# If three dice are rolled, what is the probability that all three are the same number?

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

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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Well, the probability would be $1\cdot$$1\over 6$$\cdot{1\over 6}$ because for three dice, there are six outcomes for the same number because there are six numbers, so put in a 1 for the first factor and every other outcome will be different numbers and you're talking about six-sided dice, so use $1\over 6$s for the next two factors. Also, the odds is 1 in 36 because of this. Hope this helps! Good luck also trying to beat the odds (if you have any dice, that is...)!

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