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)


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.


bumped to the homepage by Community yesterday

This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.


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

  • $\begingroup$ 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) $\endgroup$ – Georgia Nov 11 '14 at 4:43
  • $\begingroup$ Yes, but you can simplify the difference, which is the profit. $\endgroup$ – Ross Millikan Nov 11 '14 at 4:44

You describe a Martingale.


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 $1(1-\frac{20}{38}^n) + (2^n-1)(\frac{20}{38})^n$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.