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I have difficulty with set theory definitions, especially with the problem of throwing 4 dice and calculating the probability of having each pair of these four dice showing the same face. I choose the set incorrectly. The answer to this question is to choose 2 faces out of 6 and arrange the dice with repetition for the 4 dice, and the number of arrangements is 3. So here there are two sets: the first is the set of faces, and the second is the number of dice. The total result is 90. I wanted to reason differently by choosing 1 face out of 6 and then choosing 1 face out of the remaining 5. There I find myself wondering how to arrange them to get the 90 ways.

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  • $\begingroup$ Do the two pairs have to show different faces, eg $(4,4)$ and $(3,3)$ or can all four faces be identical ? $\endgroup$ Commented May 23 at 9:20
  • $\begingroup$ They show different faces (4;4) and (3;3) $\endgroup$
    – Mouh Kramo
    Commented May 23 at 9:52
  • $\begingroup$ Ok, it wasn't clear in the question. $\endgroup$ Commented May 23 at 10:15

1 Answer 1

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Here are two ways to think of it.

  1. First choose the number on the first die (6 options). Next, choose which other die has the same number (3 options). Finally, choose the number on the other two dice (5 options). This gives $6\times 3\times 5=90$.

  2. First choose two numbers to be rolled: there are $\binom 62=15$ ways to do this. Then choose which two dice show the higher number: there are $\binom 42=6$ ways to do this. This gives $15\times 6=90$.

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  • $\begingroup$ I have a problem seeing or imagining if all 90 arrangements are accounted for $\endgroup$
    – Mouh Kramo
    Commented May 23 at 10:10

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