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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.

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