Coin tosses until I'm out of money?

Question

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.

Answer 1

The expected value is (1/2)(+1) + (1/2)(-1) = 0 no matter how you slice it

So in the LONG RUN, you end up with nothing...

There is no way to reliably predict how many trials it will take until you become bankrupt.

EX.

You begin with N=$2.

You can lose all it after two trials.

You can lose it after 6 trials: WLWLLL

the only way to really generalize it is that there must be w+n losses to cancel the w wins.