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I have this simple probability problem that I am not sure I solved correctly. I am not interested in formulas, but rather the thought process of how to solve it.

Suppose we roll six 6-sided dice that are equal. I want to find the probability that at least two dice have the same face.

Let's consider the following roll notation: $$ [x] [y] [a][b][c][d] $$ I first started to calculate the probability of the first two dice being the same in an individual 2 die roll, which is $$\frac{1}{6^2}$$ Now considering the next 4 dice as static, the number of permutations of x and y in the 6 die roll would be: $$\frac{6!}{(6-2)!} = 6*5 = 30$$ And with the total number of possible outputs for a->d being 6^4 we would have the probability of rolling the same two dice in a 6 dice roll: $$\frac{\frac{1}{6^2} * 30 * 6^4}{6^6} = \frac{6^2 * 30}{6^6} = \frac{30}{6^4} < \frac{1}{6^2}$$ which doesn't make sense to me, since rolling 6 dice gives you a greater probability of having doubles as opposed to when rolling only 2.

Can anyone pinpoint a correct method of finding this probability?

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  • $\begingroup$ I think this needs some clarification: Are you requiring that exactly two faces be the same? So {6,5,1,5,3,2} would be a win but {6,5,1,5,3,5} or {6,5,6,5,3,2} would be losses? Or are you requiring that at least two be the same? In which case all three of my examples would be wins. $\endgroup$
    – lulu
    Aug 30, 2015 at 13:32
  • $\begingroup$ at least two faces be the same - I've edited the question. $\endgroup$
    – aandrei
    Aug 30, 2015 at 13:34
  • $\begingroup$ Ok, in that case I recommend the usual trick: figure out the probability that no faces match and subtract. $\endgroup$
    – lulu
    Aug 30, 2015 at 13:41

2 Answers 2

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Hint: The probability that at least two dices have the same face is equal to

$1-P(\texttt{None of the dices have the same face})$

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When you say "two with the same dice face" do you mean "at least two"? If so then it is simply 1 minus the probability they are all different. The first one will come up with something. The probability the second is different is $\frac{5}{6}$. The probability the third is different from either of the first two is $\frac{4}{6}$, etc, until the probability the last is different from any of the first 5 is $\frac{1}{6}$. The probability all 6 dice are different is the product of those, $\frac{5!}{6^6}$ and the probability of "at least two different" is $1-\frac{5!}{6^6}$.

If you mean "exactly two", again the first can be anything. The probability the second is the same as the first is $\frac{1}{6}$. Then the probability the third, fourth, fifth, and sixth are different from the all previous rolls is is $\frac{5}{6}$, $\frac{4}{6}$, $\frac{3}{6}$, and $\frac{2}{6}$ respectively. The probability the first two are the same and the last four different is $\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{5}{6}\cdot\frac{4}{6}\cdot\frac{3}{6}\cdot\frac{2}{6} = \frac{5!}{6^6}$.

There are $\binom{6}{2} = \frac{6!}{2!\ \cdot\ 4!} = 15$ different ways to order 6 things, two of which are the same so the probability of two rolls the same, the other 4 different in any order is $15\cdot \frac{5!}{6^6}$.

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