Sinead Mcgee

2021-08-23

A gambling book recommends the following “winning strategy” for the game of roulette:

Bet $1 on red. 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.

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?

(c) Find E[X].

timbalemX

Skilled2021-08-24Added 108 answers

(a)

Observe that possible outcomes of X are 1,-1,-3.

The gambler will win 1$ if he wins in the first bet or he loses the first bet, but he have success in the second and the third.

He will win -1$ if he loses the first and the second bet, but wins the third or loses the first, but wins the second.

Finally, he will win -3$ if he loses all the bets. Hence$P\left(X>0\right)=P(X=1)=\frac{18}{38}+\frac{20}{38}\cdot {\left(\frac{18}{38}\right)}^{2}=\frac{4059}{6859}\approx 0.5918$

(b)

No, the strategy is not the winning one.

Even though that the probability that the gambler wins in this system is greater that 1/2, we will show in (c) that the expected winnings is less than zero since there is a big risk of losing all the bucks.

(c)

See that

$P(X=-1)=2{\left(\frac{20}{38}\right)}^{2}\cdot \frac{18}{38}=\frac{1800}{6859}$

$P(X=-3)={\left(\frac{20}{38}\right)}^{3}=\frac{1000}{6859}$

so, using definition of expectation, we nave that

$E\left(X\right)=1\cdot \frac{4059}{6859}-1\cdot \frac{1800}{6859}-3\cdot \frac{1000}{6859}\approx -0.11$

which says that the gambler is expected to lose 11 cents.

Observe that possible outcomes of X are 1,-1,-3.

The gambler will win 1$ if he wins in the first bet or he loses the first bet, but he have success in the second and the third.

He will win -1$ if he loses the first and the second bet, but wins the third or loses the first, but wins the second.

Finally, he will win -3$ if he loses all the bets. Hence

(b)

No, the strategy is not the winning one.

Even though that the probability that the gambler wins in this system is greater that 1/2, we will show in (c) that the expected winnings is less than zero since there is a big risk of losing all the bucks.

(c)

See that

so, using definition of expectation, we nave that

which says that the gambler is expected to lose 11 cents.

Nick Camelot

Skilled2023-05-24Added 164 answers

Mr Solver

Skilled2023-05-24Added 147 answers

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