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A 6-sided die is rolled 5 times. What is the probability of getting one pair? I.e. What is the probability of getting an outcome in the form a,a,b,c,d...For example 4,5,5,2,6 is a an outcome where we got one pair (5's).

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Do you mean $\textit{exactly}$ one pair? – Alex Nov 26 '12 at 21:29
Yes, exactly one pair. – sonicboom Nov 26 '12 at 21:30
up vote 1 down vote accepted

If we first count the number of ways to get a distinct pair (e.g. we roll a $6$ twice and get three different numbers for the other rolls) then we can multiply this by $6$ and divide by $6^5$ to get the desired probability.

First consider the positions of the two identical rolls. Since we are rolling $5$ times there are $5\choose2$ ways to place these dice. Then since we can only have one pair, there are $5$ outcomes for the first of the remaining dice, $4$ for the second and $3$ for the last. In total there are ${5\choose2}\times5\times4\times3$ ways to get one specific pair. Finally the probability of getting any one pair and only one pair is: $$ \frac{6\times{5\choose2}\times5\times4\times3}{6^5}. $$

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What about the no. of ways to place each of the non-identical rolls? – sonicboom Nov 27 '12 at 1:14
@sonicboom: This is already accounted for in this method. In the example I provided, we are saying 6 is the number which shows up twice so when we then say there are 5 outcomes for the first non-identical die we are saying it can be 1,2,3,4 or 5 which means we are accounting for the order as well. Same thing goes for the second non-identical die and the third. – Patrick Nov 27 '12 at 3:11
@sonicboom: Another way to solve this that may help you understand is by saying that after "choosing" the number that constitutes the pair, we now have $$ \binom{5}{3}3! $$ ways of choosing the 3 non-indentical die and permuting them. Multiplying this out we see: $$\frac{5!}{3!(5-3)!}3! = 5\times4\times3 $$ which is what we had before. In total the probability is: $$\frac{\binom{6}{1}\binom{5}{2}\binom{5}{3}3!}{6^5} \approx 0.4629$$ Hope that helps – Patrick Nov 27 '12 at 3:11

There are $6^5=7776$ possible outcomes. Among these are ${6\choose 2}\cdot {6\choose 4}\cdot 4!=3600$ with exactly one pair (select the two positions making the pair, select the four values occuring, select an order of these values). Thus the probability of getting exactly one pair (and no triple, no two pairs, no full house, ...) is $\frac{3600}{7776}=\frac{25}{54}$.

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That is not the answer on this page - ? – sonicboom Nov 26 '12 at 21:39
@sonicboom: One source of discrepancy is stated on the wolfram page: "the two of a kind probability excludes rolls that are small straights." – Jason DeVito Nov 26 '12 at 21:42
Where do you account for the fact that the remaining three outcomes must be unique values? – Alex Nov 26 '12 at 21:44
@Alex because I select the four distinct values occuring (one for the pair, three for the rest) with $6\choose 4$. – Hagen von Eitzen Nov 26 '12 at 21:49

The probability of a yahtzee is 1/1296. Do you know how to work this out? If so, you do the same analysis to get the desired probability. See this for a detailed explanation.

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If the dice is fair, the probability to get any of the 6 possible pairs is the same.

Now think of the following: you need to find the probability of getting 2 same values in any of 5 experiments and then 3 different values out of 5 available. There are 6 ways of achieving this result (since each outcome is equiprobable).

Can you handle it from here?

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I would like to be able to do it by the method of counting ways of choosing the numbers and dice as that is how I am doing it for the other Yahtzee probabilities such as full house, etc... – sonicboom Nov 26 '12 at 21:52

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