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In one roll of four standard six-sided dice, what is the probability of rolling exactly three different numbers?

My answer: $6\cdot5\cdot4$ to get 3 numbers, then the 4th dice will have to be one of the previous 3 numbers, so it's $6\cdot5\cdot4\cdot3$; the total possibilities is $6^4$; the answer should then be $\frac{6\cdot5\cdot4\cdot3}{6^4}=\frac5{18}$.

But the correct answer is $\frac59$, what did I miss? how should one think about resolving this kind of problem?

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Noting that there are $6\cdot5\cdot4$ ways to select the three numbers that show up is a good start; it accounts for the order in which they get rolled too. However, instead of saying "the fourth die shows one of the previous three numbers", pick which two of the four rolls show the same number – there are $\binom42=6$ ways. You can fill in the individual rolls uniquely after this.

With this correction, you get the correct probability of $\frac59$.

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  • $\begingroup$ After some thinking, this way of logic seems more natural to me: to satisfy the requirement of just 3 different numbers, 2 of the dice must have same number, there are 4C2 ways to choosing these 2 dices, and they have 6 numbers as posssibilities; after that next dice has 5 possibilities, and next one has 4 possibilities, so there are 4C2 * 6 * 5 * 4 total possibilities; that divide by 6^4 gives 5/9 the right answer. Please let me know if anyone see anything wrong with this logic $\endgroup$ – user526427 Mar 3 '18 at 22:25
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Sample space = $(6)^4$

No of ways of selecting 3 different numbers =$\binom{6}{3}$

No of ways of selecting 4th number = 3

No.of ways of arranging these 4 numbers = $\frac{4!}{2!}$

Total = 3$\binom{6}{3}$$\frac{4!}{2!}$ = 720

Probability =$\frac{720}{1296}=\frac{5}{9}$

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Let us say that 1 is repeated twice, then the total number of different possibilities of having 1,1,a and b where a,b E {2,3,4,5,6} and a not equal to b; on four dices is given as 6x3!x5P2=6x3x2x5x4 Hence the probability of such is 6x3x2x5x4/6x6x6x6 = 5/9

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