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?
Best Answer
100%, for the same reason as the 1-D walk
In fact (again for the same reason), your chances are 100% of eventually reaching X-greater heads than tails (or tails than heads), where X is any non-negative integer.