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.
Best Answer
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:$$\frac{\binom22\frac{4!}{2!2!}}{2^4}=\frac6{16}=\frac38$$