5 ms·

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

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'.
        kevinwang 3y ago
        > 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. For what definition of "average"? Yes, the most likely outcome would be to break even. But if we mean "expected value" when we say average, then on average, that repeated wager would be massively profitable for us (in terms of wealth). Maybe a better way to view it is that it would be massively unprofitable for the casino offering it (this is more straightforward since the casino's outcomes are more ergodic than the individual's). --- Maybe a more clarifying discussion would be: what if you could accept these wagers (let's say 0.6x and 1.5x) if you can use a bet amount of your choice each time? That is, you don't have to wager your entire life savings, but you could choose to make the bet with $1, with the outcomes being $0.60 and $1.50. Then, this wager is clearly a free money machine, right? We can come to the conclusion that it is a free money machine by looking at its expected value. That's why I disagree that "you lose more than you gain" and I disagree that "this is purely a result of the distribution of returns from a single toss". --- I guess I should concede that I agree with your reasoning, if we make the reasonable assumption that utility is the logarithm of wealth. In that case, I agree that I would be indifferent to taking a 0.666 and 1.5 bet for my entire networth, and that I would not take a 0.6 and 1.5 bet for my entire networth. However, I still contend that we can analyze the value of these wagers using expected value -- we simply look at the expected value of what we actually care about (utility), not wealth.
        jacknews 3y ago
        I think the conclusion is that 'expected value' calculated this way is a bogus metric, even for a single toss (what is the 'average' of a single toss, it already makes no sense). You can't simply take the mean average of summed probability-outcome products, or at least, it does not mean 'expected value', it means the 'probable limit' of the total. In the $60/$150 bet the 'expected value' is $105, but no-one is getting $105. What is that number? It indicates a 'probable limit' to the total money in the system, but says nothing about what any individual should expect, or about the actual total size of the system, which will obviously tend towards the aggregate individual outcomes. And that is clearly exposed when you start repeating the bet. In fact 'expected value' is quite clearly 0.6 * 1.5 = 0.9, ie $90, or I think more generally: (outcome-1 * probability-1 * ... outcome-n * probability-n) / (probability-1 * ... probability-n) It's maybe counter-intuitive that the 'limit' of the system should keep growing, while the 'actual' size of the system tends toward the 'real' aggregate expected value, but I think it's much less counter-intuitive than what I think is claimed here, that repeating a bet turns it from being a 'good' bet to a 'bad' bet. It was always a bad bet, just measured wrong.
    - n2d4 3y ago
      > Over time it does not. It does! You take insane amounts of risks (that you want to avoid), but the expected value stays positive, even if you repeat it a million times. You might have a 99.9999999999...% chance of going broke, but you'll also have a 1/2^1mil chance of making an absurd amount of money; on average (and not median), you will come out ahead.
kevinwang 3y 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.
- deleted 3y ago
  [deleted]
- erik_seaberg 3y 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.
  - kevinwang 3y ago
    The comment I'm replying to says: > But this is purely a result of the distribution of returns from a single toss. Your examples contain two tosses.
    - erik_seaberg 3y ago
      You’re right, I missed the small probability of coming out ahead on a short series because losing dominates long series. Which is what the author is trying to get across to me. Thanks.
- throwaheyy 3y ago
  “current bankroll” is the issue here. 1 * 1.5 * 0.6 = 0.9 < 1.
  - rerdavies 3y ago
    Not getting it. There are FOUR possible outcomes in two consecutive rolls. HH: 2.25! HT: 0.9 TH: 0.9 TT: 0.36. Expectation: 4.41/4 = 1.1025. The losses due to HT/HT get more than offset by the massive gain from HH over TT. I can imagine there are arguments to be made about median expected value, and the effect on concentration of wealth. But whatever they are, they aren't being made.

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'.
      - kevinwang 3y ago
        > 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. For what definition of "average"? Yes, the most likely outcome would be to break even. But if we mean "expected value" when we say average, then on average, that repeated wager would be massively profitable for us (in terms of wealth). Maybe a better way to view it is that it would be massively unprofitable for the casino offering it (this is more straightforward since the casino's outcomes are more ergodic than the individual's). --- Maybe a more clarifying discussion would be: what if you could accept these wagers (let's say 0.6x and 1.5x) if you can use a bet amount of your choice each time? That is, you don't have to wager your entire life savings, but you could choose to make the bet with $1, with the outcomes being $0.60 and $1.50. Then, this wager is clearly a free money machine, right? We can come to the conclusion that it is a free money machine by looking at its expected value. That's why I disagree that "you lose more than you gain" and I disagree that "this is purely a result of the distribution of returns from a single toss". --- I guess I should concede that I agree with your reasoning, if we make the reasonable assumption that utility is the logarithm of wealth. In that case, I agree that I would be indifferent to taking a 0.666 and 1.5 bet for my entire networth, and that I would not take a 0.6 and 1.5 bet for my entire networth. However, I still contend that we can analyze the value of these wagers using expected value -- we simply look at the expected value of what we actually care about (utility), not wealth.
        jacknews 3y ago
        I think the conclusion is that 'expected value' calculated this way is a bogus metric, even for a single toss (what is the 'average' of a single toss, it already makes no sense). You can't simply take the mean average of summed probability-outcome products, or at least, it does not mean 'expected value', it means the 'probable limit' of the total. In the $60/$150 bet the 'expected value' is $105, but no-one is getting $105. What is that number? It indicates a 'probable limit' to the total money in the system, but says nothing about what any individual should expect, or about the actual total size of the system, which will obviously tend towards the aggregate individual outcomes. And that is clearly exposed when you start repeating the bet. In fact 'expected value' is quite clearly 0.6 * 1.5 = 0.9, ie $90, or I think more generally: (outcome-1 * probability-1 * ... outcome-n * probability-n) / (probability-1 * ... probability-n) It's maybe counter-intuitive that the 'limit' of the system should keep growing, while the 'actual' size of the system tends toward the 'real' aggregate expected value, but I think it's much less counter-intuitive than what I think is claimed here, that repeating a bet turns it from being a 'good' bet to a 'bad' bet. It was always a bad bet, just measured wrong.
  - n2d4 3y ago
    > Over time it does not. It does! You take insane amounts of risks (that you want to avoid), but the expected value stays positive, even if you repeat it a million times. You might have a 99.9999999999...% chance of going broke, but you'll also have a 1/2^1mil chance of making an absurd amount of money; on average (and not median), you will come out ahead.

kevinwang 3y 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.

deleted 3y ago
[deleted]
erik_seaberg 3y 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.
- kevinwang 3y ago
  The comment I'm replying to says: > But this is purely a result of the distribution of returns from a single toss. Your examples contain two tosses.
  - erik_seaberg 3y ago
    You’re right, I missed the small probability of coming out ahead on a short series because losing dominates long series. Which is what the author is trying to get across to me. Thanks.
throwaheyy 3y ago
“current bankroll” is the issue here. 1 * 1.5 * 0.6 = 0.9 < 1.
- rerdavies 3y ago
  Not getting it. There are FOUR possible outcomes in two consecutive rolls. HH: 2.25! HT: 0.9 TH: 0.9 TT: 0.36. Expectation: 4.41/4 = 1.1025. The losses due to HT/HT get more than offset by the massive gain from HH over TT. I can imagine there are arguments to be made about median expected value, and the effect on concentration of wealth. But whatever they are, they aren't being made.