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You play 100 rounds of a coin flipping game where you win \$2 for a head and lose \$1 for a tail on each round. Clearly since the coin tosses are independent the expected winnings are \$50.

Now, suppose you play at most 100 rounds of this game as before, but this time you stop early if you accumulated \$50 of losses. How does this change the expected winnings?

Naively one would think that this "stop-loss" reduces the losses leading to higher expected winnings compared to the first game, but this does not take into account scenarios where we subsequently recover from the losses: for example the stop-loss throws away the profitable scenario where we throw 50 tails followed by 50 heads ending with positive winnings of \$50.

ScarletPumpernickel
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1 Answers1

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The practical answer is almost not at all. The chance of accumulating $\$50$ in losses is almost zero, so the rule rarely changes the game.

The expected winnings will decrease. As your expectation is positive on each flip, you want to keep playing. Anything that reduces the expected number of games will reduce your expectation. In the event you lose the first $50$ flips you will now stop and have a loss of $\$50$. If you kept playing, your expectation would be $-\$25$ at the end, an improvement. The point that you are still losing in this case does not matter-what matters is that you are better off than $-\$50$. The calculation of the reduction would be the $\sum_{n=50}^{99}$(probability that you hit $-\$50$ for the first time on flip $n$)(loss of expectation on remaining plays$=\frac 12(100-n)$)

Ross Millikan
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  • Thank you for your clear answer. To calculate the "probability that you hit −$50 for the first time on flip $n$", I think that to reach -$50 in $n$ tosses would require $h$ heads and $n-h$ tails leading to winnings of $2h-(n-h)=-50$, i.e. $h=(n-50)/3$, and the likelihood of getting $h$ heads in $n$ tosses is just the Binomial probability $\left(\begin{array}{c} n\ h\end{array}\right)\frac{1}{2^n}$, is that correct? – ScarletPumpernickel Jul 25 '15 at 16:21
  • Almost. You have to exclude cases where you were already at $-50$. Your expression would count all cases with $n=54, h=2$, but if you threw $49$ tails, then HTTHT, you would have quit after $52$ throws, so the chance of quitting after $54$ is a little smaller. – Ross Millikan Jul 25 '15 at 22:19