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What is the probability of showing $3$ faces same and $3$ faces different when $6$ dice are thrown simultaneously?

Here's my approach:

$6$ dice thrown. Total number of outcomes that first $3$ dice show the same number is $6$, actually $\{(1,1,1),(2,2,2),(3,3,3),(4,4,4),(5,5,5),(6,6,6)\}$.

Total number of outcomes that next $3$ dice show all different numbers ( other than the one appeared in first $3$ dice ) are $5C3$ .

Total number of outcomes $= 6^6$. So according to me it should be $6\times 5C3 / 6^6$ .

Please guide me through the correct answer.

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    $\begingroup$ Please post your attempt with the question to indicate how you're thinking about this question. $\endgroup$ – OnceUponACrinoid Nov 17 '18 at 7:09
  • $\begingroup$ 6 dices thrown. Total number of outcomes that first 3 dices show the same number is 6 {(1,1,1),(2,2,2),(3,3,3),(4,4,4),(5,5,5),(6,6,6)} . Total number of outcomes that next 3 dices show all different numbers ( other than the one appeared in first 3 dices ) are 5C3 . Total number of outcomes = 6^6. So according to me it should be 6 x 5C3 / 6^6 $\endgroup$ – Vaibhav Sachdeva Nov 17 '18 at 7:59
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    $\begingroup$ That calculation counts the number of ways, when it's the first three dice that show the same number. $\endgroup$ – Gerry Myerson Nov 17 '18 at 8:29
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Guide

  • Work in probability space $\{1,2,3,4,5,6\}^6$ (so the dice are ordered/numbered) have $6^6$ equiprobable outcomes.
  • Choose one face out of $6$ to be the one that is shown $3$ times.
  • Choose $3$ spots/numbers out of $6$ for the chosen face.
  • Now $3$ spots are left to be filled up with other faces that are moreover distinct. For the first of these $3$ spots there are $5$ choices (of face), for the second $4$ and for the third $3$.

$$6^{-6}\cdot6\cdot\binom63\cdot5\cdot4\cdot3$$

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