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The infamous coin toss

ergodicityeconomics.com

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Re: The infamous coin toss

#41

Post by erik_seaberg »

kevinwang wrote: post #38
jacknews wrote: post #3

But this is purely a result of the distribution of returns from a single toss. If I win I get 1.5 times my money, if I lose I'm left with less than 1/1.5 times. When I lose, I lose more than I win. I agree it's counterintuitive how any individual (on average) will lose over time, yet the entire system grows.

> When I lose, I lose more than I win. I don't see how this is true. Winning adds 0.5x of your current bankroll. Losing subtracts 0.4x of your current bankroll. When you win, you win more than you lose. The gamble has positive expected value.

If you start with $1000, lose 40% ($400), and then gain 50% ($300 on a $600 stake), you end up with $900. If you first gain 50% ($500 on a $1000 stake), and then lose 40% ($600 on a $1500 stake), you again end up at $900.

You need to gain 67% ($400 on $600 or $667 on $1000) to break even before or after a 40% loss. The math is 1/(1-0.4)-1=0.67 or (1-0.4)*(1+0.67)=1.

Re: The infamous coin toss

#42

Post by phoenixreader »

wmorgan wrote: post #28

This is the St. Petersburg paradox with an extra variable. In SPP, EV approaches infinity as the bank's resources approach infinity. Put bounds on the bank's resources, and you find that even with trillions of dollars your EV is less than $50. Here, not only are we assuming that the bank's resources are infinite, we're also assuming that the population is large enough that there are always enough lucky players to com…

I don't think this is SPP. SPP highlights in the difference between the mathematically optimal choice and the choice chosen in practice. It is a difference between theory and practice. The problem proposed in this article occurs even in theory alone. Non-ergodicity means there is a mismatch between "the average of all possibilities in the next time-step" and "the long-term trend of one datapoint".

If we put bounds on the bank in SPP, the first coin toss would still have positive EV. In the new ergodicity problem, even with bounds on the bank, it is unclear whether the "first" coin toss is worth taking.

Re: The infamous coin toss

#44

Post by wmorgan »

phoenixreader wrote: post #42
wmorgan wrote: post #28

This is the St. Petersburg paradox with an extra variable. In SPP, EV approaches infinity as the bank's resources approach infinity. Put bounds on the bank's resources, and you find that even with trillions of dollars your EV is less than $50. Here, not only are we assuming that the bank's resources are infinite, we're also assuming that the population is large enough that there are always enough lucky players to com…

I don't think this is SPP. SPP highlights in the difference between the mathematically optimal choice and the choice chosen in practice. It is a difference between theory and practice. The problem proposed in this article occurs even in theory alone. Non-ergodicity means there is a mismatch between "the average of all possibilities in the next time-step" and "the long-term trend of one datapoint". If we put bounds on…

> If we put bounds on the bank in SPP, the first coin toss would still have positive EV.

Not if the bank starts with $0. 0 is just as valid a bound as a trillion. You can’t calculate the EV without knowing how much the bank has, and once you know that, you realize the naive calculation for EV is wrong.

Re: The infamous coin toss

#45

Post by kevinwang »

erik_seaberg wrote: post #41
kevinwang wrote: post #38

Earlier quoted context omitted.

> When I lose, I lose more than I win. I don't see how this is true. Winning adds 0.5x of your current bankroll. Losing subtracts 0.4x of your current bankroll. When you win, you win more than you lose. The gamble has positive expected value.

If you start with $1000, lose 40% ($400), and then gain 50% ($300 on a $600 stake), you end up with $900. If you first gain 50% ($500 on a $1000 stake), and then lose 40% ($600 on a $1500 stake), you again end up at $900. You need to gain 67% ($400 on $600 or $667 on $1000) to break even before or after a 40% loss. The math is 1/(1-0.4)-1=0.67 or (1-0.4)*(1+0.67)=1.

The comment I'm replying to says:

> But this is purely a result of the distribution of returns from a single toss.

Your examples contain two tosses.

Re: The infamous coin toss

#46

Post by throwaheyy »

kevinwang wrote: post #38
jacknews wrote: post #3

But this is purely a result of the distribution of returns from a single toss. If I win I get 1.5 times my money, if I lose I'm left with less than 1/1.5 times. When I lose, I lose more than I win. I agree it's counterintuitive how any individual (on average) will lose over time, yet the entire system grows.

> When I lose, I lose more than I win. I don't see how this is true. Winning adds 0.5x of your current bankroll. Losing subtracts 0.4x of your current bankroll. When you win, you win more than you lose. The gamble has positive expected value.

“current bankroll” is the issue here. 1 * 1.5 * 0.6 = 0.9 < 1.

Re: The infamous coin toss

#47

Post by throwaheyy »

kevinwang wrote: post #39
Robotbeat wrote: post #4

Earlier quoted context omitted.

Yeah, as soon as I realized that, I rolled my eyes at not immediately seeing that. When expressed as 150% and 60%, it became obvious to me.

Can you clarify? I think you might be mistaken, since a gamble with payoffs of 150% and 60% has positive expected value.

Over time it does not.

A related concept in finance/trading is “drawdown”. A single trade can have a positive expected value. But over time, if you take a loss you have to get a bigger win to end up back where you started, because you have less capital to work with.

Re: The infamous coin toss

#48

Post by erik_seaberg »

kevinwang wrote: post #45
erik_seaberg wrote: post #41

Earlier quoted context omitted.

If you start with $1000, lose 40% ($400), and then gain 50% ($300 on a $600 stake), you end up with $900. If you first gain 50% ($500 on a $1000 stake), and then lose 40% ($600 on a $1500 stake), you again end up at $900. You need to gain 67% ($400 on $600 or $667 on $1000) to break even before or after a 40% loss. The math is 1/(1-0.4)-1=0.67 or (1-0.4)*(1+0.67)=1.

The comment I'm replying to says: > But this is purely a result of the distribution of returns from a single toss. Your examples contain two tosses.

You’re right, I missed the small probability of coming out ahead on a short series because losing dominates long series. Which is what the author is trying to get across to me. Thanks.

Re: The infamous coin toss

#49

Post by wmorgan »

waveBidder wrote: post #36
wmorgan wrote: post #28

This is the St. Petersburg paradox with an extra variable. In SPP, EV approaches infinity as the bank's resources approach infinity. Put bounds on the bank's resources, and you find that even with trillions of dollars your EV is less than $50. Here, not only are we assuming that the bank's resources are infinite, we're also assuming that the population is large enough that there are always enough lucky players to com…

the St Petersburg paradox is also a problem of ergodicity, since every single player loses with probability 1 over time, even though the "space average" is net positive. No need to invoke messy reality to solve the paradox. the exponential example is just much more useful, since there are plenty of systems easily described by compound growth

I guess this is my point of disagreement with the article:

> We have thus arrived at the intriguing result that wealth averaged over many systems grows at 5% per round, but wealth averaged in one system over a long time shrinks at about 5% per round.

Wealth averaged over many systems doesn’t grow by 5%. It shrinks just like the average. The EV calculation is just wrong. For any finite starting wealth between the players and the bank, there is a number of iterations where the EV turns negative.

If you say, well let the starting wealth be infinite, I say, okay? If you have infinite dollars there are a lot of tricks for making infinite more dollars. It doesn’t work in the real world.

Re: The infamous coin toss

#50

Post by kevinwang »

throwaheyy wrote: post #47
kevinwang wrote: post #39

Earlier quoted context omitted.

Can you clarify? I think you might be mistaken, since a gamble with payoffs of 150% and 60% has positive expected value.

Over time it does not. A related concept in finance/trading is “drawdown”. A single trade can have a positive expected value. But over time, if you take a loss you have to get a bigger win to end up back where you started, because you have less capital to work with.

> Over time it does not.

Yes, the article shows that almost certainly, any individual's wealth will approach 0 from repeatedly taking this gamble. However, the comments I replied to say:

> But this is purely a result of the distribution of returns from a single toss.

which I don't understand.

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