# Progressive Roulette Betting

In this problem we discuss a betting strategy known as "progressive betting". Here's the setup:

Bets are repeatedly made at a roulette according to the following strategy:

• All bets are made the same way: by betting on black. i.e., the bettor has a 18/38 chance of winning each bet, and when that happens, he receives back two times his bet.

• The bettor consecutively doubles the value of any previous losing bet until he finally wins. More specifically, he starts by betting $1, and then: • If he wins that first bet he stops; • If he loses that first bet he makes a second bet of 2 *$1 = $2. The second bet works the same way, with the bettor stopping if he wins it and making a third bet of 2 *$2 = $4 otherwise. The same rule is then applied to any eventual third bet, fourth bet, fifth bet, etc. Note that this means that an eventual n-th bet will be a bet of$[2^(n-1)].

Roughly speaking, our goal will be to try to understand just how good (or not) this strategy is.

Let N (respectively B) denote the number of bets made (respectively the total amount of money that was bet) before stopping.

(a) Suppose N = n, so that B = 1 + 2 + 4 + ... + 2^(n-1) Find a simpler formula for B by evaluating the finite geometric sum.

(b) Suppose again N = n. What was the net profit the bettor made from the n bets? Interpret your answer in terms of how it depends on n.

(c) After finding the previous answer, you may be thinking that maybe this is actually a great strategy. We now present initial evidence that something quite suspicious is taking place.

i. Find the distribution of N and its expected value E(N);

ii. Show that E[B] = infinity, i.e., that the bettor needs, on average, an infinite amount of money in order to follow this strategy (Hint: Use the fact that, according to part (a), B = f(N) for a certain function f, and hence you can calculate E(B) using the distribution of N)

Attempt:

I am stuck on part a. I keep finding that the sum from i=0 to n of r^i is equal to (1-r^(n-1))/(1-r) , however since r=2 and it stops at n-1 that would simplify to (1-r^n)/(1-r) and then plugging in r=2 gives me (1-2^n)/-1 or (2^n)-1 Is this the right answer? It seems wrong to me but I am not sure how to know and can't move on until I solve part a.

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Yes, for part a, the desired answer is $B=2^n-1$

• So for part b would I find the net profit by saying that the total amount bet is 2^(n+1)-1 and then total amount made is 2(2^n) – Georgia Nov 11 '14 at 4:43
• Yes, but you can simplify the difference, which is the profit. – Ross Millikan Nov 11 '14 at 4:44

You describe a Martingale.

https://en.wikipedia.org/wiki/Martingale_(betting_system)

A better of infinite wealth almost surely has a positive return. But no better has infinite wealth.

$$1+2+2^2+\cdots 2^{n-1} = 2^n - 1$$

You will win $$1$$ dollar if any of the first $$n$$ bets hit and lose $$2^n-1$$ if none hit.

Your probability of losing n consecutive bets is $$\frac{20}{38}^n$$

If you wager until either your first win, or n consecutive losses, your expectation is $$1(1-\frac{20}{38}^n) + (2^n-1)(\frac{20}{38})^n$$