Three fair 6−sided dice are rolled and their upfaces are recorded.
Find the probability that the values showing upon rolling all three
dice again is the same as the original three values recorded.
Let $X_1$, $X_2$ and $X_3$ be the three first rolls. We observe $X_1=x$, $X_2=y$ and $X_3=z$ and we are interested in $$P\bigg((X_4, X_5,X_6) \in \sigma(x,y,z) | X_1=x,X_2=y, X_3=z\bigg) $$
Meaning that if we observe $(X_1,X_2,X_3)=(1,2,4)$ we want $(X_4, X_5,X_6)$ to be in the set of permutations of $(x,y,z)$ for example $(2,4,1)$ would be fine, right?
Now I feel that the answer depends on $x,y,z$ because if $x=y=z$ the probability we are looking for is $1/6^3=1/216$. If $x\neq y\neq z$ then that probability is now $6/6^3=1/36$ because there are 6 possible combinations.
I wanted to condition on the fact that numbers of three first rolls could be repeated or not but it gets messier, is there a way out?
Best Answer
I don't think the accepted answer is correct. It errs in handling the case where the initial roll results in two of the same value and one different value.
We consider three cases:
The first 3 dice roll the same value (has probability $\frac{1}{36}$). In this case our probability of getting the same values on the second roll is indeed $\frac{1}{216}$
If the first 3 dice all roll different values which occurs with probability $\frac{5}{36}\frac{4}{36} = \frac{20}{36}$, we have probability $\frac{3!}{216} = \frac{1}{36}$ of seeing the same values on our second roll.
In the case where the initial 3 dice have two of the same value and one different value, we have 3 possible ways of this appearing in the repeated roll. Say the first 3 dice roll (1,1,2). Our repeated roll can be (2,1,1) (1,2,1) or (1,1,2) each with equal probability. The probability in this second case should be 3/216 not 5/324.
The mistake in the accepted answer is that while the first die has 2 possibilities, and the third die always has 1 possibility, the second die is equally likely to have two possibilities as it is to have one possibility. So it should be $\frac{2}{6}\frac{1}{6}(\frac{1}{2}\frac{2}{6}+\frac{1}{2}\frac{1}{6}))$
So the overall probability is $\frac{20}{36}\frac{6}{216}+\frac{15}{36}\frac{3}{216}+\frac{1}{36}\frac{1}{216} = \frac{166}{7776} = 0.0213$