Roll five dice. What's the chance of rolling exactly one pair? If I roll five dice, what's the chance that exactly two of the die show the same number?
I know that the total number of possible outcomes is $6^5$ = 7776. I calculated the probability that at least 2 of the die will have the same outcome. To do this, I found the probability that no two die will be alike: 
$\frac{6!}{7776}$ = .0926
Then I subtract that result from 1: 1 - .0926 = .9074. 
This is the probability that there be AT LEAST one pair. However, I need to find the probability that there will be EXACTLY one pair. The only way I can think of doing this is by subtracting the prob. that there will be at least five pairs, four pairs, etc. from .9074. But is there a quicker way to solve the problem?
Thanks,
KM
 A: To have exactly one pair (no two pairs, not three of a kind, not "full house", ...) you will have a total of exactly four distinct values, one of these occuring twice. So out of the $6^5$ total possible dice rolls, there are ${6\choose 4}{4\choose 1}{5\choose 2}3!$ good possibilities (pick the four values, pick the duplicate value, pick the dice making the pair, rearrange the remaining three dice).
So the probability is 
$$ \frac{{6\choose 4}{4\choose 1}{5\choose 2}3!}{6^5}=\frac{25}{54}.$$
A: Another way to argue: You can choose the two paired dice in $\binom52$ ways, the value of the pair in 6 ways and the non-paired dice in $5\cdot4\cdot 3$ ways, leading to the same answer as Hagen's
A: According to Binomial Distribution, the number of ways, say 3, appear twice during the five rolls is given by a random variable X i.e.
$X \sim \mathrm{Bin}(5, 1/6)$
Here 5 = no. of rolls and 1/6 is probability of getting 3 during an independent roll.
Now,
$P(X = 2) = {\binom{5}{2}\cdot(1/6)^2\cdot(5/6)^3} \approx 0.1608$
Now, for any number to appear twice we have;
$6\cdot P (X = 2) \approx 96.45\%$
