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What is the minimum number of fair 6-sided dice that you have to roll in order to get 50% probability that all 6 faces (1, 2, 3, 4, 5, 6,) will show at least once?. Thank you for help.

Guest Aug 2, 2018
 #1
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There is a formula you can use to calculate the probability of all faces showing on any number of dice. From that, you can easily determine the minimum number of dice you need to roll. Here is the formula in question: Just change "n" as required and it will give you the probability.

n = 10;  1 - ((6C1) *(5/6)^n - (6C2)*(4/6)^n + (6C3)*(3/6)^n - (6C4)*(2/6)^n + (6C5)*(1/6)^n).

Here are half a dozen calculations to give you an idea: 

n=10        p=0.271 812 128 5

n=11        p=0.356 206 418 6

n=12        p=0.437 815 680 6

n=13       p=0.513 858 194 0

n=14       p=0.582 845 348 9

n=15       p=0.644 212 738 6

As you can easily see that you will need to roll a minimum of 13 dice in order to get at least 50% probability that all 6 faces will show.

Guest Aug 2, 2018
 #2
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I'd really like a mathematician to talk through this question/answer.     frown

Melody  Aug 2, 2018
 #3
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A variation on this question has come up before on this forum. The most important example is discussed here in detail: https://web2.0calc.com/questions/probability_882

Guest Aug 2, 2018
 #4
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Thanks Guest :)

Melody  Aug 5, 2018

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