A gambling book recommends the following “winning strategy” for the game of roulette: Bet $1 on red.

Sinead Mcgee

Sinead Mcgee

Answered question

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 1838,then take the $1 profit and quit.
If red does not appear and you lose this bet ( which has probability 2038 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

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(X>0)=P(X=1)=1838+2038(1838)2=405968590.5918
(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(2038)21838=18006859
P(X=3)=(2038)3=10006859
so, using definition of expectation, we nave that
E(X)=1405968591180068593100068590.11
which says that the gambler is expected to lose 11 cents.
Nick Camelot

Nick Camelot

Skilled2023-05-24Added 164 answers

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[X]=P(Red)*1+(P(NotRed)*P(Red)*(P(Red)+P(Red)))*2+P(NotRed)3*3=(18/38)*1+(20/38)*(18/38)*(18/38+18/38)*2+(20/38)3*3.
Mr Solver

Mr Solver

Skilled2023-05-24Added 147 answers

Answer:
(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=(20/38)*(18/38)*(18/38).
Now, we can calculate the total probability of winning:
PX>0=PX=1(Case1)+PX=1(Case2)
=18/38+(20/38)*(18/38)*(18/38).
(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[X]=1*PX=1=1*(18/38).
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=(20/38)*(18/38)*(18/38).
Now, we can calculate the total expected value:
E[X]=1*PX=1(Case1)+3*PX=3(Case2)
=1*(18/38)+3*(20/38)*(18/38)*(18/38).

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