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In a game with $n+1$ players, who each player has, independently, $p$ chance to win. The winners share a price B. Let's A be a player and X the gain of A.

EDIT 01 : If two players win, each player gain $B/2$.

$(1)$ What's the expected total gain shared by the players.

$(2)$ Demonstrate that $$\mathbb{E}[X]=\frac{1-(1-p)^{n+1}}{n+1}.$$

$(3)$ Calculate $\mathbb{E}[X]$ by conditionning on the fact that A win or lose and deduce that $$\mathbb{E}\left[(1+\textbf{B})^{-1}\right]=\frac{1-(1-p)^{n+1}}{(n+1)p}$$

EDIT 02 : With the help of this community, I come up with the answer to $(1)$, $$\begin{align} \mathbb{E}[Y_n] &=0 \cdot \mathbb{P}\left\{Y_n \mid n=0\right\} + B (1-\mathbb{P}\left\{Y_n \mid n=0\right\}) \\ &= B(1-p)^{n+1} \end{align}$$ where $Y_n$ is the gain won by $n$ winners.

Thanks for your help.

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  • $\begingroup$ There's a factor of $\mathbf B$ missing in the displayed equation in ($2$). $\endgroup$
    – joriki
    Apr 17, 2016 at 17:23
  • $\begingroup$ I thought the same thing but I copied the question correctly. $\endgroup$
    – hlapointe
    Apr 17, 2016 at 17:24
  • $\begingroup$ Well it's clear that the expected gain of any player will double if the prize being shared is doubled, so whatever you copied the question from is in error. $\endgroup$
    – joriki
    Apr 17, 2016 at 17:26
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    $\begingroup$ Hint: (1) let $ Y $ denote the total gain of all players. It can only take two values, what's the probability of $ Y $ taking both of them? $\endgroup$
    – Jytug
    Apr 17, 2016 at 17:27
  • $\begingroup$ @joriki Is my EDIT fix our problem? $\endgroup$
    – hlapointe
    Apr 17, 2016 at 17:30

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