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