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Re:The Infamous Coin Toss by didgeoridoo on Wednesday August 02, 2023 @09:28PM (#36980213)
The +50% / -40% is cleverly chosen, because it seems like the bet is weighted toward the gambler if you’re just using a naïve expected value.

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.

Re:The Infamous Coin Toss by olooney on Wednesday August 02, 2023 @09:43PM (#36980329)
Yes, this is called the Kelly Criterion[1] which tells us we should optimize the expected value of the log of wealth, not the expected value of wealth directly. On a log scale, it's clear that doubling and halving have the same magnitude but opposite signs so are perfectly balanced.

[1]: https://en.m.wikipedia.org/wiki/Kelly_criterion

Re:The Infamous Coin Toss by cubefox on Thursday August 03, 2023 @09:31AM (#36985628)
Yes. Though the logarithms aren't essential here. It really is about maximizing the "geometric expectation", the expected value calculated by using the geometric mean instead of the usual arithmetic mean.

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.

Re:The Infamous Coin Toss by DarkNova6 on Thursday August 03, 2023 @03:32AM (#36982530)
Confusing the equivalence of -50% and +50% is very common. In fact, many video games provide wrong bonuses based on this misunderstanding (Age of Empires 2 being a famous example).

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.

Re:The Infamous Coin Toss by n2d4 on Thursday August 03, 2023 @03:45AM (#36982623)
While I agree, that conclusion is wrong because you forgot to look at the win-win and loss-loss cases. Even with two coin tosses, as in your example, you'll win 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.

Re:The Infamous Coin Toss by DarkNova6 on Thursday August 03, 2023 @06:27AM (#36983840)
I think the problem with this is that people are too often deluded by the average. For many statistical purposes, the median is more conclusive as it is less prone to extremes.

Another way to read the 4 cases above is the following:

In 3 out of 4 cases, you are losing money.

Re:The Infamous Coin Toss by hgsgm on Thursday August 03, 2023 @07:38AM (#36984370)
Why do you agree with something plainly wrong than you disproved?
Re:The Infamous Coin Toss by svnt on Thursday August 03, 2023 @07:48AM (#36984488)
They are saying parent is correct that plus 50% and minus 50%, while appearing to be balanced, is a losing proposition for the individual.

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.

Re:The Infamous Coin Toss by n2d4 on Thursday August 03, 2023 @09:32AM (#36985640)
Yes, that's what I meant. You probably don't want to take unnecessary risk by going +50%/-50% markets, but it's not for the reason the parent mentioned in the conclusion.
Re:The Infamous Coin Toss by cubefox on Thursday August 03, 2023 @09:54AM (#36985961)
No the parent was right, I commented.
Re:The Infamous Coin Toss by cubefox on Thursday August 03, 2023 @09:53AM (#36985945)
You should maximize the geometric EV, not the (usual) arithmetic EV. In your average calculation you took the arithmetic mean, but the geometric mean is relevant.

(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.

Re:The Infamous Coin Toss by n2d4 on Thursday August 03, 2023 @10:50AM (#36986753)
Not quite; there's no such thing as a geometric EV (EV is by definition always arithmetic). Instead, you calculated the geometric growth rate, which the Kelly criterion would then maximize; but it doesn't maximize your average outcome. Try it; run a simulation, and you'll see what I mean! (Some guy already posted his results [1])

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.

[1] https://news.ycombinator.com/item?id=36982560

[2] https://news.ycombinator.com/item?id=36986147

Re:The Infamous Coin Toss by cubefox on Thursday August 03, 2023 @11:12AM (#36987009)
> Not quite; there's no such thing as a geometric EV (EV is by definition always arithmetic).

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.

Re:The Infamous Coin Toss by n2d4 on Thursday August 03, 2023 @11:17AM (#36987071)
> That depends on how you define it.

I agree, this is a semantic discussion. I'm using Wikipedia's definition. https://en.wikipedia.org/wiki/Expected_value

Re:The Infamous Coin Toss by operator-name on Thursday August 03, 2023 @03:57AM (#36982695)
For some reason this is not mentioned in the article but is mentioned in the video.
Re:The Infamous Coin Toss by hgsgm on Thursday August 03, 2023 @07:38AM (#36984365)
Do you bet your entire portfolio every time in AoE2? That's required for those paradox. It doesn't apply when bet size is constant.
Re:The Infamous Coin Toss by hgsgm on Thursday August 03, 2023 @07:36AM (#36984345)
No, the trick is less obvious. It's even more shocking that a hugely higher expectedly value upside (+100% , -60%) ((2x+.4x)/2 = 1.2x) is still losing long-term.
Re:The Infamous Coin Toss by kgwgk on Thursday August 03, 2023 @08:21AM (#36984831)
You think that anyone would be shocked with (+100%, -75%) if you put it in terms of "heads you double your amount, tails you halve it twice"?

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.)

Re:The Infamous Coin Toss by cubefox on Thursday August 03, 2023 @10:03AM (#36986086)
We need the geometric expectation here: sqrt(2x * .4x)=0.89x

The arithmetic expectation is wrong in this case.