Question

Gambles are independent, and each one results in the player being equally likely to win or lose 1 unit. Let W denote the net winnings of a gambler whose strategy is to stop gambling immediately after his first win. Find

(a) P{W > 0}

(b) P{W < 0}

(c) E[W]

Short answer

a) probability $P[W>0]=\frac{1}{2}$

b) probability $P[W<0]=\frac{1}{4}$

c)$E\left[W\right]=0$

Step-by-step solution

Part (a) - Step 1: To find

probability for$[W>0]$

Part(a) - Step 2: Explanation

Gambles are independent and there is an equal chance of winning or losing one unit.

W stands for a gambler's net winnings.

As a result, he immediately stops gambling following his first win.

If a man stops gambling after winning the first game, he will lose $N-1$times and win once if he gambles N times.

Then there's the chance that he'll play N times.

With a negative random variable $r=1$

$$\begin{gathered} P\{N=i\}=p(1-p)i-1 \\ As \\ p=1-p=\frac{1}{2} \\ Thus \\ P\{N=i\}=\frac{1}{2^i} \end{gathered}$$

If we wins the first game only then $w>0$

Hence $P\{W>0\}=P\{N=1\}=\frac{1}{2}$

Part (b) - Step 3: To find

Probability for$[w<0]$

Part (b) - Step 4: Explanation

Gambles are independent, and the chances of winning or losing one unit are both equal.

W stands for a gambler's net profit.

As a result, following his first win, he stops to gamble.

The later winnings are zero if he plays the second game.

Thus,

$$\begin{gathered} P\{W<0\}=1-\left(P\right\{W>0\}+P\{W=0\left\}\right) \\ =1 \end{gathered}$$

$$\begin{gathered} -(\frac{1}{2}+\{losefirst,winthesecondgamble\left\}\right) \\ =1 \\ -(\frac{1}{2}+p\{losefirst\}p\{winsecondgamble\left\}\right) \\ =1-(\frac{1}{2}+\frac{1}{2}.\frac{1}{2}) \\ =1-\frac{3}{4} \\ =\frac{1}{4} \end{gathered}$$

Part (c) - Step 5: To find

Expected value of$W$

Part (c) - Step 6: Explanation

Gambles are independent, and there is an equal chance of winning or losing one unit.

W stands for a gambler's net winnings.

As a result, he immediately stops gambling following his first win.

Find the formula for W first.

Let N be the number of gambles required until a win is obtained.

Then N is a geometric random variable.

With parameter

$p=\frac{1}{2}$

Of the N gambles,

There will be 1 win and N-1 loss.

Then

$$\begin{gathered} W=1-(N-1)=2-N \\ Thus \\ E\left[W\right]=2-E\left[N\right]=2-\frac{1}{p}=2-\frac{1}{\frac{1}{2}}=2-2=0 \end{gathered}$$

Hence$E\left[W\right]=0$