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The probability of throwing three dice and getting three different numbers is:

$\frac{6(5)(4)}{6^3}$

but why can't I also do: P(three different numbers) = 1 - P(3 same numbers):

P(three different numbers) = 1 - $\frac {\binom{6}{1} \binom{1}{1} \binom{1}{1}}{6^3}$

Thanks

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    $\begingroup$ The complementary event of "three different" is "three not all different." The complementary event of "three the same" is "three not all the same." $\endgroup$
    – David K
    Sep 25 at 16:25

2 Answers 2

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"Three different numbers" and "three same numbers" are not complementary. There's also "two same numbers and one different", which fills the gap between them. Thus you cannot take one minus the three-same probability and get the three-different probability.

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This is kind of coupon collector problem... with 6 different cupons (the six results in the dice). Each dice thrown is like a box of cereal bought. You want the probability of having 3 different coupons in 3 boxes of cereal bought.

You gotta think about the number of possibilities for 3 numbers in 3 dice without any constraint and the number of possibilities in which the 3 numbers are different in 3 dice. Easier to think considering order matter.

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