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bzax | 2 years ago

If you repeat this game n times (as n goes to infinity), you will have Θ(n) pairs of (heads, tails) and O(sqrt(n)) unpaired wins or losses, except for a vanishingly small fraction of the time when the results fall outside of any fixed number of standard deviations.

The point is that you as an individual playing a repeated game don't get to meaningfully sample the expected value of the distribution. You only get to sample once, and you will almost surely (i.e. with probability approaching 1 as n goes to infinity) sample a point in the distribution where you lose nearly all of your money.

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jakell|2 years ago

Absolutely. The individual is long-run guaranteed to be wiped out. But I disagree with the original author’s way of concluding that fact (ie, that it arises from “losing 5% per round”, which is just false).

bzax|2 years ago

I believe the entire point of the ergodicity question here is "If you apply this process n times, with n approaching infinity, obviously the result may depend on what point in the n-times iterated distribution you sample, but if you choose a volume of vanishingly small measure to exclude, can you make a single concrete statement about what the process is doing without taking an expected value over the different outcomes"

And the answer is yes - with probability approaching 1 as n increases (ie excluding a portion of the distribution whose measure decreases to 0), the random process matches a deterministic process which is described by "you lose 5% each round".