However, if you were to make it “double your money” (+100%), it would become clear that the only fair downside would be “halve your money” (-50%). For these values, the “trick” becomes much more obvious: that increases in repeated games need to be far greater in percentage terms than decreases (i.e. not just a 10% difference) in order to balance out.
Arithmetic mean: sum all values, then divide by the number of values.
Geometric mean: multiply all values, then take the nth root, where n is the number of values.
Even a simple calculation will show the truth:
-40% = 0.6
+50% = 1.5
0.6 * 1.5 = 0.9
Conclusion: With every coin-toss you lose 10% of wealth on average.
1.5 * 1.5 = 2.25
0.6 * 1.5 = 0.9
1.5 * 0.6 = 0.9
0.6 * 0.6 = 0.36
=> 1.1025x total
It's counter-intuitive because even though you almost always lose, you still win (linear) wealth on average (but not median). The difference is that if you have unlimited tosses available, you don't care about maximizing EV after X tosses. Instead you care about minimizing your risk of losing it all.
They are also saying parent is not correct in saying that it also loses on a population basis, showing that the average across a population is still winning.
(2.25 * 0.9 * 0.9 * 0.36)^(1/4)
=0.9
Which is below 1!
To test the geometric mean with the earlier example ("+100% is equivalent to -50%"):
(2.0 * 0.5)^(1/2)
=1, as expected.
If you're curious, I wrote a fairly lengthy explanation in another comment [2] about why your calculation is preferable in many cases, but to maximize average wealth, you really do want to go all-in.
That depends on how you define it. The arithmetic mean is usually presupposed when talking about expected values, but the arithmetic mean is not the only mean, and in fact it is inappropriate to use in some cases, so arithmetic expected value is likewise not the only "expected value", and in fact inappropriate sometimes. So yes, talking about non-arithmetic EV should be just as common as talking about non-arithmetic mean, even though it isn't.
> Instead, you maximized the geometric growth rate, which is what the Kelly criterion does; but it doesn't maximize your average outcome.
Well, it doesn't maximize the arithmetic average, but it does maximize the geometric average. :P There is no point in assuming we have only one kind of average and one kind of expectation. A suboptimal common usage of terms can in fact be a hindrance to thinking straight. You use logarithms to keep talking about the arithmetic mean/average, but in my opinion logarithms just obfuscate the fact that we are in geometric mean territory.
I agree, this is a semantic discussion. I'm using Wikipedia's definition. https://en.wikipedia.org/wiki/Expected_value
It would be kind-of obvious that if you play a sequence of games you need two heads for each tails just to break even. (If you're lucky and get a streak of heads you can win big though.)
The arithmetic expectation is wrong in this case.