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They're simply calculating expectation incorrectly. Expectation is a probability-weighted sum over every attempt. While the game of 50/40 is based on a probability-weighted product over every attempt.

Basically "average expectation" for that game is 0.95 (and they give a false name to it).

Tl;dr: it's got nothing to do with ergodicity.

More importantly, they quote "Peters coin game" which has no results in Google whatsoever.




>>> More importantly, they quote "Peters coin game" which has no results in Google whatsoever.

Maybe a reference to this paper [1] ?

[1] The time resolution of the St. Petersburg paradox, by Ole Peters

https://arxiv.org/pdf/1011.4404.pdf


> they quote "Peters coin game" which has no results in Google whatsoever.

As he says in the Twitter thread anyway, they are coining this name.


The expectation of a single flip is 'positive' ( > 1 ), the expectation of n flips is 'negative' ( < 1).

My layman reading of this is that this 'appears' to be a violation of martingale theory because this 'should' be a sub-martingale but martingale theory requires finite variance so I suspect the variance of this must go infinite as n tends to infinity?


Why do you say that the expectation of n flips is negative?

For 2 flips there are 4 equiprobable cases with gains +125%, -10%, -10%, -64% and the expectation is 10.25%. The expected gain for n flips is 1.05^n-1


AFAIK (am reading quickly) mean values of COMPOUND interest rates (which is what you are doing) are not useful.

If you earn 10% and then lose 10% you are worse off than before. Start with 10: (11-1.1 < 10).


Earning 50% and then losing 40% is equivalent to losing 10%. The naïve calculation you're arguing against would yield a gain of 10%. kgwgk didn't make that mistake.


The definition of expectation is what it is :-)

I agree that expectation may not be the most meaningful number here, in the long run is almost sure that we lose everything but there is a zero probability of infinite gains that makes up for it!


1.05^n is > 1, with positive finite expectation at every finite time step.

Why do you say almost sure to lose everything?


For any threshold (say $1, assuming you start with $1mn) and any confidence level (say 99%) there is a number of flips n such that you have over 99% probability of holding less than $1 after playing n times. On the other hand, the expected value of your wealth at that time is 1.05^n million dollars.


This is well-stated.

In fact, an even stronger statement is possible: If you start with $1MM, eventually you will fall below $1 and never go above $1 again no matter how long you keep playing.

And all of that not just on a fraction of sequences, but with probability 1.


True, as I said in a previous comment we have “almost sure” convergence to zero (i.e. the sequence will converge to zero with probability one).


You're right, I screwed up the arithmetic making it look like something it's not.

It's just a regular sub-martingale with 'positive' expectation as you say.




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