# Probability of getting $x$ dice with the same numbers of $n$ dice thrown once together

There are $$6$$ dice, where each die consists of $$6$$ values $$(1,2,3,4,5,6)$$. They are all thrown once at the same time. What is the probability of getting three dice pairs with the same numbers shown?

I understand that there are $$6^6$$ possible outcomes.

The solution from the book says:

$$\frac{_6C_3 \cdot 6!}{2! \cdot 2! \cdot 2!\cdot6^6}.$$

But I still don’t understand where this solution comes from. I’m even doubtful that it's a correct solution.

Because, let’s say that I change the scenario from dice to coins, which have $$2$$ values, head or tail, and assume that there are $$4$$ coins, thrown once together at the same time. If I’m looking for the probability of $$2$$ coins that may show the same number, the equation using the same formula as above would be

$$\frac{_4C_2 \cdot4!}{2!\cdot2!\cdot4^4}=\frac{6\cdot6}{16}=\frac{24}{16}.$$

How is this logical? This is why I’m very doubtful with the solution shown from the lecture book. Is there perhaps anything wrong from my trial? Or is there something that I missed to understand.

• Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. Commented Sep 2, 2018 at 13:56
• @JoséCarlosSantos noted, question description updated Commented Sep 2, 2018 at 14:03

Give the dice labels $1,2,3,4,5,6$ to make them distinguishable and let $D_i$ denote the face taken by die $i$.

Under the mentioned restriction the set $\{D_1,D_2,D_3,D_4,D_5,D_6\}$ can take $\binom63$ values.

(example of such a value: $\{1,4,5\}$)

Each value can be achieved on $\frac{6!}{2!2!2!}$ ways.

(example continued: the RHS of $(D_1,D_2,D_3,D_4,D_5,D_6)=(1,4,5,1,5,4)$ is one of those tuples that lead to: $\{D_1,D_2,D_3,D_4,D_5,D_6\}=\{1,4,5\}$)

Then there are $6^6$ possible outcomes if there are no restrictions.

So we get probability:$$\frac{\binom63\frac{6!}{2!2!2!}}{6^6}$$

If the same principle is correctly implied on your scenario with the $4$ coins then the result is:$$\frac{\binom22\frac{4!}{2!2!}}{2^4}=\frac6{16}=\frac38$$

• I see, so there will be C(6,3) x 6! / 2! 2! 2! possible outcome which three dices will show the same number. I understand C(6,3) which is 3 dice values taken from 6 values of a dice, is that correct? if it's from a dice, so that would be 3 different values because there are like 6 different values in a dice that there is no way there will be 3 different values in a dice. but why 6! / 2! . 2! . 2! ? it's something that the lecture book didn't explain Commented Sep 2, 2018 at 14:37
• $3$ faces out of $6$ will show up. That gives the factor $\binom63$. Now focus on e.g. the faces $1,4,5$ (used as example in my answer). $\frac{6!}{2!2!2!}$ it the number of tuples in $\{1,4,5\}^6$ such that exactly $2$ of its entries take value $1$, exactly $2$ take value $4$ and exactly $2$ take value $5$. Notice that there are $6!$ orders to write strings like $A_1A_2B_1B_2C_1C_2$. But if equal letters are not distinghuishable then there is multiple counting that must be repaired by dividing with $2!2!2!$, There are $\frac{6!}{2!2!2!}$ orders to write $AABBCC$. Commented Sep 2, 2018 at 14:54
• I understand now, got it Commented Sep 2, 2018 at 15:05