# Expected Value: Flipping A Fair Coin Until $10 Is Used Up Abraham and Blaise each have$\$10$. They repeatedly flip a fair coin. If it comes up heads, Abraham gives Blaise $\$1$. If it comes up tails, Blaise gives Abraham$\$1$. What is the expected number of flips until one of them runs out of money?

I have no idea how to start this, I'm stuck. Solutions are highly appreciated. Thanks in advance!

• This, and generalizations of it, are known as Gambler's Ruin problems.
– lulu
Commented Oct 18, 2016 at 23:10
• @lulu What I go is that both players have an equal propbability of winning from reading that article about Gambler's Ruin. How should I proceed? Commented Oct 18, 2016 at 23:17
• Well, symmetry tells you that. I thought that article went into much greater detail. Well, here's one that does. Otherwise, it is very well documented online.
– lulu
Commented Oct 18, 2016 at 23:19
• @lulu Wow! That article was very detailed! So using the formula that it provided at the end, the expected number of flips until one of the players runs out of money is $100$ flips? Commented Oct 18, 2016 at 23:25
• That's what I recall. But as you see it isn't exactly a trivial result.
– lulu
Commented Oct 18, 2016 at 23:27

Set up a recursive relationship by conditioning on the first toss. The probability that one of them, say Abraham WLOG, loses is given by $$P(A\;\text{loses game})\\ =P(A\;\text{loses game}|A\;\text{A lost first round})P(A\;\text{A loses first round})+P(A\;\text{loses game}|A\;\text{A won first round})P(A\;\text{A won first round})$$ Using the above relationship, you should be able to come up with a probability of losing in terms of initial conditions for any endowment.