Expected value of a certain game Consider a game with n players playing m rounds. Players play in the same time. Player wins k dollars with probability p and loses everything otherwise. If player loses the rest divides his money between themselves. I'd like to calculate expected value. I don't have much experience in probability theory. If this is known problem i would really appreciate any references.
 A: First, note that the problem is completely symmetric among the players. That's helpful, because it means a player's expected winnings are exactly $\frac{1}{n}$ the expected total winnings. It's a bit easier to understand the total winnings, because for total winnings, the transfers of wealth from losers to winners don't matter, since the total wealth of all $n$ players remains exactly the same during these transfers. Thus, the expected winnings are exactly the same whether or not one transfers wealth from losers to winners after each round. It's a lot easier to analyze the case with no wealth transfers, so I will assume no transfers from here on out.
There are two cases, depending on what happens if everyone loses.
Case 1: If every player loses in round $i$, the game stops and no more wealth is accumulated, but each player who was still active in round $i$ gets to keep what he already has. Alternatively, all the wealth accumulated by the players in round $i$ could be divided equally between all $n$ players. The expectation is the same either way.
A player's expected payoff in round $1$ is $kp$. His expected payoff in round $2$ is $kp^2$, because he has to win both rounds $1$ and $2$ to win $k$ in round $2$. Similarly, his expected payoff in round $r$ is $kp^r$. Thus, his total expected payoff is
\begin{eqnarray}
k(p + p^2 + \cdots + p^m) &=& k\sum_{i=1}^m p^i \\
 &=& k p \frac{1-p^m}{1-p}.
\end{eqnarray}
Case 2: If every player loses during a round, then everyone's wealth is forfeited. All winnings are then $0$.
Let $W_{i}$ be the total wealth generated by player $i$ through all $m$ rounds. (The player "generates" $k$ dollars if he wins a round and $0$ if he loses. $W_i$ is just $k$ times the player's total number of wins.)
In this case, we have to subtract out the expected loss that occurs when every player loses by the end of round $m$. This expected loss is easily seen to be
$$
\mathbb{E}(W_i\ |\ \mbox{every player loses by round } m) \mathbb{P}(\mbox{every player loses by round } m).
$$
The probability is easy to calculate. The probability that a particular player loses by round $m$ is $1-p^m$. By independence, we have
$$
\mathbb{P}(\mbox{every player loses by round } m) = (1-p^m)^n.
$$
Now to compute the conditional expectation, observe that by independence
$$
\mathbb{E}(W_i\ |\ \mbox{every player loses by round } m) = \mathbb{E}(W_i\ |\ \mbox{player $i$ loses by round } m).
$$
This makes the calculation easy. We have
$$
\mathbb{P}(\mbox{player $i$ loses in round }r\ |\ \mbox{player $i$ loses by round }m) = \frac{(1-p)p^{r-1}}{1-p^m}.
$$
And $W_i = k(r-1)$ if player $i$ loses in round $r$. Thus,
\begin{eqnarray}
\mathbb{E}(W_i\ |\ \mbox{player $i$ loses by round } m) &=& \sum_{r=1}^m k(r-1) \frac{(1-p) p^{r-1}}{1-p^m} \\
 &=& k \left(\frac{p}{1-p} - \frac{mp^m}{1-p^m}\right).
\end{eqnarray}
Thus, the expected loss due to everyone losing is
$$
k \left( \frac{p(1-p^m)}{1-p} - mp^m\right) (1-p^m)^{n-1}.
$$
Hence, a player's expected winnings for this case are
\begin{eqnarray}
k p \frac{1-p^m}{1-p} - k \left( \frac{p(1-p^m)}{1-p} - mp^m\right) (1-p^m)^{n-1}.
\end{eqnarray}
