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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 w

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.

olooney 3y ago
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 https://en.m.wikipedia.org/wiki/Kelly_criterion
- cubefox 3y ago
  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.
DarkNova6 3y ago
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.
- n2d4 3y ago
  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.
  - deleted 3y ago
    [deleted]
  - hgsgm 3y ago
    Why do you agree with something plainly wrong than you disproved?
    - svnt 3y ago
      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.
      - n2d4 3y ago
        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.
        cubefox 3y ago
        No the parent was right, I commented.
  - cubefox 3y ago
    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.
- operator-name 3y ago
  For some reason this is not mentioned in the article but is mentioned in the video.
- hgsgm 3y ago
  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.
hgsgm 3y ago
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.
- kgwgk 3y ago
  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.)
- cubefox 3y ago
  We need the geometric expectation here: sqrt(2x * .4x)=0.89x The arithmetic expectation is wrong in this case.

olooney 3y ago

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 https://en.m.wikipedia.org/wiki/Kelly_criterion

cubefox 3y ago
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.

DarkNova6 3y ago

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.

n2d4 3y ago
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.
- deleted 3y ago
  [deleted]
- hgsgm 3y ago
  Why do you agree with something plainly wrong than you disproved?
  - svnt 3y ago
    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.
    - n2d4 3y ago
      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.
      - cubefox 3y ago
        No the parent was right, I commented.
- cubefox 3y ago
  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.
operator-name 3y ago
For some reason this is not mentioned in the article but is mentioned in the video.
hgsgm 3y ago
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.

hgsgm 3y ago

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.

kgwgk 3y ago
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.)
cubefox 3y ago
We need the geometric expectation here: sqrt(2x * .4x)=0.89x The arithmetic expectation is wrong in this case.