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I think you're misinterpreting the author's point about roulette. She never claimed that you could profit off of roulette in the long run. You can, however, a

by splat 17y ago

I think you're misinterpreting the author's point about roulette. She never claimed that you could profit off of roulette in the long run. You can, however, almost certainly profit off of it in the short run. The strategy she describes has one double one's bet if one loses. If you bring enough money to cover 7 losses (which means you would have to have roughly 100 times your initial bet), then you have a >98% chance of coming out ahead. You won't be ahead by much, and the strategy won't work in the long run (i.e. you can only do it a couple of times--when you inevitably end up with 7 consecutive losses it will more than wipe out any earnings you make), but it is nevertheless a legitimate strategy to "beat" roulette in the short term.

jvdh 17y ago
True, I did misinterpret that. And she does own up to it that after 7 streaks you'd need 1280 pounds if you initially start with 10. Still, the article states that there's a 50:50 chance of landing on black or red, which is simply not true.
- splat 17y ago
  Fair enough. The article also claims that the outcome of a coin flip is 50:50, which is also not true.
  - req2 17y ago
    You're mixing theoretical odds and practical odds, which might be considered sloppy. Additionally, unfair coins can trivially be used as fair coins, but unbalanced roulette wheels cannot easily be used as fair roulette wheels.
    - splat 17y ago
      A fair coin has a roughly 51% chance of landing with the same side up as it started with. See the paper by Diaconis et al. "Dynamical Bias in the Coin Toss."
      - req2 17y ago
        A fair coin is one on which p = 0.5. A coin that has a 51% chance is clearly not a fair coin. This isn't even what I was talking about, but is still equally easy to turn into a fair coin. In this case, caller calls in the air after flipper hides the side that starts up. 0.5 * 0.51 + 0.5 * 0.49 = 0.5 For any other case, flip a coin twice. Assign one person HT and the other person TH, reflipping TT and HH. Since p * (1 - p) = (1 - p) * p, the dual flip is a fair flip. Please don't mention the possibility of sleight of hand next.
bigwill 17y ago
I hear what you're both saying. But I wouldn't call beating roulette in the short term a legitimate strategy exactly because it only works in the short run--strategies are supposed to work for longer than that! The only legitimate value I can see is that if you actually enjoy the experience of playing roulette that it's worth paying for it. Anyone who's willing to buy volatility can win in the short term. Otherwise, stick to a game of skill and get some money on the table while you do it. If you're good you can get some vol. and some positive EV.
Cantdog 17y ago
Splat, Would you pick out of this hat? -- There are a hundred tickets in a hat. 99 of the tickets win you $1. The last ticket loses $10,000 dollars. the sentence "it is nevertheless a legitamte strategy to "beat" roulette in the short term" is absolutely false. The expected value of each roll of the dice is negative, summing a bunch of negative expectations leads to an even large negative expectation. EDIT: Also, some people will talk about infinite bankrolls, there is no such thing! even if there was, infinity + X = inifinity, why are you gambling?
- splat 17y ago
  I would pick out of the hat once, maybe even twice. The point is that I'm never going to sum over a bunch of negative expectations. I beat the probability of losing down to something negligible, play once, and then quit forever.
  - Cantdog 17y ago
    There is a difference between being willing to gamble and beating a game. It is obvious to see that each time you play this game you are losing $9901 dollars (.99x1+.01x-10000). That is the expected outcome each time you play the game. Probability is often not intuitive. It is easy to think you "beat" the system because you won a couple bucks. This game may better illustrate my point. There is a hat with 100 tickets. 99 of the tickets produce a $1 win. The last ticket requires you kill yourself. Would you play this game once or twice?
    - splat 17y ago
      I'll agree with your semantics. To me, "beating a game" implies that you can consistently come out ahead over a long period of time. Nevertheless, playing an "unbeatable" game over a short period of time is not necessarily irrational. Although your expected value in your first version of the hat game is -$9901, you are not losing $9901 every time you play. If you play long enough it will average out to be this, but every time you play you will either win $1 or lose $10,000. If the chance of a devastating loss is negligible, it's hardly unwise to make a gamble, even if it has a negative expected value--this is why we occasionally do risky things that have a small chance of killing us, like driving or eating sushi. The problem in thinking in terms of an expected value is that an expected value presupposes many trials--given a single experiment, it's a meaningless concept. Indeed, given a single experiment, both the classical and frequency interpretations of probability are meaningless with regards to what the probability of a given outcome really is. In the case of a single experiment, the only interpretation that makes sense is the Bayesian interpretation. I would not play your second game, but the reason I wouldn't has nothing to do with expected values (notwithstanding the fact that it's technically impossible to calculate an expected value for an unquantifiable outcome like death). In your first game, I feel that I would come out ahead, and in the second game I feel like it's not worth it.
deleted 17y ago
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by splat 17y ago

jvdh 17y ago

True, I did misinterpret that. And she does own up to it that after 7 streaks you'd need 1280 pounds if you initially start with 10. Still, the article states that there's a 50:50 chance of landing on black or red, which is simply not true.

splat 17y ago
Fair enough. The article also claims that the outcome of a coin flip is 50:50, which is also not true.
- req2 17y ago
  You're mixing theoretical odds and practical odds, which might be considered sloppy. Additionally, unfair coins can trivially be used as fair coins, but unbalanced roulette wheels cannot easily be used as fair roulette wheels.
  - splat 17y ago
    A fair coin has a roughly 51% chance of landing with the same side up as it started with. See the paper by Diaconis et al. "Dynamical Bias in the Coin Toss."
    - req2 17y ago
      A fair coin is one on which p = 0.5. A coin that has a 51% chance is clearly not a fair coin. This isn't even what I was talking about, but is still equally easy to turn into a fair coin. In this case, caller calls in the air after flipper hides the side that starts up. 0.5 * 0.51 + 0.5 * 0.49 = 0.5 For any other case, flip a coin twice. Assign one person HT and the other person TH, reflipping TT and HH. Since p * (1 - p) = (1 - p) * p, the dual flip is a fair flip. Please don't mention the possibility of sleight of hand next.

bigwill 17y ago

I hear what you're both saying. But I wouldn't call beating roulette in the short term a legitimate strategy exactly because it only works in the short run--strategies are supposed to work for longer than that! The only legitimate value I can see is that if you actually enjoy the experience of playing roulette that it's worth paying for it. Anyone who's willing to buy volatility can win in the short term. Otherwise, stick to a game of skill and get some money on the table while you do it. If you're good you can get some vol. and some positive EV.

Cantdog 17y ago

Splat, Would you pick out of this hat? -- There are a hundred tickets in a hat. 99 of the tickets win you $1. The last ticket loses $10,000 dollars. the sentence "it is nevertheless a legitamte strategy to "beat" roulette in the short term" is absolutely false. The expected value of each roll of the dice is negative, summing a bunch of negative expectations leads to an even large negative expectation. EDIT: Also, some people will talk about infinite bankrolls, there is no such thing! even if there was, infinity + X = inifinity, why are you gambling?

splat 17y ago
I would pick out of the hat once, maybe even twice. The point is that I'm never going to sum over a bunch of negative expectations. I beat the probability of losing down to something negligible, play once, and then quit forever.
- Cantdog 17y ago
  There is a difference between being willing to gamble and beating a game. It is obvious to see that each time you play this game you are losing $9901 dollars (.99x1+.01x-10000). That is the expected outcome each time you play the game. Probability is often not intuitive. It is easy to think you "beat" the system because you won a couple bucks. This game may better illustrate my point. There is a hat with 100 tickets. 99 of the tickets produce a $1 win. The last ticket requires you kill yourself. Would you play this game once or twice?
  - splat 17y ago
    I'll agree with your semantics. To me, "beating a game" implies that you can consistently come out ahead over a long period of time. Nevertheless, playing an "unbeatable" game over a short period of time is not necessarily irrational. Although your expected value in your first version of the hat game is -$9901, you are not losing $9901 every time you play. If you play long enough it will average out to be this, but every time you play you will either win $1 or lose $10,000. If the chance of a devastating loss is negligible, it's hardly unwise to make a gamble, even if it has a negative expected value--this is why we occasionally do risky things that have a small chance of killing us, like driving or eating sushi. The problem in thinking in terms of an expected value is that an expected value presupposes many trials--given a single experiment, it's a meaningless concept. Indeed, given a single experiment, both the classical and frequency interpretations of probability are meaningless with regards to what the probability of a given outcome really is. In the case of a single experiment, the only interpretation that makes sense is the Bayesian interpretation. I would not play your second game, but the reason I wouldn't has nothing to do with expected values (notwithstanding the fact that it's technically impossible to calculate an expected value for an unquantifiable outcome like death). In your first game, I feel that I would come out ahead, and in the second game I feel like it's not worth it.

deleted 17y ago

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