# probability that exactly three of them shows the same face and remaining three shows different faces

## Question

$6$ dice are thrown simultaneously ,find the probability that exactly three of them shows the same face and remaining three shows different faces.

## My Approach

Total outcome=$$6^{6}$$ Number of ways to select an outcome which will be same$$=\binom{6}{1}$$

so selected outcome will be $$=\frac{ \binom{6}{1}\times 1 \times 1 \times 1 \times 5 \times4 \times 3}{6^{6}}=\frac{6\times5 \times4\times 3 }{6^6}$$

Am i correct?

Let's throw one by one.

If the first die has been thrown then the probability that the second and the third show the same face as the first, and the others show different faces is:$$\frac16\frac16\frac56\frac46\frac36$$

So this gives the probability that the first $3$ faces are the same. But there more possibilities. Also e.g. the last $3$ faces can be the same. In total there are $\binom63$ possibilities so the probability equals:$$\binom63\frac16\frac16\frac56\frac46\frac36$$

• thanks for your response.But i made a mistake in my solution ,it will be $\binom{6}{1}$ instead of $\binom{6}{3}$ .Can you please check now ? – laura May 15 '18 at 8:36
• There are $\binom63$ possibilities for the $3$ dice that show the same face. oooabc is one of them, but also aooboc is one them. Here o,a,b,c stand for distinct faces. – drhab May 15 '18 at 8:39
• okk so $\binom{6}{3}$ for which $3$ dice and $\binom{6}{1}$ for which value among 1 to 6.right ? – laura May 15 '18 at 8:41
• @drhab I may be wrong but for sets of 6 elements of which 3 are equal shouldn't the possible permutations be $\frac{6!}{3! \times 1! \times 1! \times 1!} = \frac{6!}{3!}$? – Dav2357 May 15 '18 at 8:44
• there is a misunderstanding among us. By $\binom63$ I am not thinking about $3$ different faces of dice but of $3$ spots for the dice that have equal faces. If e.g. the outcome is 233534 then these spots are $2,3$ and $5$. – drhab May 15 '18 at 8:44

One may see it as follows:

• There are 6 options for the equal faces
• $\binom{6}{3}$ choices, which 3 dice show the same face
• $5\cdot 4 \cdot 3$ choices for the mutually different faces of the other three
• all together there are $6^6$ different throws $$\frac{6\cdot \binom{6}{3}\cdot 5\cdot 4\cdot 3}{6^6}= \frac{25}{162}$$