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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

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    $\begingroup$ Surely if you run out of money, your profit is $-50$? $\endgroup$
    – jlammy
    Commented Dec 28, 2020 at 19:58
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    $\begingroup$ @jlammy no, because you do not keep the $\$50$. $\endgroup$
    – Jason Born
    Commented Dec 28, 2020 at 20:00
  • $\begingroup$ Well, before the game started you had $\$50$. Now after playing the game a bit you have nothing. Sounds like $-\$50$ profit to me... $\endgroup$
    – jlammy
    Commented Dec 28, 2020 at 20:02
  • $\begingroup$ @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 \$-$50$ $\endgroup$
    – Joe
    Commented Dec 28, 2020 at 20:02
  • $\begingroup$ @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. $\endgroup$
    – Jason Born
    Commented Dec 28, 2020 at 20:04

1 Answer 1

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(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\$.

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