Probability of repeating dice

Question

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?

Answer 1

As pointed out in the comments above, there are 3 cases to take care of here:

1. All die rolls are different
2. 2 die rolls are the same
3. All die rolls are the same

Let's deal with case 1 first. If all 3 rolls are different then the first die roll has a $\frac{3}{6}$ chance of being one that we need, the second a $\frac{2}{6}$ and then the last a $\frac{1}{6}$ chance. Multiply these out and do some simplification to get $\frac{1}{36}$.

Your original question wanted to condition this on the fact that they were all different so if we are doing that then the probability is simply $\frac{1}{36}$ as shown above.

Now, let's do the second case. The first roll always has a $\frac{2}{6}$ chance of being one that we need and the last one is always $\frac{1}{6}$ but the second one's probability is based on whether or not the first one was the one that occurred once or twice which means our equation is $\frac{2}{6}\frac{1}{6}(\frac{2}{3}\frac{2}{6} + \frac{1}{3}\frac{1}{6}) = \frac{5}{324}$.

As already pointed out, the probability for the third case is $\frac{1}{216}$.

Now let's find the total probability of this happening without the conditioning:

$\frac{4}{6}\frac{5}{6}\frac{1}{36} + $ $(\frac{1}{6}\frac{5}{6} + \frac{5}{6}\frac{2}{6})\frac{5}{324} +$ $\frac{1}{216}^2$

Which gives us (assuming I did not simplify incorrectly lol) a probability of $\frac{1021}{46656}\approx0.02188$