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7.59

Expert-verifiedFound in: Page 356

Book edition
9th

Author(s)
Sheldon M. Ross

Pages
432 pages

ISBN
9780321794772

There are $n+1$ participants in a game. Each person independently is a winner with probability $p$. The winners share a total prize of 1 unit. (For instance, if 4 people win, then each of them receives $\frac{1}{4}$ whereas if there are no winners, then none of the participants receives anything.) Let$A$denote a specified one of the players, and let $X$ denote the amount that is received by $A.$

(a) Compute the expected total prize shared by the players.

(b) Argue that $E\left[X\right]=\frac{1-(1-p{)}^{n+1}}{n+1}$

(c) Compute $E\left[X\right]$ by conditioning on whether$A$is a winner, and conclude that $E\left[(1+B{)}^{-1}\right]=\frac{1-(1-p{)}^{n+1}}{(n+1)p}$

When $B$ is a binomial random variable with parameters $n$ and $p.$

a) The expected total prize shared by the player is$E\left[W\right]=1-(1-p{)}^{n+1}$

b) The required expression is, $E\left(X\right)=\frac{1-(1-p{)}^{n+1}}{(n+1)}$

c) Computing $E\left[X\right]$by conditioning on whether $A$ is a winner, and conclude that $\left.E\left[(1+B{)}^{-1}\right]=\frac{1-(1-p{)}^{n+1}}{(n+1)p}\right]$.

The probability of each person winning the game is, $p.$

The number of players in the game is, $(n+1).$

The probability of the no winner of the game is, $(1-p{)}^{n+1}$

Let $W=0$ denote the no winner in the game

Let $W=1$ denote the at least one winner in the game.

Find the expected total prize shared by the players,

$E\left(W\right)=P(W=1)$

$=P\left(\text{At least one winner of the game}\right)$

$=1-P\left(\text{No winner of the game}\right)$

$=1-(1-p{)}^{n+1}$

Hence, the required value is $1-(1-p{)}^{n+1}.$

The probability of each person wins the game is, $p.$

The number of players in the game is, $(n+1)$.

The probability of the no winner of the game is, $(1-p{)}^{n+1}$

Let $W=0$ denote the no winner in the game

Let $W=1$ denote the at least one winner in the game.

Let ${W}_{i}$ denote the prize of the ${i}^{\text{th}}$player.

Each and every player wins independently and has equal probabilities.

From part (a),

$\begin{array}{r}1-(1-p{)}^{n+1}=E(W)\end{array}$

$1-(1-p{)}^{n+1}=\sum _{i=1}^{n+1}\u200a{W}_{i}$

localid="1647500883344" $E\left({W}_{i}\right)=E\left(X\right)$ is the expected prize of player localid="1647500893021" $A,$

localid="1647500947723" $1-(1-p{)}^{n+1}=(n+1\left)E\right(X)$

localid="1647500955566" $\frac{1-(1-p{)}^{n+1}}{(n+1)}=E\left(X\right)$

Hence, the required expression is, $E\left(X\right)=\frac{1-(1-p{)}^{n+1}}{(n+1)}.$

The probability of each person winning the game is, $p.$

The number of players in the game is, $(n+1).$

The probability of the no winner of the game is, $(1-p{)}^{n+1}$

Let $W=0$ denote the no winner in the game

Let $W=1$ denote the at least one winner in the game.

Find the $E\left(X\right)$ by conditioning on whether $A$ is winner.

Let $B$ follows a Binomial random variable with parameter$(n,p)$.

Let $Y=\left\{\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}A\text{wins}\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{Otherwise}\end{array}\right.$

$\begin{array}{r}E\left(X\right)=E\left[E\right[X\mid Y\left]\right]\end{array}$

$\begin{array}{r}E\left(X\right)=E[X\mid Y=0]P(Y=0)+E[X\mid Y=1]P(Y=1)\end{array}$

$\begin{array}{r}E\left(X\right)=0+E\left[(1+B{)}^{-1}\right]\left(p\right)\end{array}$

$E\left(X\right)=E\left[(1+B{)}^{-1}\right]\left(p\right)$

$\frac{E\left(X\right)}{p}=E\left[(1+B{)}^{-1}\right]$

From part (b) Calculation,

$\frac{1-(1-p{)}^{n+1}}{(n+1)p}=E\left[(1+B{)}^{-1}\right]$

Hence, the required expression is, $E\left[(1+B{)}^{-1}\right]=\frac{1-(1-p{)}^{n+1}}{(n+1)p}.$

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