# Working out expected values of coin flipping games when starting with an initial value

Suppose I start with $$\50$$ and I flip a fair coin, where flipping heads rewards me with $$\2$$ and flipping tails loses me $$\1$$, and I either flip until I run out of money, or I flip 100 times otherwise. What is the expected value of this game, considering that my profits do not include the initial $$\50$$?

Clearly, if I consider the more basic game of starting with $$\0$$ and flip 100 times, then the expected value can be calculated to be $$\50$$. Moreover, the case of running out of money yields an expected value of $$0$$, but I don't know how to put this all together

• Surely if you run out of money, your profit is $-50$? Commented Dec 28, 2020 at 19:58
• @jlammy no, because you do not keep the $\$50$. Commented Dec 28, 2020 at 20:00 • Well, before the game started you had$\$50$. Now after playing the game a bit you have nothing. Sounds like $-\$50$profit to me... Commented Dec 28, 2020 at 20:02 • @JasonBorn You said that the expected value of the game was based on how much money you had relative to what you had at the start (\$$50$). If you lose the money, then the profit would be \$-$50$– Joe Commented Dec 28, 2020 at 20:02 • @jlammy sorry, I think I didn't explain the game properly. The$\$50$ is a starting point once you begin playing the game. You either walk away with a minimum of $\$0$(if you deplete the aforementioned money) or you walk away with a maximum of whatever the maximum is from the maximum number of 100 flips. Commented Dec 28, 2020 at 20:04 ## 1 Answer (3rd edit...) You either finish 100 rounds, or you don't. If you finish 100 rounds, you end up with an expected profit of 50\$, but if you don't, it means you have lost your initial 50\$, i.e., -50\$ profit. You lose if you flip 50 tails in a row, or 52 tails and one head, or 54 tails and 2 heads, etc. The only condition in each case is that your last three outcomes should be tails, which makes all losing scenarios disjoint. The probability of losing being:

$$P_L = \sum\limits_{k=0}^{16} {47+3k \choose k} \left(\frac{1}{2}\right)^{47+3k} \approx 1\times 10^{-10}$$

So, keep playing! You will end up with +50\\$.