Rebrained!

Let’s call p the chances of winning a dollar on any given toss, and (1-p) the chances of losing a dollar. Also, lets call P the chances of eventually going bust from our starting $1 position.

Imagine you managed not to go bust on the first toss, and are about to take the second toss. At this point, there are two possibilities – that at some point in the future you’ll move back to having $1, or that you’ll never move back to $1.

Here’s where a simple but clever insight can lead to this problem magically solving itself: when you’re at the $2 position, the probability of eventually moving back to $1 is the same as the probability you had of eventually moving to $0 (and going bust) when you started the game. You can see that it’s the same situation, it’s effectively just been transposed.

So the probability of eventually going bust at the beginning of the game (which we’ve defined as P) is equal to the probability on the first throw of tossing a tail (1-p), or tossing a head (p) and then going bust from the $2 state. The probability of going bust from the $2 state is the probability of eventually going back to the $1 state (which we worked out is P) and then eventually going bust from the $1 state (also P).

We can write the above like this:

P = (1-p) + p * P^2

We can solve this algebraically to have two solutions for P: 1 and (1-p)/p.

There’s some interesting conclusions from this:
- Without even the formulas we can see that if p = 1, we’ll keep winning $1 each time and the chances of ever going bust (P) will be 0. By the same token, if p = 0, then we’ll directly go bust so P will be 1.
- For p = 1/2, the formula will give 1. This means that for a fair coin, you’re guaranteed to go bust if you play long enough.
- For p < 1/2, the formula gives a result greater than 1, which is invalid. But clearly you'll be guaranteed to go bust in this case as well.
- For p > 1/2, the formula will give a probability ranging from 1 to 0. When p = 2/3 (as in our question above) then we have a 50/50 chance of eventually going bust.

What we worked out in the previous section is that the probability of eventually getting to $1 less than what you started with is 1 if p is smaller than or equal to 1/2, or (1-p)/p if p > 1/2.

So if you start with $x, and p is greater than 1/2, then your chances of going bust will be ((1-p)/p)^x.