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Probability and Roulette

  1. Aug 24, 2010 #1
    I was reading an introduction to probability, and it discussed the game roulette.

    The book recommended the strategy of always betting on a color or odd/even, which gives you 18/38 odds, and to double your bid after every loss. This way you will continue to gain--assuming that you had infinite money to gamble. :)

    However, in real roulette there are two constraints on this: 1. there is a maximum bid limit. 2. you don't have unlimited funds.

    Based on this, I was trying to figure out how to determine the:

    1. Optimum amount of money a person should begin with to maximize his chances.

    2. The best amount to wager per round (low amounts allows for longer losing strings, higher amount increases rate of growth).

    3. What is the optimum amount of profit to shoot for before walking away. (Because the more you play, the greater your chance of seeing a loosing streak that you can't afford.)

    I started the problem by assuming that I had about 1000 units to bet, so that I could double it 10 times. I calculated (perhaps incorrectly) that the chance of wining at least once in 10 rolls (ensuring that a profit, assuming that you start over after each win) is about 99.9%

    But that also means that in order to double my money, betting 1 unit a round, will take over 1000 rounds, which makes a 10 losing streak a significant chance (I think).

    How would I figure out the best balance of betting amount and the amount of profit that is optimal to call it quits, before the chances of a long losing streak becomes significant?

    (By the way, I'm not planning to really gamble or anything, I was just intrigued by the problem. I also have nothing beyond high school level mathematics training, although I was good at math.)

    To start with a simple example: Someone has $1,000 to bet. He can bet in denominations of 1, 5, 25, or 100. The maximum bid amount is 500. What is the biggest profit (largest winning string) that he can expect to see before seeing a losing string too long. And what denomination would he use to bid. Again, this is assuming that he doubles his bid after every loss.

    You could also simplify further, by assuming an even probability of 50/50 like a coin toss.

    I tried looking for equations for some of this, but got stuck and didn't understand the math.

    Last edited: Aug 24, 2010
  2. jcsd
  3. Aug 24, 2010 #2


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    That strategy is called a martengale, for what it's worth. In the real world it's somewhere between a bad idea and a disaster.

    With a finite amount of starting money, no ability to borrow, no maximum bet, and 18/38 odds at even money, it's a losing proposition to make any bets.

    What you can do is shuffle the probabilities around. If you decide that you want to make just $1 with a martengale, starting with $1023 (minimum bid: $1) you have very good odds: 1 - (20/38)^9 ≈ 99.83%. But the remaining 0.17% of the time you lose $1023. Expected gain: -$0.67.

    So if there's a gangster who's going to shoot you unless you pay him the $1024 you owe him, this can make sense: you don't mind the 0.17% chance that you lose $1023, since the $1023 is worthless to you without that last dollar. But in most situations it's not a good gamble. (Of course if a person enjoys being at a casino, the expected 'losses' may be a fair price for the entertainment.)
    Last edited: Aug 24, 2010
  4. Aug 24, 2010 #3


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    If the odds are 50/50, this changes things. You don't gain or lose money on average. In the example of $1023 above, you have about 99.9% odds of gaining $1 and about 0.1% odds of losing $1023, with expected winnings of $0.
  5. Aug 24, 2010 #4
    You don't need an equation, you need a system. I do gamble, blackjack and craps, and casinos have the law of large numbers in their favor. Even if it were a true 50/50 proposition they will always win.

    Roulette is a bad game, period. With craps you can get the odds to 49.99% against the casino, if you know how to bet. And if you know how to play blackjack you can get the odds to even.

    To really put it in their favor they do two things: table limits and minimums. They use minimums in several ways, but the biggest thing is that it forces a player to have a higher bankroll.

    A hint on blackjack, that chart in all the books and sell in the casino is wrong. Always split 8's.. how about rarely. And never split facecards, how about occasionally split them. Using those two adjustments, at the right time, and you will increase your odds in blackjack.

    And when the pit boss and dealer say they don't care, it is not thier money. they are lying. The majic number that gets them nervous is about 2,500. I spent a winter in Vegas when working on a contract in San Jose. ( it was cheaper to rent a hotel room in vegas, and commute by plane to San Jose every day than it was to actually stay in San Jose ). Besides that, finding a hotel room anywhere near my client was impossible from week to week.

    Sorry, my posting was more about gambling than the actual question, but what the heck
  6. Sep 5, 2010 #5
    I've actually tried this. And from experience I can tell you it's the table maximums that make this a horrible strategy. I bet black, doubling my bet every time I lost, everything was going swell until red/green came up ~8 times in a row. The bet I needed to place was over the maximum and the system crumbled. Fortunately this all took place over in the Philippines so it was a cheaper lesson than I would have received from any casino in the US.

    Roullette is a fun game though, I went to an indian casino in Cali 6 days in a row specifically to play it; the first five days I walked out with over $500 each time, but the sixth day I walked out down $1500. So the real lesson here is: no matter what game you play don't be a dumbass and lose $1500 playing it.

    Hope this helps.
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