# A game with $n+1$ players and a price $\textbf{B}$

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.

• There's a factor of $\mathbf B$ missing in the displayed equation in ($2$). Apr 17, 2016 at 17:23
• 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? Apr 17, 2016 at 17:27