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Re:The Infamous Coin Toss by didgeoridoo on Wednesday August 02, 2023 @10:15PM (#36980548)
You’re taking arithmetic averages of percentages… I don’t think that calculates anything meaningful.

Try 225% * 90% * 90% * 36% to get the expected value.

Re:The Infamous Coin Toss by jakell on Wednesday August 02, 2023 @10:23PM (#36980613)
You can substitute $ for % in my comment if it helps. If you start with $100, your expected wealth after two throws is the average of $225, $90, $90 and $36.
Re:The Infamous Coin Toss by didgeoridoo on Wednesday August 02, 2023 @10:40PM (#36980741)
That average is still greater than $100, because you haven’t yet hit the Kelly point beyond which the downside catastrophe dominates. Play it out a few more rounds and see where the average heads to.

[Edit: delete bad math]

Re:The Infamous Coin Toss by bzax on Wednesday August 02, 2023 @10:59PM (#36980846)
The expected value of this distribution goes up with every iteration, there is no such Kelly point. You could try this with

heads: double your money tails: lose all your money

in which case the expected value is always $1, as you have a 1/2^n chance of having $2^n dollars after n rounds, and 0 otherwise.

The point of discussing ergodicity here, however, is whether you can describe the behavior of the iterated distribution deterministically if you exclude a portion of that distribution which has measure zero.