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Thread: Roulette’s chips distribution: Wins vs Losses

  1. #1
    Oct 2009

    Roulette’s chips distribution: Wins vs Losses

    Hi guys, my first post here and yes, it’s about roulette yet again, but I promise you this time will be a different question. And also, due to the nature of my question, I would expect helps from someone who are familiar with roulette.

    I am fully realised that mathematically, roulette pays out has been designed to favour the house or break even if not player loss. I will not be asking question about what is the chance of what number comes up in whatever spins. What I’m trying to do is to find the most effective chips distribution to cover as much as possible of the numbers (I am disregarding the 0) so it could provide more chance to win, (1 profit is enough) and mitigate the loss as much as possible (I know that it will be an inevitable loss somewhere, but as long as it is not more than 4, it would be ideal).

    For example:
    - One could distribute bets on: 1 chip on each black and white (1:1 pay out). This will result in break even, because one wins 2 but wagered 2, so no profit.

    - One could distribute bets on: 3 chips on the first dozen (2:1 pay out), 3 chips on the second dozen (2:1 pay out) and 2 chip on the second half (1:1 payout). This combination will create a chance for one to wins 1, if the ball lands on either first or second dozen (9 wins, 8 wagered, 1 wins). But it will also creates a “loss area” of 4 chips if the ball lands on the ¾ of the second half (4 wins, 8 wagered, 4 losses) even though it also provides more wins of 3 wins, if the ball lands on the overlap area between the second dozen and the second half (6 numbers) (11 wins; 9 from the second dozen+ 2 from the second half, 8 wagered, hence 3 wins)

    - The losses of above combination, could be mitigated if the last 2 chips are placed on the third dozen (6 wins, 8 wagered, 2 losses instead of 4). But it still have the same amount of “loss area” (12 numbers)

    I have been trying to arrange this chips distribution so it could cover as much numbers on the board therefore reducing the loss area, and if loss, it wont be greater than 5 chips (again I am ignoring the 0) and none of them is good enough (always end up with scenarios with 5 loss with 12 numbers of loss area.

    Now, obviously, I am not a math expert or anything like that, so I am asking you guys, if you could come up with better arrangement that could accommodates balance between minimum loss area and profit(s). I will be using the following pay out rules:

    - Straight bet: 1:35
    - 2 Splits bet: 17:1
    - 4 Splits bet: 8:1
    - Row (1st, 2nd and 3rd) bet: 2:1
    - Dozen (1st, 2nd, and 3rd) bet: 2:1
    - Colours (black and red) bet: 1:1
    - Odd/Even bet: 1:1
    - Half (1st and 2nd) bet: 1:1

    I reckon, by reducing the loss area as much as possible and to limit the amount of loss of that area as much as possible, would make a very effective playing system which force the favour to player rather than the house, and, we can all make money together J,…

    Thanks guys.
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  2. #2
    Dec 2009

    Does this make sense?

    there are tree sections divided into the 1st 12 and 2nd 12 and 3rd 12. PLace and equal amount on 2 of the 3 sections ( say 5) so i place 5 on 1st 12 and 5 on the 2nd 12. The pay out on these is 2:1 so technically you have about a 64% chance of winning (taking into account the 0 and 00). Then if you lose you have to multiply your previous bet by 3 then 9 then 27 then 81. The best cover your loss of previous bets and your loss on that bet.
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