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Not an easy question. Here's a long but logical answer.

The chance of not getting a 3 on any throw is 5/6, so the chance of not getting any threes in the first 10 throws is (5/6)^10.

Now, having reached that point, the chance of throwing three threes in the last three throws is 1/6 x 1/6 x1/6 or 1/216. So the probability of throwing three threes in any chosen three throws is (5/6)^10 x 1/216. This should seem logical because the probability of throwing three threes in the first three throws followed by ten failures to throw threes, should be the same as just getting threes in the last three throws. The probability is the same for getting threes in only the second, fourth and sixth throw, or any other combination of three throws.

Now, there are a set number of combinations of 13 things taken in groups of three. This is given by the formula for C(13,3) =

(13x12x11x10x9x8x7x6x5x4)/((10x9x8x7x6x5x4x3x2x1)x(3x2x1)). That cancels down to (13x12x11)/(3x2x1), and equals 286.

So, there are this many ways you can pick out three throws which you want to be the threes, and the probability of them actually being threes is the first calculation we did.

The answer to the question is 286 x (5/6)^10 x 1/216 which is 0.2138453, about.

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βˆ™ 2013-04-04 13:29:11
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Q: What is the probability that 13 rolls of a fair die will show three threes?
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