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I roll 6 fair dice. What is the probability that at least one pair shows up?

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up vote 8 down vote accepted

The number of possible output$=6^6$ (each dice can have 6 possible outputs)

If we don't allow pair, then the 1st can have $6$ possible outputs, the 2nd can have $(6-1)$ possible outputs(excluding the previous result), the 3rd $(6-2)$ possible outputs(excluding the previous two results) and so on.

So, the number of possible output with no pair is $6!$

The probability of no pair is $\dfrac{6!}{6^6}=\frac{5}{324}$

The probability of at least one pair will be $1-\dfrac{6!}{6^6}=1-\frac{5}{324}=\frac{319}{324}$.

(as the probability of at least one $= 1- \text{the probability of none)}$.

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$1-6!/6^6\approx 0,984567901$. – Sigur Aug 29 '12 at 14:32
Contrast the typography in (6-1) with that in $(6-1)$. Your minus signs looked like hyphens since you didn't put them in $\TeX$. (I changed that.) (+1, however). – Michael Hardy Aug 29 '12 at 17:03

If I understand you correctly, you have six dice which you are rolling, and you want to know the probability of at least two dice of the same number appearing in that roll. One way to solve this is to subtract the probability of there not being a pair in your six dice from unity. There are $6!$ permutations of the six dice where each die is unique out of a total of $6^{6}$ possible permutations. This means that there are $6^{6} - 6!$ possible rolls of your dice that contain at least a pair. Therefore, $$1 - 6!/6^{6} \approx 0.98.$$

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