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Hi. Author here. It is shown here on different in the browser executable and editable programs with Monte Carlo methods that one cannot win at roulette if one p

by chkas 3y ago

Hi. Author here. It is shown here on different in the browser executable and editable programs with Monte Carlo methods that one cannot win at roulette if one plays many times. Many know this, but many do not, and therefore this can be helpful. After all, there are more than a few who lose their financial existence in the casino.

iamcreasy 3y ago
Thank you for writing. Very clearly explained. Can you explain how do you derive this? print 10 * 10000 * (18 / 37 - 19 / 37)
- alexeldeib 3y ago
  Per round expected value of +/- 10, times 10k rounds, 18/37 lose probability 19/37 win probability.
- Jtsummers 3y ago
  That's an expected value calculation. It breaks down into three parts: 1. The wager per round ($10). This is, in the game being simulated, the amount won or lost based on the game's outcome. 2. The number of rounds (10000). 3. The probability of winning minus probability of losing. Forgetting the number of rounds for a moment, the way to calculate the expected value of a single wager like this is: E[X] = x1*p1 + x2*p2 + ... + xn*pn Where each `xN` is the value (amount won or amount lost since we're talking about wagers) of an event and `pN` is the probability of that event. In this case there is an 18/37 chance of winning and 19/37 chance of losing. The value of winning is +10 and the cost of losing is -10. So the expected value of a $10 wager ends up being: 10 * 18/37 - 10 * 19/37 Multiply that by 10,000 for the number of rounds played and you get the expected value of a series of games.
  - sokoloff 3y ago
    Note that's the calculation for a European (single green 0) wheel, not the American (green 0 and green 00) wheel, in which you'd replace with +18/38 (or 9/19) and -20/38 (or 10/19). Yes, the American version is twice as bad for the player, but it doesn't matter; people are still plenty eager to play it...
  - listenallyall 3y ago
    I'll note that if you can deterministically calculate an EV, a Monte Carlo simulation isn't necessary. The fact that the EV of roulette is so easy to calculate defeats the purpose of this exercise. Perhaps the MC can help you conceptualize volatility and distribution of results, but once you know that the EV is negative, there's no good (mathematical) reason to participate.
    - User23 3y ago
      > but once you know that the EV is negative, there's no good (mathematical) reason to participate. Unless of course you know that the casino is willing to pay you out as part of their marketing budget (based on your ev to them); usually in the form of compensated room, food, and beverage, and sometimes even transportation. At that point your choice to participate becomes vacation planning for the mathematically literate.
      - listenallyall 3y ago
        That has nothing to do with the example presented in this article.
    - chkas 3y ago
      It is also about the "law of large numbers". If you go to the casino with $1000 and want to go home with $2000, the chance is more than 48% if you bet the $1000 on red. But if you play often and always bet $10 on red until you have $2000, your chance is less than 1%.
      - listenallyall 3y ago
        ...which is why I stated "the MC can help you conceptualize distribution of results", however again this doesn't really require an MC simulation. To achieve $2000 while betting $1000 requires you to win one bet more than you lose. To get there while betting $10 requires you to win 100 more. Common sense should tell you which is significantly easier to achieve.
guhcampos 3y ago
The last method, where you always double the bet (called Martingale) "works" and can guarantee you are at least even - if you have infinite time and money.
- maxbond 3y ago
  And if you're allowed to make bets of an infinite size, which casinos restrict in part to defeat this strategy (as well as manage their risk).
  - wan23 3y ago
    Though even if you are allowed to make infinite bets, most people have finite money to bet with.
- Nursie 3y ago
  Thanks for the name "Martingale", I had wondered what this was called since an old gambler I met on a contract about 10 years ago described 'his' system. After a quick wikipedia rabbithole, I now grok the St Petersburg paradox, which is fascinating - https://en.wikipedia.org/wiki/St._Petersburg_paradox https://en.wikipedia.org/wiki/St._Petersburg_paradox - but mostly tells me that the 'expected value' is not a good measure of whether you should bet on something, because games can be constructed in such a way that the expected value is infinite, but the chances of you even getting your stake back even at a $16 buy in are 1/8...
  - beaugunderson 3y ago
    the St. Petersburg paradox also reminds me of Pascal's mugging: https://en.wikipedia.org/wiki/Pascal's_mugging https://en.wikipedia.org/wiki/Pascal's_mugging
    - Nursie 3y ago
      So much so that I think they are likely to be different narratives around more or less the same phenomenon!

by chkas 3y ago

iamcreasy 3y ago

Thank you for writing. Very clearly explained. Can you explain how do you derive this? print 10 * 10000 * (18 / 37 - 19 / 37)

alexeldeib 3y ago
Per round expected value of +/- 10, times 10k rounds, 18/37 lose probability 19/37 win probability.
Jtsummers 3y ago
That's an expected value calculation. It breaks down into three parts: 1. The wager per round ($10). This is, in the game being simulated, the amount won or lost based on the game's outcome. 2. The number of rounds (10000). 3. The probability of winning minus probability of losing. Forgetting the number of rounds for a moment, the way to calculate the expected value of a single wager like this is: E[X] = x1*p1 + x2*p2 + ... + xn*pn Where each `xN` is the value (amount won or amount lost since we're talking about wagers) of an event and `pN` is the probability of that event. In this case there is an 18/37 chance of winning and 19/37 chance of losing. The value of winning is +10 and the cost of losing is -10. So the expected value of a $10 wager ends up being: 10 * 18/37 - 10 * 19/37 Multiply that by 10,000 for the number of rounds played and you get the expected value of a series of games.
- sokoloff 3y ago
  Note that's the calculation for a European (single green 0) wheel, not the American (green 0 and green 00) wheel, in which you'd replace with +18/38 (or 9/19) and -20/38 (or 10/19). Yes, the American version is twice as bad for the player, but it doesn't matter; people are still plenty eager to play it...
- listenallyall 3y ago
  I'll note that if you can deterministically calculate an EV, a Monte Carlo simulation isn't necessary. The fact that the EV of roulette is so easy to calculate defeats the purpose of this exercise. Perhaps the MC can help you conceptualize volatility and distribution of results, but once you know that the EV is negative, there's no good (mathematical) reason to participate.
  - User23 3y ago
    > but once you know that the EV is negative, there's no good (mathematical) reason to participate. Unless of course you know that the casino is willing to pay you out as part of their marketing budget (based on your ev to them); usually in the form of compensated room, food, and beverage, and sometimes even transportation. At that point your choice to participate becomes vacation planning for the mathematically literate.
    - listenallyall 3y ago
      That has nothing to do with the example presented in this article.
  - chkas 3y ago
    It is also about the "law of large numbers". If you go to the casino with $1000 and want to go home with $2000, the chance is more than 48% if you bet the $1000 on red. But if you play often and always bet $10 on red until you have $2000, your chance is less than 1%.
    - listenallyall 3y ago
      ...which is why I stated "the MC can help you conceptualize distribution of results", however again this doesn't really require an MC simulation. To achieve $2000 while betting $1000 requires you to win one bet more than you lose. To get there while betting $10 requires you to win 100 more. Common sense should tell you which is significantly easier to achieve.

guhcampos 3y ago

The last method, where you always double the bet (called Martingale) "works" and can guarantee you are at least even - if you have infinite time and money.

maxbond 3y ago
And if you're allowed to make bets of an infinite size, which casinos restrict in part to defeat this strategy (as well as manage their risk).
- wan23 3y ago
  Though even if you are allowed to make infinite bets, most people have finite money to bet with.
Nursie 3y ago
Thanks for the name "Martingale", I had wondered what this was called since an old gambler I met on a contract about 10 years ago described 'his' system. After a quick wikipedia rabbithole, I now grok the St Petersburg paradox, which is fascinating - https://en.wikipedia.org/wiki/St._Petersburg_paradox https://en.wikipedia.org/wiki/St._Petersburg_paradox - but mostly tells me that the 'expected value' is not a good measure of whether you should bet on something, because games can be constructed in such a way that the expected value is infinite, but the chances of you even getting your stake back even at a $16 buy in are 1/8...
- beaugunderson 3y ago
  the St. Petersburg paradox also reminds me of Pascal's mugging: https://en.wikipedia.org/wiki/Pascal's_mugging https://en.wikipedia.org/wiki/Pascal's_mugging
  - Nursie 3y ago
    So much so that I think they are likely to be different narratives around more or less the same phenomenon!