Coin toss game

Problem

You and your friend play a game involving coin-flips. Each time the coin lands head, your friend pays you $1. But each time the coin lands tail, you pay your friend $1. The game goes on until either you win $100 or you lose $50. What is the probability of you winning?

Solution

3 solutions, two of them are algebraic, one of them is argumentative.

Algebraic Solution 1

Let $p_i$ be the probability of winning $100 given that we already accumulated $i$ dollars (can be negative, meaning we are losing money).

Then we have $p_i = \frac{1}{2} p_{i-1} + \frac{1}{2} p_{i+1}$, with special value $p_{100} = 1$ and $p_{-50} = 0$.

It can be shown that $p_i = \frac{i + 50}{150}$ the only solution that satisfies the system of linear equations.

This approach can be extended to a more general problem: what is the probability of winning if we stop at either winning $n or losing$m? It should be $\frac{m}{n+m}$.

Algebraic Solution 2

This approach is inspired by the idea of reflecting the -$50 to $50, and symmetry.

Suppose we know the probability q of going from $0 to either $50 or -$50 for the first time.

And suppose we know the probability p of going from $0 back to $0 again for the first time without touching $50 and -$50.

Then we can say that $p + q = 1$. (From $0, you either eventually end up crossing $50 or - $50, or you end up back to $0. Wait, are you sure? Actually it's possible that e.g. you got stuck in between 0 and 50 forever. What is the probability of that?)

So $p_0 = p \cdot p_0 + \frac{q}{2} p_{50}$, and $p_{50} = p \cdot p_{50} + \frac{q}{2} + \frac{q}{2} p_0$.

Solving the above will give us $p_0 = \frac{1}{3}$.

Argumentative solution

If you believe that there is no strategy that can earn you expected value > 0 (otherwise you should actually invest in the stock market and become really rich), then you can say that the expected value of the bet is 0. If p is probability of winning, then we have 100 * p - 50 * (1-p) = 0.