5 ms·

> 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 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
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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.
  - deleted 3y ago
    [deleted]
  - HelloNurse 3y ago
    In the long time limit, there are about the same number of heads and tails, and since all changes are multiplicative, the coin tosses can be permuted. We can thus divide the game into two sets of coin tosses: excess heads or tails that represent a negligible amount of good or bad luck, and many HT pairs, each of which shrinks EV by 0.9. Put another way, for each HH pair you should expect a TT pair, and a HHxTT or TTxHH sequence is worth the same as HTxHT, losing the usual 10% every two tosses. 2.250.36=0.81=0.90.9.
    - arc-in-space 3y ago
      Thanks, putting it this way does help me understand a bit better. So a large number of repeated fair tosses can be broken down into a sequence of win-loss pairs, which are always =0.9. The thing that still confuses me is, why the heck is the EV 1.05? It seems to be expressing something true - if you were to split your money into a thousand piles and "play" each individually, you make money overall.
      - sushibowl 3y ago
        The gamble is set up in such a way that there is a fairly large chance that you will lose money, coupled with a tiny chance of an incredibly large gain. The article, just like the poster above you, characterizes a series of bets into win/loss pairs that add up to a 0.9 return per pair. There are lots of sequences that can be characterized this way. However there will be a few sequences that contain many, many heads and win a lot of money. There are of course also a few sequences with many tails, but their loss cannot decrease below zero so it is contained. So it's a little bit like a lottery ticket, where the positive gains are extremely concentrated into a very small lucky group. The more rounds of the gamble you play, the smaller the lucky winning group gets, and the larger their wealth.
jacknews 3y 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.
- kevinwang 3y ago
  > Your stake should grow or shrink by the same factor if you win or lose. > If you lose, it should then be divided by 1.5 (multiplied by 0.66..). In what sense "should" this be true?
  - jacknews 3y ago
    So that a winning gain would be the 'opposite' of a loss, they are balanced. Your situation is enhanced by some factor if you win, and diminished by the same factor if you lose. Of course the actual amounts, percentages, proportions look bigger for the win compared to the loss, but that's just maths.

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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.
- deleted 3y ago
  [deleted]
- HelloNurse 3y ago
  In the long time limit, there are about the same number of heads and tails, and since all changes are multiplicative, the coin tosses can be permuted. We can thus divide the game into two sets of coin tosses: excess heads or tails that represent a negligible amount of good or bad luck, and many HT pairs, each of which shrinks EV by 0.9. Put another way, for each HH pair you should expect a TT pair, and a HHxTT or TTxHH sequence is worth the same as HTxHT, losing the usual 10% every two tosses. 2.250.36=0.81=0.90.9.
  - arc-in-space 3y ago
    Thanks, putting it this way does help me understand a bit better. So a large number of repeated fair tosses can be broken down into a sequence of win-loss pairs, which are always =0.9. The thing that still confuses me is, why the heck is the EV 1.05? It seems to be expressing something true - if you were to split your money into a thousand piles and "play" each individually, you make money overall.
    - sushibowl 3y ago
      The gamble is set up in such a way that there is a fairly large chance that you will lose money, coupled with a tiny chance of an incredibly large gain. The article, just like the poster above you, characterizes a series of bets into win/loss pairs that add up to a 0.9 return per pair. There are lots of sequences that can be characterized this way. However there will be a few sequences that contain many, many heads and win a lot of money. There are of course also a few sequences with many tails, but their loss cannot decrease below zero so it is contained. So it's a little bit like a lottery ticket, where the positive gains are extremely concentrated into a very small lucky group. The more rounds of the gamble you play, the smaller the lucky winning group gets, and the larger their wealth.

jacknews 3y 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.

kevinwang 3y ago
> Your stake should grow or shrink by the same factor if you win or lose. > If you lose, it should then be divided by 1.5 (multiplied by 0.66..). In what sense "should" this be true?
- jacknews 3y ago
  So that a winning gain would be the 'opposite' of a loss, they are balanced. Your situation is enhanced by some factor if you win, and diminished by the same factor if you lose. Of course the actual amounts, percentages, proportions look bigger for the win compared to the loss, but that's just maths.