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He actually picked bad numbers. That is a losing bet even on average. You can see that pretty easily by just assuming you get exactly the opposite result each t

He actually picked bad numbers. That is a losing bet even on average. You can see that pretty easily by just assuming you get exactly the opposite result each time. 1.5 * 0.6 = 0.9, so no matter how you start, you're behind with an equal number of wins and losses. Naively, you might think "50% is more than 40%," but that isn't how percentages work. You need win and loss proportions with a geometric average of 1 for a breakeven expectation (percentages are multiplicative, not additive). In this case, 3/2 and 2/3 is what you'd need. This happens to not show after only 100 trials just because some tiny number of people get really lucky and draw up the ensemble average, but if you keep going, somewhere between 200 and 500 trials, the ensemble average pretty quickly drops below the starting average wealth and stays there, asymptotically approaching 0.

alkonaut 6y ago
It’s just a really weird way of expressing the bet. This bet either has a positive EV or a negative one and if it’s positive it’s going to tend to infinity when repeated. There is one assymmetry at the zero point (assuming people can’t recover from bankruptcy by borrowing another dollar) but that’s easily fixed by adding a simple bet strategy e.g “bet at most 1/10 your bankroll on each bet”.
- jsweojtj 6y ago
  > This bet either has a positive EV or a negative one and if it’s positive it’s going to tend to infinity when repeated. This is wrong. The bet as described has a positive EV and the time average for a single player tends to zero as the bet is repeated. > There is one assymmetry at the zero point (assuming people can’t recover from bankruptcy by borrowing another dollar) ... The result is not due to zero being an absorbing value. In the setup you can go arbitrarily small and come back without issue. The result is the same.
  - alkonaut 6y ago
    For a 50% chance you need odds of over 2.0 (regardless of bet size) to have positive EV. This is a bet of 0.4. With 2.0 odds the payout would be 0.8, and here it is 0.9? That looks like odds over 2.0 for 50/50 chance. So the individual bet is positive EV but the EV of an infinite sequence of such bets is still zero and that’s without the zero limit? There is just nothing in the article that begins to explain how that works. Maybe the video is better
    - alkonaut 6y ago
      If I just stay away from the 0 point (by betting a fraction of the bankroll) then regardless of sequence length the ensemble stats positive. I must simply not be understanding the rules of the simulation. https://dotnetfiddle.net/5r4ARk https://dotnetfiddle.net/5r4ARk
maximilianroos 6y ago
This is wrong and misses the point of the post. The median outcome is indeed negative, for the reason you give. But the mean outcome is positive, because some players become exceedingly rich. You can try it at home, here's some Julia which runs it over 1M people, each with 1K flips: using Distributions n = 1000 d = Binomial(n, 0.5) to_wealth(heads) = 1.5^heads * 0.6^(n - heads) rand(d, 1_000_000) .|> to_wealth |> mean You can keep running that, it's above 1 almost all the time. To look at this another way — would you take the other side of the bet? Someone on average has to be making money, and the other side is clearly losing money.
- nonameiguess 6y ago
  I did run this many times by adapting the OP's own code (and vectorizing it, which he asked someone to do). Here it is: import numpy as np import pandas as pd n_subjects = 100 n_trials = 1000 start = 10.0 win = 1.5 loss = 0.6 prob = 0.5 results = np.ones((n_trials, n_subjects)) * start for trial in range(1, n_trials): wins = np.random.binomial(1, prob, n_subjects) == 1 results[trial, wins] = results[trial - 1, wins] * win results[trial, ~wins] = results[trial - 1, ~wins] * loss ax = pd.DataFrame(results).plot(legend=False, figsize=(18,10), logy=True, linewidth=0.5) ax.plot(results.mean(axis=1), color='red', lw=2, linestyle='--') The mean always trends to 0 and every single player eventually loses. There are never any winners at all past around 500 trials or so. Not sure how you're getting a different result as I have never used Julia and can't tell what your code is doing (except apparently something different).
  - pcmonk 6y ago
    I think your n_subjects is too low. You need that to be high enough or you'll miss those low-probability winners that bring up the average.
    - nonameiguess 6y ago
      I did it with more subjects and it doesn't make a difference. The only reason I reduced to 100 is because the plot is unreadable otherwise. Looking at the Julia code, I think what he is doing wrong is making all wins worth $.50 and all losses worth $.40, but the bet computes a win or loss based on your current wealth, not your starting wealth. His formula would work if you were always betting $1 no matter what your bankroll was, but that isn't what the actual post stipulates.
tryptophan 6y ago
OOf, yeah, expected value is a very misleading metric. Hard for me to wrap my head around it. After 2 tosses, the probable outcomes are 25% 2.25, 50% .9, and 25% .36, giving an expected value of 1.1025 interestingly enough. Overall a 75% chance of losing money.
jsweojtj 6y ago
The ensemble average is the expected value, and the expected value is positive. For a bet of $X, the expected value is: (1.5 * X) * 0.5 + (0.6 * X) * 0.5 => 1.05 * X. The ensemble average per round is positive (1.05) and over multiple rounds smoothly tends to infinity with the number of bets. (Definition here: https://en.wikipedia.org/wiki/Expected_value https://en.wikipedia.org/wiki/Expected_value). The time average for any specific person betting in this game is 0.95 * X (for the reasons you mention) and tends to zero with the number of bets. So let's go through a few specifics of your comment: > He actually picked bad numbers. That is a losing bet even on average. The point of this article is that "on average" is trickier than people tend to assume. There are different ways of taking averages. If you do the expected value calculation and get a positive number, you might (as other comments have said explicitly) expect that a participant repeatedly engaging such a bet would have his wealth trend toward infinity. But, they are wrong (as shown in the article). > This happens to not show after only 100 trials just because some tiny number of people get really lucky and draw up the ensemble average, but if you keep going, somewhere between 200 and 500 trials, the ensemble average pretty quickly drops below the starting average wealth and stays there, asymptotically approaching 0. The ensemble average is positive and monotonically increases w/ the number of rounds of betting.
- nonameiguess 6y ago
  I'm getting too into the weeds below with these simulations, but I think it maybe speaks to this author's point. I can see you are clearly right analytically. With a positive expectation on each round, the expectation itself clearly trends to infinity. But I can also clearly see that never actually happens, and I do think I can explain it. The measure of people with any positive expected return at all after a large enough number of trials is so small that they eventually drop out of any actual simulation just because nobody ever gets that lucky, even though theoretically it is possible. You eventually reach a point after a large enough of trials (apparently about 500) where if anybody at all was actually hitting the 400+ heads out of 500 trials requires to still be above water, enough of them would be so fabulously wealthy that they'd draw up the entire average. But the probability of these trials ever happening is so low that we can run simulations for thousands of years and never see it happen, so what we see instead is nothing but common cases. Everyone after 500 trials is overwhelmingly within +/- 50 of 250 heads and 250 tails, and if you're in that range, you're a loser. It also may be the case that the way these pseudorandom number generators work makes it completely impossible to ever see 400 heads out of 500 trials no matter how many simulations you run since they aren't actually random, but even if they were, practically speaking, I wouldn't be the least bit surprised if an any trial ever run of 500 consecutive fair coin tosses, nobody for as long as humans have existed has ever hit 400+ heads. If you try to analytically compute the probability and store it in a floating point number, it just rounds to 0 because we can't store a probability that low.
jerry80 6y ago
Yeah. OP might be conflating arithmetic mean with geometric mean. This is a common problem that many investors make. I see it all the time. People mistakenly think that if their investment made 50% and then lost 50% that they broken even, but they're actually down 25% (1.5*0.5=0.75). However, if you invest $100 and make 50%, then invest a second $100 and lose 50%, you do indeed break even. When OP stated that the bet has a positive EV, it's for a flat $1 bet. Indeed, if you always bet $1 (or any fixed amount), it does have a positive return of $+0.05, and you should take the bet. It's only when you change it from a $1 bet amount to an "entire bankroll" amount that you're looking at a geometric mean of 0.949 return per bet. That number is simply the geometric mean of the two possible returns, 1.5 and 0.6. So the actual EV is sqrt(1.5*0.6)-1.00 = $-0.051 per bet (normalized to $1.00). I don't understand the point this article is trying to make about anything else. The entire effect here is explained either by misstating the problem or by using the wrong type of mean for the EV calculation. The idea that some lucky people will make money while the rest lose is explained by simple luck. Run the simulation longer and they will all lose out to the law of large numbers.

alkonaut 6y ago

It’s just a really weird way of expressing the bet. This bet either has a positive EV or a negative one and if it’s positive it’s going to tend to infinity when repeated. There is one assymmetry at the zero point (assuming people can’t recover from bankruptcy by borrowing another dollar) but that’s easily fixed by adding a simple bet strategy e.g “bet at most 1/10 your bankroll on each bet”.

jsweojtj 6y ago
> This bet either has a positive EV or a negative one and if it’s positive it’s going to tend to infinity when repeated. This is wrong. The bet as described has a positive EV and the time average for a single player tends to zero as the bet is repeated. > There is one assymmetry at the zero point (assuming people can’t recover from bankruptcy by borrowing another dollar) ... The result is not due to zero being an absorbing value. In the setup you can go arbitrarily small and come back without issue. The result is the same.
- alkonaut 6y ago
  For a 50% chance you need odds of over 2.0 (regardless of bet size) to have positive EV. This is a bet of 0.4. With 2.0 odds the payout would be 0.8, and here it is 0.9? That looks like odds over 2.0 for 50/50 chance. So the individual bet is positive EV but the EV of an infinite sequence of such bets is still zero and that’s without the zero limit? There is just nothing in the article that begins to explain how that works. Maybe the video is better
  - alkonaut 6y ago
    If I just stay away from the 0 point (by betting a fraction of the bankroll) then regardless of sequence length the ensemble stats positive. I must simply not be understanding the rules of the simulation. https://dotnetfiddle.net/5r4ARk https://dotnetfiddle.net/5r4ARk

maximilianroos 6y ago

This is wrong and misses the point of the post. The median outcome is indeed negative, for the reason you give. But the mean outcome is positive, because some players become exceedingly rich. You can try it at home, here's some Julia which runs it over 1M people, each with 1K flips: using Distributions n = 1000 d = Binomial(n, 0.5) to_wealth(heads) = 1.5^heads * 0.6^(n - heads) rand(d, 1_000_000) .|> to_wealth |> mean You can keep running that, it's above 1 almost all the time. To look at this another way — would you take the other side of the bet? Someone on average has to be making money, and the other side is clearly losing money.

nonameiguess 6y ago
I did run this many times by adapting the OP's own code (and vectorizing it, which he asked someone to do). Here it is: import numpy as np import pandas as pd n_subjects = 100 n_trials = 1000 start = 10.0 win = 1.5 loss = 0.6 prob = 0.5 results = np.ones((n_trials, n_subjects)) * start for trial in range(1, n_trials): wins = np.random.binomial(1, prob, n_subjects) == 1 results[trial, wins] = results[trial - 1, wins] * win results[trial, ~wins] = results[trial - 1, ~wins] * loss ax = pd.DataFrame(results).plot(legend=False, figsize=(18,10), logy=True, linewidth=0.5) ax.plot(results.mean(axis=1), color='red', lw=2, linestyle='--') The mean always trends to 0 and every single player eventually loses. There are never any winners at all past around 500 trials or so. Not sure how you're getting a different result as I have never used Julia and can't tell what your code is doing (except apparently something different).
- pcmonk 6y ago
  I think your n_subjects is too low. You need that to be high enough or you'll miss those low-probability winners that bring up the average.
  - nonameiguess 6y ago
    I did it with more subjects and it doesn't make a difference. The only reason I reduced to 100 is because the plot is unreadable otherwise. Looking at the Julia code, I think what he is doing wrong is making all wins worth $.50 and all losses worth $.40, but the bet computes a win or loss based on your current wealth, not your starting wealth. His formula would work if you were always betting $1 no matter what your bankroll was, but that isn't what the actual post stipulates.

tryptophan 6y ago

OOf, yeah, expected value is a very misleading metric. Hard for me to wrap my head around it. After 2 tosses, the probable outcomes are 25% 2.25, 50% .9, and 25% .36, giving an expected value of 1.1025 interestingly enough. Overall a 75% chance of losing money.

jsweojtj 6y ago

The ensemble average is the expected value, and the expected value is positive. For a bet of $X, the expected value is: (1.5 * X) * 0.5 + (0.6 * X) * 0.5 => 1.05 * X. The ensemble average per round is positive (1.05) and over multiple rounds smoothly tends to infinity with the number of bets. (Definition here: https://en.wikipedia.org/wiki/Expected_value https://en.wikipedia.org/wiki/Expected_value). The time average for any specific person betting in this game is 0.95 * X (for the reasons you mention) and tends to zero with the number of bets. So let's go through a few specifics of your comment: > He actually picked bad numbers. That is a losing bet even on average. The point of this article is that "on average" is trickier than people tend to assume. There are different ways of taking averages. If you do the expected value calculation and get a positive number, you might (as other comments have said explicitly) expect that a participant repeatedly engaging such a bet would have his wealth trend toward infinity. But, they are wrong (as shown in the article). > This happens to not show after only 100 trials just because some tiny number of people get really lucky and draw up the ensemble average, but if you keep going, somewhere between 200 and 500 trials, the ensemble average pretty quickly drops below the starting average wealth and stays there, asymptotically approaching 0. The ensemble average is positive and monotonically increases w/ the number of rounds of betting.

nonameiguess 6y ago
I'm getting too into the weeds below with these simulations, but I think it maybe speaks to this author's point. I can see you are clearly right analytically. With a positive expectation on each round, the expectation itself clearly trends to infinity. But I can also clearly see that never actually happens, and I do think I can explain it. The measure of people with any positive expected return at all after a large enough number of trials is so small that they eventually drop out of any actual simulation just because nobody ever gets that lucky, even though theoretically it is possible. You eventually reach a point after a large enough of trials (apparently about 500) where if anybody at all was actually hitting the 400+ heads out of 500 trials requires to still be above water, enough of them would be so fabulously wealthy that they'd draw up the entire average. But the probability of these trials ever happening is so low that we can run simulations for thousands of years and never see it happen, so what we see instead is nothing but common cases. Everyone after 500 trials is overwhelmingly within +/- 50 of 250 heads and 250 tails, and if you're in that range, you're a loser. It also may be the case that the way these pseudorandom number generators work makes it completely impossible to ever see 400 heads out of 500 trials no matter how many simulations you run since they aren't actually random, but even if they were, practically speaking, I wouldn't be the least bit surprised if an any trial ever run of 500 consecutive fair coin tosses, nobody for as long as humans have existed has ever hit 400+ heads. If you try to analytically compute the probability and store it in a floating point number, it just rounds to 0 because we can't store a probability that low.

jerry80 6y ago

Yeah. OP might be conflating arithmetic mean with geometric mean. This is a common problem that many investors make. I see it all the time. People mistakenly think that if their investment made 50% and then lost 50% that they broken even, but they're actually down 25% (1.5*0.5=0.75). However, if you invest $100 and make 50%, then invest a second $100 and lose 50%, you do indeed break even. When OP stated that the bet has a positive EV, it's for a flat $1 bet. Indeed, if you always bet $1 (or any fixed amount), it does have a positive return of $+0.05, and you should take the bet. It's only when you change it from a $1 bet amount to an "entire bankroll" amount that you're looking at a geometric mean of 0.949 return per bet. That number is simply the geometric mean of the two possible returns, 1.5 and 0.6. So the actual EV is sqrt(1.5*0.6)-1.00 = $-0.051 per bet (normalized to $1.00). I don't understand the point this article is trying to make about anything else. The entire effect here is explained either by misstating the problem or by using the wrong type of mean for the EV calculation. The idea that some lucky people will make money while the rest lose is explained by simple luck. Run the simulation longer and they will all lose out to the law of large numbers.