The question I think is a simple one, but I've been unable to answer or find an answer for it yet:
There's a simple game: if you flip heads you win a dollar (from the house), but if you flip tails you lose a dollar (to the house).
If I start with n dollars (and the house has infinite money), how many flips can I expect to do before I've lost all my money? This is different than the common question of how many flips can I do before I have a run of length 'n'. In this case you can lose your money by never having a run of length more than 2, for example, simply by repeating win 1, lose 2, win 1, lose 2, etc…
I can write out a decision tree on this, but I haven't been able to generalize it into a formula yet.
Best Answer
Here is a solution where one computes nothing.
Let us consider $t_k$ the expected number of flips until one is out of money when one's initial fortune is $k$ and let us compute $t_1$. After the first flip, either one is out of money or one's fortune is $2$, thus $t_1=\frac12(1)+\frac12(1+t_2)$. To compute $t_2$, note that, to be out of money when one's initial fortune is $2$, one needs to lose $1$, which takes in the mean $t_1$ flips, and then to lose $1$ again, which again takes in the mean $t_1$ flips. Thus $t_2=2t_1$ and $t_1=1+t_1$.
The equation $t_1=1+t_1$ has a unique solution, which is $t_1=+\infty$.