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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?



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up vote 5 down vote accepted

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 $$ \frac{{6\choose 4}{4\choose 1}{5\choose 2}3!}{6^5}=\frac{25}{54}.$$

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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

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Indeed, this is how I first thought about it. – mixedmath Sep 11 '13 at 21:28

The probability would be 7,055.9424/7,776 and I'm sorry about the Yahtzee thing. That's 90.74% of the time. Also, got it! I'm talking about the answer dfeuer posted. You happy now?! Now I'm talking about what I said was wrong, you know, about the 1-in-1,296-chance-of-rolling-a-Yahtzee thing with five dice. Do you like my answer? If you do, please vote up! If not, vote down. I have -4 votes and I'm disappointed. Thanks, dfeuer! Also, why do I have a lot of negative votes? By: Vlad Taylor

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This is wrong. Exactly two must match, and none of the others may. – dfeuer Nov 25 '13 at 2:02
Now THIS is probably right. – user111236 Nov 26 '13 at 1:31
This is still wrong, but more importantly, it's useless. You haven't explained where that absurd number you decreed to be the answer came from. – dfeuer Nov 27 '13 at 3:32

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