Question

4.20. A gambling book recommends the following "winning strategy" for the game of roulette: Bet $\backslash (1$on red. If red appears (which has probability $\frac{18}{38}$), then take the $\backslash )1$profit and quit. If red does not appear and you lose this bet (which has probability $\frac{20}{38}$of occurring), make additional $\$1$bets on red on each of the next two spins of the roulette wheel and then quit. Let $X$denote your winnings when you quit.

(a) Find $P\{X>0\}$.

(b) Are you convinced that the strategy is indeed a "winning" strategy? Explain your answer!

(c) Find $E\left[X\right]$.

Short answer

(a) $0.5918$

(b) No

(c) $0.108$(Loss)

Step-by-step solution

Given information Part (a)  

If red appears (which has probability $\frac{18}{38}$), then take the $\$1$profit and quit. If red does not appear and you lose this bet (which has probability $\frac{20}{38}$of occurring), make additional $\$1$ bets on red on each of the next two spins of the roulette wheel and then quit.

Explanation Part (a)  

$\mathrm{P}(\mathrm{X}>0)=\mathrm{P}(\mathrm{X}=1)$

Win on first spin $+$(Loose on first spin)(win both the next two spins)

$\frac{18}{38}+\frac{20}{38}\left(\frac{18}{38}\times \frac{18}{38}\right)=0.4737+0.1181$

$=0.5918$

Given information Part (b)  

If red appears (which has probability$\frac{18}{38}$), then take the $\$1$profit and quit. If red does not appear and you lose this bet (which has probability $\frac{20}{38}$of occurring), make additional$\$1$ bets on red on each of the next two spins of the roulette wheel and then quit.

Explanation Part (b)  

No, because one is betting on Red whose winning probability is:

$\frac{18}{38}<\frac{1}{2}$

Given information Part (C)  

If red appears (which has probability $\frac{18}{38}$), then take the $\$1$profit and quit. If red does not appear and you lose this bet (which has probability $\frac{20}{38}$of occurring), make additional $\$1$ bets on red on each of the next two spins of the roulette wheel and then quit.

Explanation Part (c)  

$\mathrm{X}=1\mathrm{p}=0.5918$

$\mathrm{X}=-1(\mathrm{L},\mathrm{W},\mathrm{L})(\mathrm{L},\mathrm{L},\mathrm{W})$

Where L: loose

W: win

$P(X=-1)$=( Loose on first spin )( loose exactly one in the next two spins )

$=\left(\frac{20}{38}\times \frac{18}{38}\times \frac{20}{38}\right)+\left(\frac{20}{38}\times \frac{20}{38}\times \frac{18}{38}\right)$

$=2\times \left(\frac{20}{38}\times \frac{20}{38}\times \frac{18}{38}\right)$

$=0.2624$

Calculation Part (c)  

$\mathrm{X}=-3(\mathrm{L},\mathrm{L},\mathrm{L})$

$\mathrm{P}(\mathrm{X}=-3)$=( Loose on first spin )( Also loose both the next two spins )

$=\frac{20}{38}\times \left(\frac{20}{38}\times \frac{20}{38}\right)$

$=0.1458$

Final answer Part (c)  

$E\left(X\right)=(1\times 0.5918)+(-1\times 0.2624)+(-3\times 0.1458)$

$=-0.108$