30 ms·

The infamous coin toss

206 points by beefman 3y ago

wodenokoto 3y 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?
- kgwgk 3y ago
  Because long ago people started using a word he doesn't like. https://web.universiteitleiden.nl/fsw/verduin/stathist/1stword.htm 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
greatfilter251 3y ago
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jacknews 3y 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 3y 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.
  - kevinwang 3y ago
    Can you clarify? I think you might be mistaken, since a gamble with payoffs of 150% and 60% has positive expected value.
    - throwaheyy 3y ago
      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.
      - kevinwang 3y ago
        > 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.
        jacknews 3y ago
        It will only approach zero because you lose more than you gain. If the loser got 0.6666c instead of 0.6c, and the winner got $1.50, then over time you'd break even, on average. And yet the expected return would apparently be 1.08333. If think the conclusion is that 'expected return' is a fallacy, you just can't add proabability-outcomes in this way to get an 'expected outcome'.
didgeoridoo 3y 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 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.
jakell 3y 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 3y 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.
  - jakell 3y ago
    Exactly
  - bzax 3y 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.
- bzax 3y 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.
  - jakell 3y 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).
jstanley 3y 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!
- didgeoridoo 3y 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.
- selcuka 3y 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.
  - jstanley 3y 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.
wmorgan 3y 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 3y 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
  - wmorgan 3y 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 3y 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.
falsandtru 3y ago
What percentage of people will lose money is also very important but does not seem to be written. What percentage?
- civilized 3y ago
  Almost every person who plays the game will see their money decay towards zero as time goes to infinity.
  - falsandtru 3y ago
    Knowing the time scale is very important.
  - chronolitus 3y 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 https://math.stackexchange.com/a/493446
    - chronolitus 3y ago
      if true, an interesting question is - on average - how many tosses are needed for you to reach some arbitrary value (e.g. $1M)?
    - civilized 3y ago
      It's exactly like a random walk... with downward drift. If you take the log of the wealth, the steps are: log(1.5) = 0.18, log(0.6) = -0.22. So at each step, there's a 50% chance you go up 0.18, and a 50% chance you go down 0.22. Random walks with downward drift don't inevitably go up arbitrarily high like ones without drift.
- falsandtru 3y ago
  If it follows the median, the percentage of losers increases exponentially from 50%.
KRAKRISMOTT 3y ago
No mention of Martingales?
daxfohl 3y 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.
rerdavies 3y 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 3y 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 3y ago
    Like a large(r) tax on upper incomes and capital gains??
- drekipus 3y 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
deleted 3y ago
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inopinatus 3y 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.
- shove 3y ago
  applause
soVeryTired 3y 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 3y 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.
- soVeryTired 3y 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.
  - kevinwang 3y 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 3y 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
- kevinwang 3y 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.
kgwgk 3y 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 3y ago
If I owned a casino, should I offer this game?
operator-name 3y 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%.
coryfklein 3y 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.

206 points by beefman 3y 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?

kgwgk 3y ago
Because long ago people started using a word he doesn't like. https://web.universiteitleiden.nl/fsw/verduin/stathist/1stword.htm 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

greatfilter251 3y ago

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jacknews 3y 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 3y 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.
- kevinwang 3y ago
  Can you clarify? I think you might be mistaken, since a gamble with payoffs of 150% and 60% has positive expected value.
  - throwaheyy 3y ago
    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.
    - kevinwang 3y ago
      > 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.
      - jacknews 3y ago
        It will only approach zero because you lose more than you gain. If the loser got 0.6666c instead of 0.6c, and the winner got $1.50, then over time you'd break even, on average. And yet the expected return would apparently be 1.08333. If think the conclusion is that 'expected return' is a fallacy, you just can't add proabability-outcomes in this way to get an 'expected outcome'.

didgeoridoo 3y 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 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.

jakell 3y 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 3y 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.
- jakell 3y ago
  Exactly
- bzax 3y 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.
bzax 3y 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.
- jakell 3y 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).

jstanley 3y 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!

didgeoridoo 3y 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.
selcuka 3y 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.
- jstanley 3y 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.

wmorgan 3y 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 3y 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
- wmorgan 3y 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 3y 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.

falsandtru 3y ago

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

civilized 3y ago
Almost every person who plays the game will see their money decay towards zero as time goes to infinity.
- falsandtru 3y ago
  Knowing the time scale is very important.
- chronolitus 3y 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 https://math.stackexchange.com/a/493446
  - chronolitus 3y ago
    if true, an interesting question is - on average - how many tosses are needed for you to reach some arbitrary value (e.g. $1M)?
  - civilized 3y ago
    It's exactly like a random walk... with downward drift. If you take the log of the wealth, the steps are: log(1.5) = 0.18, log(0.6) = -0.22. So at each step, there's a 50% chance you go up 0.18, and a 50% chance you go down 0.22. Random walks with downward drift don't inevitably go up arbitrarily high like ones without drift.
falsandtru 3y ago
If it follows the median, the percentage of losers increases exponentially from 50%.

KRAKRISMOTT 3y ago

No mention of Martingales?

daxfohl 3y 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.

rerdavies 3y 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 3y 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 3y ago
  Like a large(r) tax on upper incomes and capital gains??
drekipus 3y 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

deleted 3y ago

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inopinatus 3y 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.

shove 3y ago
applause

soVeryTired 3y 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 3y 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.

soVeryTired 3y 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.
- kevinwang 3y 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 3y 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
kevinwang 3y 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.

kgwgk 3y 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 3y ago

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

operator-name 3y 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%.

coryfklein 3y 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.