Andy, Barbara and Chrisy play a game. Their probabilities of winning are 1/2, 1/3, and 1/6 respectively. What is the probability that each wins exatly one of the first three games played? Express your answer as a common fraction.

To find the probability that each player wins exactly one of the first three games, we can use the principle of inclusion-exclusion.

Let's consider the possible outcomes for the first game:
1. Andy wins
2. Barbara wins
3. Chrisy wins

For each outcome, we need to calculate the probability that the other two players don't win the remaining games.

1. If Andy wins the first game, the probability that Barbara wins the second game (1/3) and Chrisy wins the third game (1/6) is:
(1/3) * (1/6) = 1/18

2. If Barbara wins the first game, the probability that Andy wins the second game (1/2) and Chrisy wins the third game (1/6) is:
(1/2) * (1/6) = 1/12

3. If Chrisy wins the first game, the probability that Andy wins the second game (1/2) and Barbara wins the third game (1/3) is:
(1/2) * (1/3) = 1/6

Now, we need to calculate the probability of the intersection (both players winning on a different game) for each case:

1. Andy wins first, Barbara wins second, and Chrisy wins third:
(1/18)

2. Barbara wins first, Andy wins second, and Chrisy wins third:
(1/12)

3. Chrisy wins first, Andy wins second, and Barbara wins third:
(1/6)

Finally, to find the probability that each player wins exactly one game, we need to sum up these probabilities:
(1/18) + (1/12) + (1/6) = 1/18 + 1/12 + 1/6 = 1/9 + 1/12 + 1/6

To combine these fractions, we need to find a common denominator, which is 36:
1/9 = (4/4) * (1/9) = 4/36
1/12 = (3/3) * (1/12) = 3/36
1/6 = (6/6) * (1/6) = 6/36

Adding them up:
4/36 + 3/36 + 6/36 = 13/36

Therefore, the probability that each player wins exactly one of the first three games played is 13/36.