# Given enough time, what are the chances I can come out ahead in a coin toss contest?

Assuming I can play forever, what are my chances of coming out ahead in a coin flipping series?

Let's say I want "heads"...then if I flip once, and get heads, then I win, because I've reached a point where I have more heads than tails (1-0). If it was tails, I can flip again. If I'm lucky, and I get two heads in a row after this, this is another way for me to win (2-1).

Obviously, if I can play forever, my chances are probably pretty decent. They are at least greater than 50%, since I can get that from the first flip. After that, though, it starts getting sticky.

I've drawn a tree graph to try to get to the point where I could start see the formula hopefully dropping out, but so far it's eluding me.

Your chances of coming out ahead after 1 flip are 50%. Fine. Assuming you don't win, you have to flip at least twice more. This step gives you 1 chance out of 4. The next level would be after 5 flips, where you have an addtional 2 chances out of 12, followed by 7 flips, giving you 4 out of 40.

I suspect I may be able to work through this given some time, but I'd like to see what other people think...is there an easy way to approach this? Is this a known problem?

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Here's a hint: "Forever" is a really, really long time. – Justin L. Jul 23 '10 at 21:51
The answers are (reasonably) assuming that the coin is fair, but note that the question becomes more interesting -- and more relevant to real-world gambling -- if you consider the more general case of a biased coin which comes up heads with probability $p < .5$. – Pete L. Clark Aug 7 '10 at 20:23

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