5 ms·

The post doesn't really shed much light on it, it feels like a missed opportunity to explain things well. There's two main effects: - Negative expected value

by jstx1 5y ago

The post doesn't really shed much light on it, it feels like a missed opportunity to explain things well. There's two main effects: - Negative expected value (EV)[1]. The games are set up in a way that on average you're likely to lose money. This is true for every game where you play against the house, otherwise the game wouldn't be there. Poker is different because your EV depends on how you play relative to other people. - Bankroll management and risk of ruin[2]. Even in a fair game, if you start with a short bankroll, you're likely to go broke. For example, let's say two of us flip a completely fair coin and bet $1 on each flip. The expected value for both players is $0. Now if one of us starts with $5 and the other $1000 and we play repeatedly, the person with the smaller amount is almost guaranteed to go broke in a long series of flips. This what happens when you play against the house and it's the main reason why people lose money when they gamble in casinos. [1] https://en.wikipedia.org/wiki/Expected_value https://en.wikipedia.org/wiki/Expected_value [2] https://en.wikipedia.org/wiki/Risk_of_ruin https://en.wikipedia.org/wiki/Risk_of_ruin

confidantlake 5y ago
[Edit] As has pointed out I have misread the parent comment and agree with it. Leaving my original response below. For number 2 it depends on strategy. The strategy you outlined has you risk $5 to win $1000. Of course in a fair game you will go broke most of the time. For every time you win you should lose 200 times to come out even. But you can easily take the reverse strategy. Bet $5 if you win, leave. If you lose double your bet, if you go down $1000 dollars leave. In this case you are almost guaranteed to win. But the times you do lose you lose enough to make up for all of your wins. EV is EV, it doesn't change based on who has the larger bankroll. The casino doesn't magically get an edge because they have more money. They get the edge because the game odds are in their favor.
- deleted 5y ago
  [deleted]
- kqr 5y ago
  Well... that's bordering on oversimplifying it. The key distinction is whether you count the EV of a single wager, or the expected growth of compound returns. As you say, the single-wager EV is independent of bankroll provided you can afford it in the first place. This is also what most people mean when they say EV, and indeed the common mathematical definition of it. However, skilled risk takers know that the arithmetic EV isn't what matters more generally. What matters in the long run is geometric EV, or expected growth of compound returns. And for geometric EV, bankroll absolutely matters. The larger your bankroll, the better your geometric EV. I suspect this is what the parent comment referred to. (However, when the wager shrinks in comparison to your total wealth, the arithmetic EV approaches the geometric EV by Taylor series approximation. For "everyday affairs" (as I think Bernoulli put it) you can think of them as equal.)
  - confidantlake 5y ago
    I would have to disagree that I am oversimplifying it. We are talking about a casino vs a player. Most players betting $5 at a time are not losing because of bankroll considerations. They are losing because the game they are playing is negative ev for each individual bet. A casino would much rather play against a player with $1,000,000,000 than a player with $5. Sure the player with $5 is going to go broke nearly 100%. And the player with more money will never go broke betting $5 at a time, they will die from old age way before that happens. The fact that a player with only $5 has almost a 100% chance of ruin is absolutely not why a casino is making money. A casino is always going to want a player with a lower risk of ruin than a higher one. When a player goes broke they can no longer make money from them. Do people really think a casino would prefer someone going broke 100% of the time with $5 then a player going broke 0% of the time with $1,000,000,000? That they are making money because of players losing their bankrolls and unable to bet?
    - kqr 5y ago
      Aha, I see what you're saying. You're right, but I still think you're reading something into the top-level comment that isn't there. The way I understand it, they're not saying "casinos make money because their counterpart is going broke" (and you're right that they don't.) They're saying "in the casino vs. player situation, the player will almost always run out of bankroll before the casino does, and therefore the casino does not risk ruin, even in highly volatile games."
- stavros 5y ago
  I don't think you're talking about the same thing as the GP. If you bet $5 and lose, you can't double your bet, you have no money (which is exactly the GP's point). It's a random walk, and it's much more likely to take you to -$5 way before it takes you to +$1000.
  - confidantlake 5y ago
    Yes exactly, completely agree with you. I am seeing this does not matter. It is not why a casino makes money.
User23 5y ago
Powerball tickets have negative EV after taxes, but I’m willing to spend a grand total of about $20 a year on them by playing when the jackpot is north of half a billion. Because the $20 loss has absolutely no effect on my quality of life, but the win would.
- HWR_14 5y ago
  That's just measuring the EV in Utils as opposed to Dollars.
- fallingknife 5y ago
  Typically, when the jackpot is up in that range, the tickets actually have a positive EV.
  - wbc 5y ago
    Super unlikely, the main reason is at that point many people buy tickets and it becomes more and more likely the jackpot is split between multiple winners [0]. Actually the largest Powerball at $1.586 billion was split between 3 people [1] [0] https://chance.amstat.org/2020/02/lottery-ticket/jackpots-ever-get-big-enough-that-playing-has-a-positive-expected-value https://chance.amstat.org/2020/02/lottery-ticket/jackpots-ev... [1] https://apnews.com/article/archive-6d98eff765c9e2f4b683671311fe5b8b https://apnews.com/article/archive-6d98eff765c9e2f4b68367131...
    - fallingknife 5y ago
      Odds of winning are 1 in 292 million. So unless the odds were that the jackpot would be split 6 ways or more the tickets have a positive EV.
      - listenallyall 5y ago
        Except that Powerball tickets cost $2 each, not one dollar as you were assuming. (And of course, taxes)
chkas 5y ago
> Negative expected value Yes. Also briefly addressed in the post. The simulation already comes close to the calculated value. print 10 * 10000 * (18 / 37 - 19 / 37) > Bankroll management and risk of ruin No. The loss results purely from the EV (in roulette -0.027) . And that is my total bet * -0.027. No matter if Bezos is playing or I am.
- jstx1 5y ago
  You ignored the rest of the comment where I explained why EV isn't the only thing that matters. In a hypothetical neutral EV casino people will still lose money. > The loss results purely from the EV This is true for a single spin, it is not true for a sequence where once you run out of money, you stop playing.
  - chkas 5y ago
    I also read the rest. You wrote that this is the main reason people lose at the casino. This hypothetical casino does not exist, and if it did, it would eventually go bankrupt (costs aside) because the players have more money in total. > > The loss results purely from the EV > This is true for a single spin, it is not true for a sequence where once you run out of money, you stop playing. If you have to stop gambling because you lost your million dollars, you didn't lose because you don't have any more money. You couldn't turn it around with another 100 million either (in a non-hypothetical casino).

by jstx1 5y ago

confidantlake 5y ago

[Edit] As has pointed out I have misread the parent comment and agree with it. Leaving my original response below. For number 2 it depends on strategy. The strategy you outlined has you risk $5 to win $1000. Of course in a fair game you will go broke most of the time. For every time you win you should lose 200 times to come out even. But you can easily take the reverse strategy. Bet $5 if you win, leave. If you lose double your bet, if you go down $1000 dollars leave. In this case you are almost guaranteed to win. But the times you do lose you lose enough to make up for all of your wins. EV is EV, it doesn't change based on who has the larger bankroll. The casino doesn't magically get an edge because they have more money. They get the edge because the game odds are in their favor.

deleted 5y ago
[deleted]
kqr 5y ago
Well... that's bordering on oversimplifying it. The key distinction is whether you count the EV of a single wager, or the expected growth of compound returns. As you say, the single-wager EV is independent of bankroll provided you can afford it in the first place. This is also what most people mean when they say EV, and indeed the common mathematical definition of it. However, skilled risk takers know that the arithmetic EV isn't what matters more generally. What matters in the long run is geometric EV, or expected growth of compound returns. And for geometric EV, bankroll absolutely matters. The larger your bankroll, the better your geometric EV. I suspect this is what the parent comment referred to. (However, when the wager shrinks in comparison to your total wealth, the arithmetic EV approaches the geometric EV by Taylor series approximation. For "everyday affairs" (as I think Bernoulli put it) you can think of them as equal.)
- confidantlake 5y ago
  I would have to disagree that I am oversimplifying it. We are talking about a casino vs a player. Most players betting $5 at a time are not losing because of bankroll considerations. They are losing because the game they are playing is negative ev for each individual bet. A casino would much rather play against a player with $1,000,000,000 than a player with $5. Sure the player with $5 is going to go broke nearly 100%. And the player with more money will never go broke betting $5 at a time, they will die from old age way before that happens. The fact that a player with only $5 has almost a 100% chance of ruin is absolutely not why a casino is making money. A casino is always going to want a player with a lower risk of ruin than a higher one. When a player goes broke they can no longer make money from them. Do people really think a casino would prefer someone going broke 100% of the time with $5 then a player going broke 0% of the time with $1,000,000,000? That they are making money because of players losing their bankrolls and unable to bet?
  - kqr 5y ago
    Aha, I see what you're saying. You're right, but I still think you're reading something into the top-level comment that isn't there. The way I understand it, they're not saying "casinos make money because their counterpart is going broke" (and you're right that they don't.) They're saying "in the casino vs. player situation, the player will almost always run out of bankroll before the casino does, and therefore the casino does not risk ruin, even in highly volatile games."
stavros 5y ago
I don't think you're talking about the same thing as the GP. If you bet $5 and lose, you can't double your bet, you have no money (which is exactly the GP's point). It's a random walk, and it's much more likely to take you to -$5 way before it takes you to +$1000.
- confidantlake 5y ago
  Yes exactly, completely agree with you. I am seeing this does not matter. It is not why a casino makes money.

User23 5y ago

Powerball tickets have negative EV after taxes, but I’m willing to spend a grand total of about $20 a year on them by playing when the jackpot is north of half a billion. Because the $20 loss has absolutely no effect on my quality of life, but the win would.

HWR_14 5y ago
That's just measuring the EV in Utils as opposed to Dollars.
fallingknife 5y ago
Typically, when the jackpot is up in that range, the tickets actually have a positive EV.
- wbc 5y ago
  Super unlikely, the main reason is at that point many people buy tickets and it becomes more and more likely the jackpot is split between multiple winners [0]. Actually the largest Powerball at $1.586 billion was split between 3 people [1] [0] https://chance.amstat.org/2020/02/lottery-ticket/jackpots-ever-get-big-enough-that-playing-has-a-positive-expected-value https://chance.amstat.org/2020/02/lottery-ticket/jackpots-ev... [1] https://apnews.com/article/archive-6d98eff765c9e2f4b683671311fe5b8b https://apnews.com/article/archive-6d98eff765c9e2f4b68367131...
  - fallingknife 5y ago
    Odds of winning are 1 in 292 million. So unless the odds were that the jackpot would be split 6 ways or more the tickets have a positive EV.
    - listenallyall 5y ago
      Except that Powerball tickets cost $2 each, not one dollar as you were assuming. (And of course, taxes)

chkas 5y ago

> Negative expected value Yes. Also briefly addressed in the post. The simulation already comes close to the calculated value. print 10 * 10000 * (18 / 37 - 19 / 37) > Bankroll management and risk of ruin No. The loss results purely from the EV (in roulette -0.027) . And that is my total bet * -0.027. No matter if Bezos is playing or I am.

jstx1 5y ago
You ignored the rest of the comment where I explained why EV isn't the only thing that matters. In a hypothetical neutral EV casino people will still lose money. > The loss results purely from the EV This is true for a single spin, it is not true for a sequence where once you run out of money, you stop playing.
- chkas 5y ago
  I also read the rest. You wrote that this is the main reason people lose at the casino. This hypothetical casino does not exist, and if it did, it would eventually go bankrupt (costs aside) because the players have more money in total. > > The loss results purely from the EV > This is true for a single spin, it is not true for a sequence where once you run out of money, you stop playing. If you have to stop gambling because you lost your million dollars, you didn't lose because you don't have any more money. You couldn't turn it around with another 100 million either (in a non-hypothetical casino).