Finding the expected value when there are too much cases to consider Imagine a game consisting of two stages.
In the first stage, contestants attempt the challenge one by one. Each contestant has a p probability of advancing to the 2nd stage, independent of other players.
If no constestants pass the first stage, then no one gets the money.
Otherwise, the contestants compete in the second stage to share a cash prize of M = 1,000,000 dollar.
Each 2nd-stage contestant draws a number independently and uniformly at random from {1, 2, 3, 4, 5}, and the money is distributed proportionally. 
For instance, if there are 3 contestants in 2nd stage and they drew the numbers 1, 2, 4, then the prize is shared among them in 1:2:4 ratio.

In this particular scenario, you are one of the 5 contestants at the beginning of the game and you have 0.2 probability of advancing to 2nd stage.
What is the expected amount you will win by the end of the game?

Problem:
Right of the bat, the sheer number of potential cases I need to consider is overwhelming.


*

*Not all contestants will advance to 2nd stage.

*There is no way of knowing how many people will be in 2nd stage.

*There is no way of knowing what number you will get in 2nd stage.

*I can roughly list out the # of permutations there can be, but it takes a lot more to get the expected value of the prize.


I was told that there is actually a simpler way to reach the conclusion, but so far this is stupendously confusing.
 A: The situations in which you do not pass to the second stage contribute zero to the expected value of your winnings. So ignore those cases.
If exactly $k$ contestants reach the second stage, then each of those contestants can expect $\frac1k$ of the prize money, as each is equally likely to be fortunate as unfortunate.
Suppose exactly $k$ contestants, including you, reach the second stage. The probability of that is $(\frac15)^k(\frac45)^{5-k}$ times the number of ways to distribute the $5-k$ losses among people who aren't you, or $\binom{4}{5-k}$. Then that case contributes $\frac1k$ of the prize money to the expected value of your winnings. Sum over $k$.
A: There is a simpler way: suppose there are $n$ people, and each has probability $p$ of making it to the second round.  Let $A$ be the event that at least one person makes it to the second stage, and note that $P[A] = 1 - (1 - p)^n$.  Now, for each $j = 1,\ldots,n$, let $X_j$ be the amount won by person $j$.  Then note that $$(\sum_{j = 1}^n X_j \, |\, A) = M\,.$$
In particular, this implies that $$\sum_{j = 1}^n E[X_j \, | \, A] = M\,.$$
But since the people are indistinguishable, we must have that each of these conditional expected values are the same, implying $$E[X_j \, | \, A] = \frac{M}{n}\,.$$
Now we just need to use the law of total expectation:
$$E[X_j] = P[A]\cdot E[X_j \, | \, A] + P[A^c] \cdot E[X_j \, | \, A^c] = (1 - (1 - p)^n)\frac{M}{n}\,.$$
