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 $1 bets on red on each of the next two spins of the roulette wheel and then quit. Let X denote your winnings when you quit. (a) Find P{X>0}. (b) Are you convinced that the strategy is indeed a “winning” strategy? (c) Find E[X]. ### Answer & Explanation timbalemX Skilled2021-08-24Added 108 answers (a) Observe that possible outcomes of X are 1,-1,-3. The gambler will win 1$ 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 $P\left(X>0\right)=P\left(X=1\right)=\frac{18}{38}+\frac{20}{38}\cdot {\left(\frac{18}{38}\right)}^{2}=\frac{4059}{6859}\approx 0.5918$
(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
$P\left(X=-1\right)=2{\left(\frac{20}{38}\right)}^{2}\cdot \frac{18}{38}=\frac{1800}{6859}$
$P\left(X=-3\right)={\left(\frac{20}{38}\right)}^{3}=\frac{1000}{6859}$
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.

Nick Camelot

To solve this problem, let's analyze the given strategy for the game of roulette step by step.
(a) We want to find P{X > 0}, which represents the probability of ending the game with a positive winning.
Let's consider the different scenarios:
1. Red appears on the first spin: In this case, you win 1 and quit. The probability of this event is P(Red) = 18/38.
2. Red does not appear on the first spin, but appears on the second or third spin: In this case, you lose 1 on the first spin and place an additional 1 bet on red on each of the next two spins. If red appears on either the second or third spin, you win 2 (since you placed a 1 bet each time) and quit. The probability of this event is P(Not Red) * P(Red) * (P(Red) + P(Red)) = (20/38) * (18/38) * (18/38 + 18/38).
Therefore, P{X > 0} = P(Red) + (P(Not Red) * P(Red) * (P(Red) + P(Red))) = 18/38 + (20/38) * (18/38) * (18/38 + 18/38).
(b) To determine if the strategy is a 'winning' strategy, we need to compare the expected value of winnings (E[X]) with the initial bet amount (1).
(c) Now let's find E[X], the expected value of winnings.
We can express the winnings X as a random variable with the following values:
1. If red appears on the first spin, X = 1.
2. If red appears on the second or third spin after a loss on the first spin, X = 2.
3. If red does not appear on any of the spins, X = -3 (the total loss).
Therefore, we can calculate E[X] as:
$E\left[X\right]=P\left(Red\right)*1+\left(P\left(NotRed\right)*P\left(Red\right)*\left(P\left(Red\right)+P\left(Red\right)\right)\right)*2+P\left(NotRed{\right)}^{3}*-3=\left(18/38\right)*1+\left(20/38\right)*\left(18/38\right)*\left(18/38+18/38\right)*2+\left(20/38{\right)}^{3}*-3$.

Mr Solver

(a) P{X > 0} = 18/38 + (20/38) * (18/38) * (18/38).
(b) To determine whether the strategy is a winning strategy, compare E[X] with the initial bet.
(c) E[X] = 1 * (18/38) + 3 * (20/38) * (18/38) * (18/38).
Explanation:
(a) Finding P{X > 0}:
The winnings X will be greater than 0 if red appears on the first spin or if red appears on the second or third spin after losing the initial bet. Lets calculate the probability for each case separately.
Case 1: Red appears on the first spin (probability = 18/38):
In this case, the winnings X will be 1, so the probability is P{X = 1} = 18/38.
Case 2: Red appears on the second or third spin after losing the initial bet (probability = 20/38):
To calculate the probability of winning on the second spin, we need to lose the initial bet and then win twice in a row. The probability of losing the initial bet is (20/38), and the probability of winning twice in a row is (18/38) * (18/38). So, the probability for this case is $PX=1=\left(20/38\right)*\left(18/38\right)*\left(18/38\right)$.
Now, we can calculate the total probability of winning:
$PX>0=PX=1\left(Case1\right)+PX=1\left(Case2\right)$
$=18/38+\left(20/38\right)*\left(18/38\right)*\left(18/38\right)$.
(b) Evaluating the strategy:
To determine whether the strategy is indeed a winning strategy, we compare the expected value of the winnings with the initial bet.
(c) Finding E[X]:
To calculate the expected value E[X], we need to consider all possible outcomes and their corresponding probabilities.
Case 1: Red appears on the first spin:
In this case, the winnings X will be 1, so the expected value is $E\left[X\right]=1*PX=1=1*\left(18/38\right)$.
Case 2: Red appears on the second or third spin after losing the initial bet:
The winnings X will be 3 in this case, as we lose 1 on the initial bet and win 1 on each of the next two spins. The probability for this case is $PX=3=\left(20/38\right)*\left(18/38\right)*\left(18/38\right)$.
Now, we can calculate the total expected value:
$E\left[X\right]=1*PX=1\left(Case1\right)+3*PX=3\left(Case2\right)$
$=1*\left(18/38\right)+3*\left(20/38\right)*\left(18/38\right)*\left(18/38\right)$.

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