Martingale probability

One popular strategy for gambling on roulette, the "martingale," gives a very high probability of a positive outcome. Here's how it works, for a gambler who starts with a stake of $15. In the US roulette wheels have 38 equally-likely outcomes, 18 each Redand Black and two Green (European wheels have 37 outcomes, with only one Green). Themartingale strategy is to double the bet after each loss:

• Bet $1 on Red. If Red appears (US probability 18/38), quit with winnings $1. Otherwise,
• Bet $2 on Red. If Red appears (US probability 18/38 again), quit with winnings -$1+$2=$1. Otherwise,
• Bet $4 on Red. If Red appears, quit with winnings -$1-$2+$4=$1. Otherwise,
• Bet $8 on Red. If Red appears, quit with winnings -$1-$2-$4+$8=$1. Otherwise,
• You're broke, so you have to quit anyway with total winnings -$1-$2-$4-$8= -$15. Hitchhike home.

Let X denote the gambler's winnings (which could be negative!) when she quits.

• Find the probability distribution of X, i.e., find p(x)=P[X=x] for every real number x.
• Find P[X>0].
• Are you convinced that the strategy is indeed a "winning" strategy? Explain your answer.
• The queen of England is said to have a net worth of about 400 million dollars. If she commits to play the martingale strategy until she either wins $1, betting on Red every time and doubling her bet each time Red does not appear (as in steps 1-4 above, continuing indefinitely), or loses everything, what is the probability that she wins $1? What is the probability she loses $400,000,000? What is the expectation of her gain?

For #1, I found that for x = $1, P(X=x) is 1 - (20/38)^4 and that for x = -$15, P(X=x) is (20/38)^4.

For #2, is the P[X>0] just the same as P(X=1)?

I need help getting started with 3 and 4.

Thanks!

sillybuffalo said:

One popular strategy for gambling on roulette, the "martingale," gives a very high probability of a positive outcome. Here's how it works, for a gambler who starts with a stake of $15. In the US roulette wheels have 38 equally-likely outcomes, 18 each Redand Black and two Green (European wheels have 37 outcomes, with only one Green). Themartingale strategy is to double the bet after each loss:

• Bet $1 on Red. If Red appears (US probability 18/38), quit with winnings $1. Otherwise,
• Bet $2 on Red. If Red appears (US probability 18/38 again), quit with winnings -$1+$2=$1. Otherwise,
• Bet $4 on Red. If Red appears, quit with winnings -$1-$2+$4=$1. Otherwise,
• Bet $8 on Red. If Red appears, quit with winnings -$1-$2-$4+$8=$1. Otherwise,
• You're broke, so you have to quit anyway with total winnings -$1-$2-$4-$8= -$15. Hitchhike home.

Let X denote the gambler's winnings (which could be negative!) when she quits.

• Find the probability distribution of X, i.e., find p(x)=P[X=x] for every real number x.
• Find P[X>0].
• Are you convinced that the strategy is indeed a "winning" strategy? Explain your answer.
• The queen of England is said to have a net worth of about 400 million dollars. If she commits to play the martingale strategy until she either wins $1, betting on Red every time and doubling her bet each time Red does not appear (as in steps 1-4 above, continuing indefinitely), or loses everything, what is the probability that she wins $1? What is the probability she loses $400,000,000? What is the expectation of her gain?

For #1, I found that for x = $1, P(X=x) is 1 - (20/38)^4 and that for x = -$15, P(X=x) is (20/38)^4.

For #2, is the P[X>0] just the same as P(X=1)?

I need help getting started with 3 and 4.

Thanks!

Click to expand...

P(red) on any single spin is p=18/38,
and P(not red) = q = 20/38.

You show the probability of "no red" in four trials is q^4 = 0.0767, in which case you suffer a loss of $15.
The probability of having a red before going broke is 1 - q^4 = 0.923, in which case you gain $1.

The expectation value is the sum of those two possibilities: $1 - (q^4)($16) = - $0.23

If you have a large enough bankroll that you are willing to risk, then the number of trials increases logarithmically. If you have $400,000,000, then n=28 (2^28 = 268,435,456 = largest power of 2 in 4×10^8). The probability og having 28 non-red in a row is q^28 = 1.57×10^-8, so you are "almost sure to win a $. BUT because of the huge amount risked, the expectation value is
E = $1 - (q^n) × $1(2^n) = - $3.20

Another question, for #2, I found that P[X>0] to just be 1 - q^4 = 0.923. Is this correct? The answer is just a repeat of #1, which is making me wary.

sillybuffalo said:

Another question, for #2, I found that P[X>0] to just be 1 - q^4 = 0.923. Is this correct? The answer is just a repeat of #1, which is making me wary.

Click to expand...

I don't fully understand what your variable X is. It appears tht X can hve only two values, either +$1 or -$15. If that is the case, then P(X>0) = P(X=1)

Are you using a negative binomial distribution? If so, you can look at the exact distribution of number of trials before the first red.