# Can you profit from a game where you toss a coin infinitely many times?

Let's say that you have a fair coin, and someone offers to pay you a fixed amount for every time the coin shows heads, and you will lose the same amount if the coin shows tails.

Once you have the coin, you're allowed to toss it as many times as you want. When you stop, the amount you receive or owe is determined based on the difference between the number of heads and tails.

My first question is, is this game rigged against you, or does it favor you, or is it a fair game? Next, Is this game profitable, as in, are you guaranteed to make money?

Now, let's say the other party adds a condition that you must stop on a tails. Is the game still profitable?

• Sounds like homework to me. Commented Jun 6, 2017 at 5:55
• it's a question out of my own interest but I can't really prove that to you... In any case, my thoughts are that in infinite expectation you're going to have some point in which the number of heads exceeds the number of tails, and so it's a profitable game if you have all the time in the world. I'm not sure if the same argument can be made for stopping on a tails, because at that point if you can show that you'll reach a point where you have at least 1 more heads than tails in expectation, then it will be profitable for you. Commented Jun 6, 2017 at 6:19
• en.wikipedia.org/wiki/Law_of_large_numbers Commented Jun 6, 2017 at 10:04
• If you mean "infinitely many times", then that is how you should phrase it. Commented Jun 6, 2017 at 14:49
• Name the amount of money you want to end up with, say $U$. Stop at the first tail observed after reaching total winnings of $U+1$. Since the chance of this sequence of events is 100%, you're fine. You need an arbitrarily large amount of capital and time to play this game, though.
– whuber
Commented Jun 6, 2017 at 14:55

If you can get to any number doing this, then it's easy to see that ending on Tails doesn't make things worse. This is because you will come back infinitely many times to the same number $x$ with probability 1, and so with probability one there will be a time when you land on $x$ by flipping tails (i.e. coming down from $x+1$).