The problem that I solved is: given a tetrahedron (see below figure for reference), for each face, you choose 1 of its 3 edges at random. What is the probability that an edge is picked twice?
The way I approached this problem is as follows:
I started with the bottom face (the one you can't see in the figure below). The probability of choosing a unique edge is 1. Then without loss of generality, say that we choose the edge shared with the dark green face so now the dark green face has a 2/3 probability of choosing a unique face. Again, without loss of generality, I said it chose the face shared with the brown face. So far, we have 1*2/3 as the probability for choosing unique edges for the first 2 faces. Now for the brown face, it could either choose the edge shared with the bottom face with probability 1/3 or it could choose the face with probability 1/3 with the lighter green face.
If it chooses the edge shared with the bottom face, then the lighter green face has a probability of 1 of choosing a unique edge. If it chooses the edge shared with the lighter green face, then the lighter green face has a probability of 2/3 of choosing a unique edge so the probability of all 4 faces choosing a unique edge is
$$ 1*\frac{2}{3}\frac{1}{3}(\frac{2}{3} + 1) = \frac{10}{27} $$ The probability of choosing an edge twice is thus the complement of this. I believe this is the correct answer, but I am having a hard time writing it out formally. By formally, I mean in terms of variables/events and in terms of total/conditional probabilities. The numbers that I wrote above are just things that made sense to me intuitively, but there's obviously no formality behind it.
How would you wrote this out formally?
I tried to define events $U1, U2, U3$ and $U4$ as the events that face 1, face 2, face 3, and face 4 choose a unique edge, respectively. So the probability of all 4 faces choosing a unique edge is $P(U1, U2, U3, U4)$.
Then I tried to write this as $$ P(U1, U2, U3, U4) = P(U1)P(U2|U1)P(U3|U1,U2) P(U4|U1, U2, U3) $$
But it's not clear to compute these probabilities. I think $P(U1) = 1$ for the same reason I stated in my above solution (since it's the first face that picks an edge, all 3 of its edges have obviously never been picked before). Now, $P(U2|U1)$ is the probability that the second face chooses a unique edge given that the first edge is unique. $U1$ eliminates 1 possible edge to choose from for face 2, so we have $\frac{2}{3}$. But I'm kind of confused on how to compute $P(U3|U1,U2)$ and $P(U4|U1, U2, U3)$. It seems to compute $P(U3|U1,U2)$, we should apply the law of total probability and condition on another event, in this case, I think it should be conditioned on which face we choose. But I am lost here on how to write this out formally, and was hoping someone can assist.
Your calculation and explanation thereof are correct and clear; what you have there is good. I am not a fan of unnecessary formalism, and in all honesty I wouldn’t try for anything more formulaic; if you absolutely need such a write-up, I hope that someone will provide one.
At most I might have taken a different approach from the start, reducing it to a purely combinatorial problem. Since there are $4$ faces, each of which has $3$ edges, there are $3^4$ equiprobable ways to choose our $4$ edges. If $e$ is one of the edges, exactly $9$ of these ways involve choosing $e$: we must choose $e$ for each of the two faces adjacent to it, and we have $3^2=9$ choices of edges from the other two faces. Since there are $6$ edges, at first sight this means that there are $6\cdot9=54$ ways to choose an edge twice. However, this counts a few possibilities twice: for each edge $e$ there is exactly one edge $e'$ that does not share a face (or even a vertex) with $e$, and it is possible to choose both $e$ and $e'$. There are just $3$ such pairs of ‘opposite’ edges, so we’ve overcounted by $3$, and there are really only $54-3=51$ different ways to choose some edge twice. The probability of doing so is therefore $\frac{51}{81}=\frac{17}{27}$.