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The dice are fair.

You have a $1\over6$ chance of getting the first number. A $1\over6$ chance of the second and so on. Is it just $({1\over6})^3$ (1/216) or is that not accounting for the second and third roll properly?

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I think the error you are making is that you're not accounting for the first roll properly... –  David Mitra Mar 8 '12 at 3:30

2 Answers 2

up vote 8 down vote accepted

It's just $({1\over6})^2$ It's the probability that the second roll is the same as the first (1/6) multiplied by the probability that the third roll is the same as the second (1/6).

Or, think of it this way. The desired outcomes are $(1,1,1)$, $(2,2,2)$, ... ,$(6,6,6)$. Each of these outcomes has probability $({1\over6})^3$. Sum these the probabilities of these mutually exclusive outcomes to get $6\cdot({ 1\over6})^3 =({1\over6})^2$.

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Makes sense, thanks for clearing it up very succinctly! –  switz Mar 8 '12 at 3:33

total no of outcomes will be 6x6x6=216.....favourable otcomes are (1,1,1), (2,2,2), ... ,(6,6,6)...i.e. 6 favouracle outcomes.....so probability will be 6/216=1/36

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