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I've been trying to figure this on my own and look on the web but I couldn't find a clear explanation, so I'm looking for a way to calculate the probability to have $N$ identical dice or more out of $x$ rolls.

I think that out of three rolls I have a $\frac{21}{216}$ probability to have at least $2$ identical dice but I'm not really sure how to get there.

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To have exactly $N$ dice the same out of $x$ rolls it is easy if $N \gt \frac x2$ because you can't have more than one set of them. For $N=2, x=3$ out of $216$ ordered rolls you have ${3 \choose 2}=3$ ways to choose the matching dice, $6$ choices for the number to match, and $5$ for the other number, for a total of $90$ ways. This gives a chance of $\frac {90}{216}=\frac 5{12}$. There are $6$ ways to get three the same for three dice. With the restriction $N \gt \frac x2$ there are ${x \choose N}6\cdot 5^{x-N}$ ways following the same logic.

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  • $\begingroup$ Thank you but I realized I misstititled my question I actually meant N sixes out of x rolls. $\endgroup$ – 8ilO Jan 21 '17 at 18:16
  • $\begingroup$ Then you divide these numbers by $6$ because that factor was to choose the value of the identical dice. The chance for exactly two sixes is then $\frac {15}{216}$ and $\frac 1{216}$ for three. $\endgroup$ – Ross Millikan Jan 21 '17 at 18:18

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