The article concludes that a +50%/-40% coin toss on average loses 10% every two tosses because 150% * 60% = 90%, but that ignores the two heads/two tails outcomes. Including those outcomes, ie AVERAGE(225%, 90%, 90%, 36%) = 110.25%, recovering the more intuitive result that the coin toss gains on average.
The author seems to be confusing mode and mean; the modal path does approach zero.
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.
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).
This was my reaction too, I feel like there is something I'm not getting. I get that the bet is virtually guaranteed to go negative if the number of rounds is high enough (100+ rounds), but the opportunity for huge payouts still pulls the average up.
The "average" of the distribution goes up as you increase the number of rounds, but the probability that you get an average or above value when you sample that distribution once goes to zero as the number of rounds increases.
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.
bzax|2 years ago
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.
jakell|2 years ago
bogtog|2 years ago
bzax|2 years ago
jakell|2 years ago
didgeoridoo|2 years ago
Try 225% * 90% * 90% * 36% to get the expected value.
jakell|2 years ago