# Chance of getting $3$ same faces and $3$ different faces when $6$ dice are thrown

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.

• Please post your attempt with the question to indicate how you're thinking about this question. – OnceUponACrinoid Nov 17 '18 at 7:09
• 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 – Vaibhav Sachdeva Nov 17 '18 at 7:59
• That calculation counts the number of ways, when it's the first three dice that show the same number. – Gerry Myerson Nov 17 '18 at 8:29

• 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 $$3$$.
$$6^{-6}\cdot6\cdot\binom63\cdot5\cdot4\cdot3$$