A Coin Tossing Game

Here's the rule: With a fair coin, Helen gets a point when it flips head and Tim gets a point when it flips tail. Helen wins when she becomes 3 points ahead. Tim wins when he becomes 10 points ahead. How much more likely is Helen to win? And if we use a weighted coin, what should the weight be to make this into a 50/50 fair game?

This problem is just a Brownian motion for the score of the game: number of head minus number of tail after n flips. The only added twist are the two "walls" that the score can "walk" into to stop the game. Let:

be the probability that the game will stop at 3 (and thus Helen would win) when it is started at score x. Clearly:

This is just saying that if we start the game with Helen 3 points ahead, she wins automatically; if we start the game with Tim 10 points ahead, he wins automatically. Further, it is easy to see that:

So satisfy the discrete Laplace's equation, since the Laplacian on a grid is:

Such functions are called Discrete Harmonic Functions. One could imagine that when we go to the continuum limit, that it becomes Harmonic Function. So Harmonic Functions are probability measures for a bounded random walk problem.

In any case, it is just a difference equation for with boundary conditions given earlier. Its (unique) solution is:

Therefore, is the probability that Helen will win. And so she's more likely to win.

For the second part, suppose we have a weighted coin that has probability of flipping head, the only thing that change from the analysis of the last paragraph is that the recursion equation becomes:

Consider general boundary condition:

with . The solution is:

Notice that with

as expected. So to make the game fair, we need . There is no algebraic way to solve for easily. I will just state that for , .