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Suppose you are playing a gambling game. There is a $50/50$ chance of losing $1$ dollar, and winning $1$ dollar. Your starting money is $1$ dollar, and you keep playing until you either lose all your money, or you finish $1000$ rounds.

Using a computer, with a sample size of $100,000$ players, the average player ends with about $1$ dollar. My gut tells me that the average person should lose money, since hitting $0$ dollars locks you out from the game. Why does kicking players with $0$ dollars not affect the average significantly?

An observation is, when looking at an individual's money at the end, there are many people with $0$ dollars, but also a few big winners.

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    $\begingroup$ Linearity of expectation. $\endgroup$ – Lord Shark the Unknown Feb 15 '18 at 5:47
  • $\begingroup$ Thanks. I now realize that no matter what balance you have, the expected change is always 0. Sum up all the games played, and you should still have a net expected value of 0. $\endgroup$ – Allan Chang Feb 15 '18 at 5:55
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    $\begingroup$ This is an example of a martingale in the probability rather than gambling sense $\endgroup$ – Henry Feb 15 '18 at 8:08
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On each round of the game, one's expected loss is zero, whether or not one has been eliminated. If one is active, one's expected loss on a round is $\frac12(1+(-1))=0$. If one is eliminated, one always loses zero. By linearity of expectation, one's expected loss in $10^4$ rounds is $10^4\times 0$.

This problem is essentially random walk with one absorbing barrier.

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