote the color Alice picks as C. Then, Bob's bag now has six balls: his original five plus C. Then Bob selects a ball from his six to put back. The probability that he selects C is 1/6. Therefore, the probability that the two bags are the same is the probability that Alice picks C and then Bob picks C. Since Alice's choice is independent of Bob's, we can compute it as (1/5) * (1/6) = 1/30. But wait, that seems low. Let me check again.

Wait, no. Because regardless of which color Alice picks, the probability that Bob picks the same color is 1/6. So, since Alice has five possible colors to pick, each with probability 1/5, and for each color, the probability Bob picks the same is 1/6. So, the total probability is sum over each color of (1/5)*(1/6) = 5*(1/30) = 1/6. Wait, that contradicts my previous thought.

Wait, maybe I need to think differently. Let's consider the entire process.

The total number of possible outcomes is the number of choices Alice can make times the number of choices Bob can make. Alice has 5 choices, Bob has 6 choices. So total possible outcomes: 5*6 = 30. Now, how many of these outcomes result in the bags being the same?

For the bags to be the same after the transfers, we need that the ball Alice moved to Bob's bag is the same one Bob moved back. So, for each color C, if Alice moves C to Bob's bag, and Bob then moves C back, then the bags are the same. So, for each color, there's exactly one outcome where Alice moves C and Bob moves C. Therefore, there are 5 favorable outcomes (one for each color). Therefore, the probability is 5/30 = 1/6.

Wait, so that would be 1/6. But let me verify again. Let's take an example. Suppose Alice picks blue. Then Bob's bag has blue plus the original four. Bob then has to pick blue to get the bags back to original. The probability that Bob picks blue is 1/6. Similarly, if Alice picks green, same thing. So for each color, the probability is 1/5 (Alice picks it) * 1/6 (Bob picks it). So total probability is 5*(1/5)*(1/6) = 1/6. So that seems correct.

But wait, let me think of another way. Suppose we fix the color that Alice moves. Let's say she moves color C. Then Bob has to move C back. The probability that Bob does this is 1/6. Since Alice has 5 choices, each equally likely, the total probability is 1/5 * 1/6 summed over all 5 colors, which is 5*(1/5*1/6) = 1/6. So, that's consistent.

Alternatively, think of the process as a permutation. The net effect is that two balls are swapped: Alice gives one to Bob, and Bob gives one back. For the bags to be the same, the net effect must be that the same ball is swapped back. So, the probability is equal to the probability that Alice and Bob swap the same ball. The number of possible swaps is 5 (each ball can be swapped), and the total number of possible exchanges is 5*6=30. So, probability is 5/30=1/6.

Therefore, the answer is 1/6. But wait, let me check once more to be sure.

Alternatively, let's compute it step by step. Let's fix the color Alice moves. Let's say she moves red. Then Bob's bag now has red, blue, green, orange, violet, and the original five. Wait, no. Wait, Bob originally has five balls, and Alice moves one into his bag. So, Bob's bag now has six balls: his original five plus the one from Alice. Then Bob randomly picks one of those six to put back into Alice's bag. The probability that he picks the same one Alice moved is 1/6. Therefore, regardless of which color Alice moved, the probability that Bob moves the same color back is 1/6. So, the overall probability is 1/6. Therefore, the answer is 1/6.

But wait, another way to think about it is to consider the entire process as a permutation of the balls. The original set is A: {B, G, O, R, V} for Alice and B: {B, G, O, R, V} for Bob. After Alice moves one to Bob, Alice has four, Bob has six. Then Bob moves one back. The probability that the net result is that Alice's bag is back to five original, and Bob's bag is back to five original. Which only happens if Alice moved a ball to Bob, and Bob moved the same back. So, the probability is the chance that the two moves are inverses of each other. Since each time, the selection is random, the probability is 1/5 * 1/6 = 1/30 for a specific color, times 5 colors, so 5*(1/30)=1/6. So, same result.

Alternatively, consider the problem as a Markov chain. The state is the set of balls in Alice's bag. Initially, it's {B, G, O, R, V}. After Alice moves one to Bob, her bag has four. Then Bob moves one back. The probability that Bob moves the same one Alice moved is 1/6. Therefore, the probability that Alice's bag is restored is 1/6. Therefore, the answer is 1/6.

Alternatively, think of all possible transfers. The total number of possible first moves is 5 (Alice's choices). For each of these, the number of possible second moves is 6 (Bob's choices). So total 30 possibilities. The number of favorable outcomes is 5 (each color being swapped back). So, 5/30=1/6.

Therefore, I think the answer is 1/6. But let me check with another approach to make sure.

Let's think of the two transfers as a single operation. The first transfer is Alice moving a ball to Bob, and the second transfer is Bob moving a ball to Alice. The net effect is that Alice has her original set minus the ball she moved plus any ball Bob moved. For the contents to be the same, the ball Bob moved must be the same as the one Alice moved. Therefore, the probability is equal to the probability that Bob moves the same ball Alice moved. Since Bob has six balls, one of which is the one Alice moved, the probability is 1/6. Therefore, regardless of which color Alice moved, the chance is 1/6. Therefore, the probability is 1/6.

Another way: suppose we fix the color that Alice moves. Let's say she moves red. Then Bob's bag has six balls: red and the original four. Bob has to pick red to put back. The chance is 1/6. Similarly, if she moves green, same thing. Since there are five colors, each with equal probability, the total probability is 5*(1/5)*(1/6) = 1/6. So, that's consistent.

Alternatively, using conditional probability. Let’s denote:

- Event A: Alice picks color C and puts it into Bob's bag.

- Event B: Bob picks color C from his bag and puts it into Alice's bag.

We need P(A and B) = P(A) * P(B|A). Since A and B are independent? Wait, no. Once Alice has picked C and put it into Bob's bag, Bob's bag has six balls, one of which is C. So, P(B|A) is 1/6. Therefore, P(A and B) = P(A) * 1/6 = (1/5)*(1/6) = 1/30 for each color. Summing over all five colors, we get 5*(1/30) = 1/6.

Therefore, all approaches lead to the same answer of 1/6. Therefore, I think the correct probability is 1/6.
Therefore, the probability that after this process the contents of the two bags are the same is $\boxed{\frac{1}{6}}$.
</think>