The infamous coin toss

https://ergodicityeconomics.com/2023/07/28/the-infamous-coin-toss/

by beefman • 3 years ago

206 245 3 years ago

ergodicityeconomics.com

jacknews5• 3 years ago

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.

Robotbeat• 3 years ago

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.

kevinwang10• 3 years ago

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

kevinwang10• 3 years ago

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.

erik_seaberg2• 3 years ago

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.

throwaheyy2• 3 years ago

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

jacknews5• 3 years ago

Percentages etc are deceiving here. A 50% bonus is not the opposite of a 50% discount.

Your stake should grow or shrink by the same factor if you win or lose.

If you win, your stake is multiplied by 1.5.

If you lose, it should then be divided by 1.5 (multiplied by 0.66..).

But is in fact multiplied by only 0.6.

didgeoridoo4• 3 years ago

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• 3 years 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

cubefox7• 3 years 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• 3 years 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.

n2d46• 3 years 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.

operator-name2• 3 years ago

For some reason this is not mentioned in the article but is mentioned in the video.

hgsgm4• 3 years 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.

hgsgm4• 3 years 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.

kgwgk11• 3 years 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.)

cubefox7• 3 years ago

We need the geometric expectation here: sqrt(2x * .4x)=0.89x

The arithmetic expectation is wrong in this case.

jakell5• 3 years ago

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.

bogtog• 3 years ago

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.

jakell5• 3 years ago

Exactly

bzax5• 3 years ago

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.

bzax5• 3 years ago

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.

jakell5• 3 years ago

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).

didgeoridoo4• 3 years ago

You’re taking arithmetic averages of percentages… I don’t think that calculates anything meaningful.

Try 225% * 90% * 90% * 36% to get the expected value.

jakell5• 3 years ago

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.

jstanley5• 3 years ago

I think the part about the misalignment between the individual and the collective is basically wrong.

In the given game, the collective loses money just as the individual does. The collective wealth is the summation of the individuals' wealths, and both the collective and individual wealths drop over time. Write a simulation and try it if you don't believe me. I did, because I couldn't work out how the collective wealth could grow while the individual wealths drop, and the answer is that it does not.

The only way it looks like the collective makes a gain is if you try to say that x_i(t) = x_i(t-1)*1.05, but that's a simplification that doesn't hold in either the individual or the collective case.

But if you forget the part about the supposed misalignment between individual and collective, then the idea that the expected value of 1 iteration can be positive while the expected value of repeated iteration can be negative is very fascinating!

didgeoridoo4• 3 years ago

Take a simplified system: two players, both with $10. Flip a coin. Let’s say one comes up heads, the other tails. Total is now $15 + $6 = $21. A $1 increase for the whole system.

I agree the distinction is not at all between individual and collective, but rather between one-off and repeated games.

selcuka5• 3 years ago

In the given game, the collective loses money just as the individual does.

No, they don't. If 100 people starts with $100 (i.e. a total wealth of $10,000), and half of them win as expected, the total wealth at the end of first round will be $10,500.

Granted it won't be a 5% increase for the next round, and eventually collective wealth will also approach 0, but that's the point. 100 people playing 1 round is not the same as 1 person playing 100 rounds.

jstanley5• 3 years ago

Yeah, that's after 1 round. Both the individuals and the collective expect to gain if they only play 1 round.

The interesting part is that they expect to lose money over time even though they expect to gain money if they only play once. But that holds for both the individuals and the collective.

A less confusing game with the same mechanism is "flip a coin, if it's heads I give you 1000x your initial investment, if it's tails I wipe you out completely" - you expect to make a massive gain on any given turn, but if you play enough rounds you'll eventually lose everything.

petermcneeley3• 3 years ago

This simulation is the collective wealth over N players for M times. It is increasing. https://coin-n.streamlit.app/?utm_medium=oembed

selcuka5• 3 years ago

If every individual in a population is definitely doomed to lose all their wealth, how could the collective wealth increase?

ActivePattern• 3 years ago

No, your expected value is indeed positive over repeated iterations: (1.05^i)*w, where i is the number of iterations and w is the starting wealth.

The intuition for why this happens is that the losses of the majority are made up by the big gains of a minority. You can even see that after two iterations:

Case 1: two heads -- 225% wealth Case 2: heads, tails -- 90% wealth Case 3: tails, heads -- 90% wealth Case 4: two tails -- 36% wealth

So, most people are losing money but a single lucky guy makes enough money that the expected value is still positive.

If you run a simulation with a low sample size, you might not see this effect because the "lucky guy" case is rare (and gets more rare with more iterations).

petermcneeley3• 3 years ago

Note a single lucky guy is not enough. P of N heads is (1/2)^N but payout is (1.5)^N for this single lucky guy.

(The question can be reformulated such that this could be the case.)

mizzao• 3 years ago

Mmm, seems similar to the game of venture capital!

jstanley5• 3 years ago

Wow, great point! I had ran my simulation several times to check that it was reasonably stable and it gave me less than $1 average wealth each time, so I thought that was that. But just now I re-ran it with a much larger population and indeed the average wealth grows now instead of shrinking. Sadly I can't edit my comment above any more, but:

that the expected value of 1 iteration can be positive while the expected value of repeated iteration can be negative is very fascinating!

is indeed untrue. Thanks!

rcxdude• 3 years ago

Yeah, which implies for any finite N eventually the average will collapse, once the lucky guy's luck runs out. The expected value prediction is only true over a long enough timescale for actually infinite N (I do wonder how this scales: I would intuitively expect that the time window scales with log(N), which implies even very large N will not last drastically longer, but I don't know if that's the case)

D_Alex• 3 years ago

I disagree, and furthermore I think the entire article is basically wrong - in a way. Specifically, the expected average wealth of both the collective and an individual will increase every round. However the median wealth will fall.

The reason for this is that the very rare outliers will accumulate truly enormous wealth, while the rest of the players will never lose more than they start off with. You can see this in the first graph after say 250 rounds - the richest player has around 10,000x his initial wealth, even as the other 1000(or so) have very little - so on average they have 10x what they started with. And the reason no one is rich after 1000 rounds is simply that there were not enough samples. If there were say a billion samples, you might expect one to come in with say a trillion times its starting wealth, overcompensating for the other 999,999,999 losers - on average.

dongobread• 3 years ago

I tried this and collective wealth does increase: https://imgur.com/a/Zc2aApj

The median person does get poorer over time though, a small percent get extrmeely wealthy: https://imgur.com/a/v2AVyrv https://imgur.com/a/7pPfKk6

jstanley5• 3 years ago

Thanks, it turns out to really matter how big the population is, otherwise you aren't likely to have enough lucky members to bring the mean up.

civilized3• 3 years ago

Any finite collective eventually loses almost surely, but the larger the collective, the longer it takes, and the greater the inequality in the meantime, before everyone loses. In the limit, an infinite collective exhibits the paradoxical behavior from the article: expected value goes up forever, but almost every individual eventually loses.

An ever smaller slice of huge winners is pulling up the average, but in any finite population this slice disappears in finite time. Only in an infinite population can it persist indefinitely.

All of this is fairly straightforward for a mature student of probability. If enough folks want, maybe I could write up the details to prove these things.

wmorgan3• 3 years ago

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 compensate for the unlucky ones. Put bounds on the size of the population, and you see that everyone goes bust in all but a tiny fraction of cases. Put bounds on the bank, and even that tiny fraction can't compensate for all the losers, and EV is negative.

waveBidder• 3 years ago

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

wmorgan3• 3 years ago

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.

phoenixreader• 3 years ago

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.

wmorgan3• 3 years ago

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.

wodenokoto2• 3 years ago

It also made me think of the St. Petersburg paradox and I think the authors did too, as they call their toss “Peter”

geminger• 3 years ago

He called it "Peters", after himself, which is off-putting.

falsandtru3• 3 years ago

What percentage of people will lose money is also very important but does not seem to be written. What percentage?

civilized3• 3 years ago

Almost every person who plays the game will see their money decay towards zero as time goes to infinity.

falsandtru3• 3 years ago

Knowing the time scale is very important.

chronolitus2• 3 years ago

but, since it is somewhat like a random walk, isn't it guaranteed that if you go on for infinitely many tosses, at some point you will be very rich?

[1] https://math.stackexchange.com/a/493446

falsandtru3• 3 years ago

If it follows the median, the percentage of losers increases exponentially from 50%.

KRAKRISMOTT• 3 years ago

No mention of Martingales?

daxfohl• 3 years ago

Isn't this just E(f(x)) != f(E(x))?

The first part shows the expected value of the return function, which is positive (from the gambler's perspective).

The second part uses f(E(x)): the gambling function applied to expected number of heads and tails (50/50), which is a loss on average. But that's disingenuous because it silently thwacks off the outliers where a gambler wins almost every time or loses almost every time, which is hugely positive on average.

Essentially any nonlinear return function would have the same inequality. It's just stated trickily here and the analysis is deliberately misleading.

rerdavies3• 3 years ago

Huh. So I wrote the code, and ran the simulation. Now I get it.

Investors: 100,000 Iterations: 100

Average worth after 100 iterations: $83.923

Average net worth increases. However the distribution of wealth is skewed dramatically.

Winners: 13,704 (net worth of more than $1 at the end) Investors worth < $0.01: 53,935

What That Guy was worth (the investor who made the most money):

$1,171,830.00

He flipped 71 heads and 29 tails.

Median wealth of all investors: $0.00515378

What could have been made more clear in the presentation. Average wealth increases; but average log(wealth) decreases. Investors on the long improbable upper tail make huge amounts of money; investors on the lower tail just don't make much difference to the average, when they lose 40% of $0.01. What causes the wild income distribution: the fact that log(wealth) decreases on average. This doesn't decrease the total money supply, but it does increase the skewed distribution of wealth.

It has nothing whatsoever to do with anything related to flipping consecutive heads and tails. And a great deal to do with the fact that the log(gain) is 0.405465 and the log loss is -0.510826. So the average investors log(wealth) decreases over time even though the average total wealth increases.

Most figures are wildly variable from trial to trial. Most especially, what That Guy makes is highly variable. If he rolls 72 heads and 28 tails instead of 71 heads and 29 tails, then he's worth $2,565,000 more, and single-handedly adds $25.65 to the average net worth! The number of people who lose money is surprisingly, depressingly consistent.

The obvious conclusion: increasing wealth concentration is an unavoidable consequence of capitalism that probably has very little to do with merit.

The only solution: eat the rich.

ta1243• 3 years ago

Or you could redistribute wealth to even out the winner-takes-all part while still encouraging merit of some sort by having winner-takes-some

waldothedog• 3 years ago

Like a large(r) tax on upper incomes and capital gains??

drekipus• 3 years ago

increasing wealth concentration is an unavoidable consequence of capitalism

My bad I thought you made a coin flip simulation, not a capitalism simulation.

At what point do we stop wagging the finger at the ghost of capitalism and just say it's mathematics? You don't hear people crying for the upheaval of mathematics

AdamN• 3 years ago

People do say that. Usually the critique of (perfect) capitalism is that it (structurally and mathematically) skews towards increasing wealth for winners at the expense of the rest of the population and that this is inevitable.

Then the response of advocates against this situation is to either commit to more structures to prevent that outcome towards diluted (non-perfect) capitalism or other social forms (socialism, etc...)

Swenrekcah6• 3 years ago

The obvious conclusion: increasing wealth concentration is an unavoidable consequence of capitalism that probably has very little to do with merit. The only solution: eat the rich.

This is exactly why we need progressive asset taxes. Income tax should be almost abolished, it can be something like 10% on all types of income and then nobody needs to bother with tax evasions on that.

5-15% VAT or sales tax to generate revenue for the places where people congregate and do their business, the rest of society should be funded by asset taxes.

LeonB19• 3 years ago

Vat is irredeemably regressive. I suspect you know that.

Income tax fails in its progressive goals not because it is inherently bad (like VAT) but because the tax laws are written by the richest players. (Ask a poor person how much influence they’ve had on tax law, and then ask a top 5 accountancy firm. This isn’t a controversial statement.)

LeonB19• 3 years ago

Imagine if — on top of that - we let the undeserving rich invest their wealth in things such as:

- owning media - lobbying the government - owning the means of production

It could take that dire situation and make it drastically worse. I mean, in theory.

robertlagrant25• 3 years ago

owning media

Who should own media? Private people, or the state?

lobbying the government

If lobbying the government is effective, it's either because the lobbying was correcting a wrong, or the government is allowing itself to be corrupted.

owning the means of production

The "means of production" was barely relevant when Marx found out what a factory was from his factory-owning mate. A plumber owns "the means of production" when he owns his own tools. It's far too low resolution a phrase to be useful, except for anyone who finds it useful to teach people to hate another group of people.

robertlagrant25• 3 years ago

The only solution: eat the rich.

Don't look at history when people did this, though. Just do it based on this simulation.

shove7• 3 years ago

Nom nom nom

credit_guy2• 3 years ago

The obvious conclusion: increasing wealth concentration is an unavoidable consequence of capitalism that probably has very little to do with merit.

Not really. P implies Q is not the same as Q implies P.

The fact that you can get a large gap between rich an poor with pure coin toss does not mean that when you have such a gap the gap resulted from pure randomness. It could have resulted from any combination of merit, hard work, inheritance, cheating, and luck. Your conclusion simply does not follow.

svnt3• 3 years ago

If you change the equation slightly and place a fixed resource consumption requirement (take 1-10% of the initial amount to pay to flip a coin) on each person’s continuation, and then unevenly distribute the initial balance, it will become apparent that your ability to become “that guy” is dominantly determined by your initial balance.

In this version people do not end up at some small fraction of the initial balance but instead they go negative, i.e. exit the simulation to homelessness.

The ability to keep flipping the coin without starving is absolutely the primary determinant.

ano-ther• 3 years ago

Average wealth increases; but average log(wealth) decreases

What’s the mathematical rationale (or intuition) for looking at the log values?

I don’t quite understand what they represent.

joncrocks• 3 years ago

It's reducing the impact of the outliers.

e.g.

x = 60, 10, 10, 10, 10

avg. x = 20

avg log(x) = 1.1556 (which is 10^1.1556 = 14.3)

twobitshifter2• 3 years ago

Log is useful for repeated trials, since it’s a way of looking at exponential growth. If you write out the formula for a repeated trial, you’ll get an exponential function. Log tells you the value of the exponent a in e^at.

kgwgk11• 3 years ago

Multiplication becomes addition after you take logs. That makes many things easier to understand and manipulate.

Exponential trends become linear.

The geometric random walk where you multiply by X or divide by Y with equal probability becomes an additive random walk where you add log(X) or substract log(Y) and you can represent that as a standard random walk (where the up and down steps have the same magnitude and probability) plus a trend.

n2d46• 3 years ago

The other people gave good intuition on why the logarithm is good but not on why the logarithm is the best choice. For example, max(sqrt(EV)) would also reduce outliers.

There is a more mathematical way akin to the original derivation of the Kelly criterion [1]. Let's compare two strategies, max(EV) and max(log(EV)). Our game is coin-toss with heads=+50%/tails=-40% payouts and infinite repetitions; max(EV) would always go all-in, while max(log(EV)) would only bet a fraction of its current wealth at each turn.

- After the first turn, there are two cases; if heads, max(EV) has a higher payout, if tails, max(log(EV)) has a higher payout. On average, max(EV) has a higher payout because it went all-in, and there's a positive EV per toss.

- After two turns, there are four cases; in one of them (heads-heads), max(EV) has a higher payout, and in the other three, max(log(EV)) does. The latter is hence the "safer" choice. However, in the 25% chance that max(EV) wins, it has significantly more money than max(log(EV)), so the average (not median) payout is higher for max(EV).

- This repeats, and after more and more repetitions, the chance that max(log(EV)) has a higher payout converges towards 1. Note that the payout for max(EV) in the (very!) rare case that it gets lucky skyrockets at an even faster speed, so on average, max(EV) still has a higher payout.

- Turns out this is not just true for max(EV) and max(log(EV)), but in fact max(log(EV)) always maximizes the chance of higher payouts, given enough repetitions.

So, in other words, max(EV) maximizes the average payout, while max(log(EV)) maximizes the chance of having a higher payout than any other strategy. The original Kelly paper has more precise maths, but I hope this helps to guide you.

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

nzealand• 3 years ago

If you lose half your wealth, you have to double your wealth to get back to where you were.

The log change intuitively shows this.... log (0.5) = -0.3 & log (2) = 0.3

If you lose three quarters of your wealth, you have to quadruple your wealth to get back to where you where.

Again, log intuitively shows this.... log (0.25) = -0.6 & log (4) =0.6

kevinwang10• 3 years ago

I think my explanation is more intuitive than every other comment so far.

An intuition is that log(wealth) represents utility. An intuition for why log(wealth) represents utility is that wealth has decreasing marginal returns. The logarithm is a function that fits this. If you play around with it a bit you might see that it conforms to your intuitions: increasing your net worth from $100k to $200k increases your utility significantly, maybe about as much as increasing your net worth from $1m to $2m, and much more than increasing your net worth from $1m to $1.1m.

Fredkin5• 3 years ago

Except, in your simulation you only consider accumulation of dollar amounts. Pure gambling, wealth in a vacuum. Calling these dollar amounts 'wealth' is not helpful because they need to have purchasing power, there need to be goods and services to buy, otherwise they're just meaningless entries in a ledger, bits of paper (see Zimbabwe). It's wrong to dismiss capitalism on the basis of this simulation.

In the real world there are usually side effects to these 'bets'. For example, a person, who is already rich, uses their wealth to purchase a factory and produce goods at a dollar price in the hope to earn more wealth. Those goods might be rejected by the market, but if accepted, they receive dollars and the rest of the world receives goods. The factory owner can only receive the dollars if others can afford to buy them. This means there is another road to ruin - fail to produce anything people want at a price they can afford and you'll soon go bust.

The buyers who receive goods in exchange for money, particularly durable labor saving goods like dishwashers/fridges/cars/computers, are surely more wealthy in some sense? In addition, the factory owner might make tools/machinery that enables other entrepreneurs to make stuff too.

Also consider ongoing costs, that capital has to be replaced (things break!), unexpected bumps in the road, finite lifespan, waste, and so on.

Which leads me to our current system: the rich have got richer through financialization, government lobbying, zero-interest financing et.c, that has allowed them to acquire assets that appreciate without a substantial increase in production.

The solution is to make rich people take much bigger, potentially ruinous, risks with their wealth on production instead of financial chicanery. More abundant goods and technology are the payoff while failure leads to liquidation of their assets, making assets more accessible to everyone else who might need them.

Deflation, and crashes are a feature, not a bug. Inflationism, stimulus, cheap money et.c helps protect workers in the short term during a downturn, but prolonged stimulus makes assets unaffordable. Worse it causes huge misallocations of capital into unproductive endeavors at no risk to the investor.

My belief is you don't 'prop-up' the big fish using the government/central bank. Then yes, you may still have inequality, but it won't be nearly as static, but more dynamic. i.e there may always be an extreme power law wealth distribution, but what's more important is there's regular turnover at the top and nobody spends too long in the top or the tail.

twobitshifter2• 3 years ago

My somewhat controversial opinion is even if success were based on merit, this is still random chance since merit depends on one being born with ability in a supporting environment. Meritocracy can be social Darwinism.

sobriquet94• 3 years ago

The only solution: eat the rich.

The only? Maybe not bet 100% of your wealth every time? That's what Kelly criterion is about.

eli_gottlieb• 3 years ago

You have to have enough in the first place that trying to support yourself isn't betting 100% of your wealth.

coryfklein2• 3 years ago

Uh do you mean "bet 0% of your wealth every time"? Or does betting some number greater than 0% lead to net positive expected gains for an individual?

sershe• 3 years ago

The whole experiment takes as a premise that merit doesn't matter (coin toss)... and arrives to the obvious conclusion that merit doesn't matter.

Especially since, as someone indicates in another top level comment, the percentages are not even chosen properly, gambling on absolute money is a dumb choice, one should only gamble on log-money. So, a bunch of people making a dumb gamble with no merit involved produce a distribution.

I don't understand why anybody at all cares about this - it's just garbage in, garbage out. The only real-world connection one could make is... don't play lottery, I guess? Oh wait I do understand why people care - that's because this "supports" a social conclusion they already pre-supposed.

daveFNbuck2• 3 years ago

I think the key thing you're missing here is that they were talking about behavior as the number of flips goes to infinity. That long improbable upper end becomes more and more improbable over time. As time goes to infinity, the probability of being in it goes to 0. With a finite population, everyone is ruined.

I ran the same simulation as you. After 100 iterations, the richest one had $71. After another 1,000 iterations, the richest one had $78 (of a total $133). After another 10,000 iterations, the richest one had 10^-165 dollars. They were still an extreme outlier with 95% of the total wealth, but he had almost nothing.

kgwgk11• 3 years ago

With a finite population, everyone is ruined.

With high probability.

DrScientist2• 3 years ago

He mentions some solutions in the linked articles - and basically it involves pooling risk.

In concrete terms I guess that means such things as re-distributive taxes ( either on death or through life ) - pooled pension funds, insurance, family and friends etc etc.

AtlasBarfed• 3 years ago

Okay, so aside from the cool terms (ergodicity, which... well whatever), what I think is really going on here is that the results of the iterative system, which intuitively appear to be a normally distributed value distribution for "not large" iterations, actually is NOT.

Since it is actually a very not-normal distribution, the typical concepts of "mean", "median", and "std deviation" (which AFAIK isn't even really mentioned and would be meaningful) are warped to the point of meaninglessness.

Of course economics being essentially a social science with some math and therefore defaults to thinking in typical normal distribution assumptions, won't naturally deal with that sort of pattern.

And while I agree game theory (and this is a game) is relevant to economic study, the fact is that real world "games" like wealth distribution has a vast chasm of chaos/randomness/nonlinearity that blocks things like this or the "pirate distribution" game and others.

Distrubingly, the games are parroted in some either overt or subvert manner to justify the distribution of wealth in the current 30 year era.

"It's the natural way of economics and mathematics, how dare you chop off my head in the guillotine!"

And yet "Some men want to watch the world burn". The elite also love the "great man" theory of history, because they are, of course, great men. But bring up the revolutionaries, and suddenly they don't like those pet ideas.

Etherlord87• 3 years ago

Why make it about the rich. Your statistic doesn't magically change if you change the starting $1 to something else.

And to put the phenomena in much simpler words: flipping a coin once is good, having $100 bucks you either lose $40 or get $60, on average you get $10, and you win half of the time. The problem is if you have to flip a coin at least twice, because then the accumulated multiplier of your value, rather than .6 / 1.5 becomes:

.6 / .9 / .9 / 2.25

So while on average you gain 16.25% of value, you are a winner only ¼ of the time, and it gets better/worse depending on average gain / chance to gain.

…And apparently, what I learned just now, since your chance to gain approaches 0, for a finite population and infinite time, the chance approaches 0, therefore rendering the average gain to also be 0.

wodenokoto2• 3 years ago

Off topic or at least sidetracked, this took me by surprise

For historical reasons, this average is also called the expected value

Why is this “for historical” reasons?

kgwgk11• 3 years ago

Because long ago people started using a word he doesn't like.

https://web.universiteitleiden.nl/fsw/verduin/stathist/1stwo...

EXPECTATION. According to A. W. F. Edwards, expectatio occurs in 1657 in Huygens's De Ratiociniis in Ludo Alae (David 1995).

According to Burton (p. 461), the word expectatio first appears in van Schooten's translation of a tract by Huygens.

The two references above point to the same text as Huygens's De Ratiociniis in Ludo Alae was a translation by van Schooten. NB The word expectatio is used quite frequently throughout the text.

This is the Latin translation by Van Schooten of the first proposition:

Si a vel b expectem, quorum utriusque aeque facile mihi obtingere possit. expectatio mea dicenda est (a+b)/2

This is the Dutch text of Huygens' Van Rekeningh in Spelen van Geluck. This text was published in 1660 but already written in 1656.

Als ick gelijcke kans hebbe om a of b te hebben, dit is my so veel weerdt als (a+b)/2

The literal translation of the Dutch text is: If I have an equal chance to get either a or b, this to me is worth as much as (a+b)/2. There is no explicit mention of expectation only of value, but as the rest of the explanation of the first proposition is concentrated on the possible outcomes of a game of chance, expectation is implicitly around.

Expectation appears in English in Browne's 1714 translation of Huygens's De Ratiociniis in Ludo Alae (David 1995).

This is Browne's 1714 translation of the first proposition:

If I expect a or b, and have an equal chance of gaining either of them, my Expectation is worth (a+b)/2

inopinatus• 3 years ago

PHYSICIST: I offer you the following gamble. I toss a fair coin, and if it comes up heads I’ll add 50% to your current wealth; if it c—

ROSENCRANTZ: (interrupting) Do it.

shove7• 3 years ago

applause

soVeryTired2• 3 years ago

In the first regime, where we calculate at the population level, we stop and average the wealth of the whole population after every coin toss. So if E[X_t] is the expected sum of wins and losses from the coin toss,

W_t = exp(k E[X_t])

Whereas the wealth of any individual grows like

W_t = E[k exp(X_t)]

There's a result called Jensen's inequality that says that

f(E[X]) > E[f(X)]

for any random variable X and any convex function f (exp is one such function). In a sense, I think this all just falls out of Jensen's inequality.

Tarsul• 3 years ago

if I go into the casino with 67 bucks and make 1,5 times what I have, I will own 100 bucks. If I go into the casino with 100 bucks and lose a third only (not 40%!) then I will own 67 bucks.

Thus, +50% and -40% are not the right arithmetic pairs. It should've been +50% and -33%. It's even more intuitive to say 3/2 and 2/3. Waiting for a matician who can explain better, but this whole story is more of a parlor trick than anything else.

soVeryTired2• 3 years ago

but this whole story is more of a parlor trick than anything else

I don't think it is really. You're right in saying that 2/3 and 3/2 are the right numbers to ensure that the population wealth stays the same. But the point of the exercise is to show that the population wealth can increase while the wealth of almost any individual in the population decreases.

This is the author's jumping-off point to argue that the standard arguments in economics about average welfare can be far removed from the experience of any single worker in the economy.

kevinwang10• 3 years ago

You're right in saying that 2/3 and 3/2 are the right numbers to ensure that the population wealth stays the same.

Quick correction:

The article uses 0.6 and 1.5. In this setting, the population wealth increases and the wealth of almost any individual decreases.

In the comment you replied to, they propose 0.66 and 1.5. In this setting, the population wealth doesn't stay the same, it increases even more than before.

(In order for population wealth to stay the same, the payoffs should be 0.5 and 1.5)

DougBTX• 3 years ago

It looks like those were chosen to get a opposite numbers for the comparison, +5% and -5%:

the expected value ... grows by 5% in every round ... but, crucially, a head and a tail experienced sequentially is different from two different agents experiencing them ... we have lost ... approximately 5% per round

kevinwang10• 3 years ago

Either wager (+50%/-40% or +50%/-33%) would be profitable wagers for a gambler in a casino to take. If you ever found a casino offering bets like that, you would never have to work again (unless they sued and won against you to get their money back). No casino would ever knowingly offer a wager with negative expected value for them.

kgwgk11• 3 years ago

Consider this alternative bet:

  heads +100%
  tails -75%

The expected value is positive: 12.5%

If you play five times in parallel you're as likely to make money as to lose money but when you gain you gain 45% on average and when you lose you lose 20% on average. Overall, the expected gain is 12.5%.

If you play five times in sequence you're unlikely to make money (less than 20% probability) but then you make a 300% gain (four heads) or even a 3100% gain (five heads). Overall, the expected gain is 12.5% compounded five times which is 80%.

As the number of coin flips grows it is less and less likely that you lose if you do it in parallel (the expected gain is 12.5% and the distribution of returns converges to that value) and more and more likely that you lose if you do it in sequence (the expected gain is again 12.5% compounded but the distribution of returns is an increasingly skewed lognormal).

Exercise for the reader: take this interesting but simple and well-known fact and make a career out of it from repeating it again, and again, and again, and again, and again, and again, and again...

ortichic• 3 years ago

If I owned a casino, should I offer this game?

operator-name2• 3 years ago

I think I've reached an understanding, but do correct me if I'm wrong.

    lim players -> inf
    mean wealth -> starting_wealth * (win_chance * (1 + win_gain) + lose_chance * (1 - lose_loss)^rounds)

So in their example case

    win_change = lose_change = 0.5
    win_gain = 1.5
    lose_loss = 0.4
    (0.5 * (1 + 0.5) + 0.5 * (1 - 0.4)) = 1.05

so on average the mean wealth increases as the number of rounds increases. Yet at the same time for a single player

    lim rounds -> inf
    wealth -> 0

As others have mentioned, this is because the win multiplier is 1 + win_gain = 1.5 whereas the loss multiplier is 1 - lose_loss = 0.6. This is more easily seen by comparing the log of the multiplier

ln(1.5) = 0.41 ln(0.6) = -0.51

For the game to be profitable, it must satisfy

ln(1 + win_gain) > -ln(1 - lose_loss) <=> 1 + win_gain > 1/(1 - lose_loss)

So for a gain of 50% the loss must be less than 33.3%.

coryfklein2• 3 years ago

Hah! Actually the first time I came across this concept was playing Stardew Valley. In the Fall season there is a festival with a spinning wheel. Visually the wheel is half green, half orange. You can bet N tokens on a color, and if the spinning arrow lands on that color then you receive N, otherwise you lose N.

The funny thing though is that despite the wheel being painted half green half orange, the actual odds are 75% green and 25% orange. But calculating how much of your money to bet at each iteration isn't super straightforward; in the end it turns out that to maximize your profit you should always bet exactly half of your money on green.

The infamous coin toss

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