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 $E\left[X\right]$.

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.

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

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

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

$=0.5918$

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.

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

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

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.

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

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

Where L: loose

W: win

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

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

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

$=0.2624$

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

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

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

$=0.1458$

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

$=-0.108$