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Roulette "system"

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  • 18-02-2019 12:40pm
    #1
    cmssjone
    Registered Users Posts: 355 ✭✭


    Hi guys,

    My father's told me about his brother who claims that he has system (yeah I know) that is making him money. I am absolutely sure it has no mathematical validity to it but thought I'd throw it out to you guys for verification. So his system is as follows:

    1) Wait for 6 blacks or 6 reds to come up consecutively
    2) Example: If 6 reds come up then place a bet on black. If bet wins, then restart system. If red, then bet on black a second time, placing enough money to cover the first bet and come out with the same winnings as if he would have won on the first spin.

    So if at 2:1 he wants to win 100, he waits until 6 of any colour (say red) is spun consecutively. He then bets 100 on black.

    If he wins then he stops with 100 profit
    If he loses he then has to bet 150 to cover the previous bet and still walk away with 100 profit.
    If he loses then he stops and he has made a 250 loss.

    Mathematically I can't see how this system can work and said as much to my father. As this is a Bernoulli trial and each spin is independent of each other I can't see how my father's brother is making a profit using this system other than pure dumb luck. Any insights to where my logic is failing (if it is) would be highly appreciated.


Comments

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    #2
    Yakuza
    Registered Users Posts: 5,141 ✭✭✭Yakuza


    The first thing I'd say is that he could be waiting for those 6 reds to appear. In a formula I found online, the expected number of trials for 6 successes (where P(success) = 18/37 (assuming the wheel is 18 red, 18 black and one green)) is 144 trials. If a spin (and all the attendant moving of chips, paying winners etc) takes (conservatively) 30 seconds, then he could be waiting over an hour. On time value alone, he'd probably be better off playing blackjack!

    (1-p^r) / ((p^r) * (1-p) gives the expected number of trials needed for r successes with probability p

    See here for the formula, it looks plausible! :
    https://stats.stackexchange.com/questions/91518/waiting-time-for-successive-occurrences-of-a-result-when-rolling-a-die

    As you correctly say, the fact that there have been 6 reds makes no difference to how the 7th trial turns out (the "memoryless" property) but your uncle could be forgiven for using "surely a black is due up now" logic but it has no mathematical basis. In the long run (on an unbiased wheel and ball), you would expect the distribution of results to be in proportion to the relative frequency of the colours but for any given individual trial, the probability of a red / black / green is unchanged.

    By the way, it's the green slot that on average gives the house its edge - people tend to bet on black or red (and only get double their stake back even though their chances are slightly less than evens). Also, the 35 to 1 payout is less than the number of slots (37 (or 38 in the US with both 0 and 00)) so your long run expected earnings are negative. Steer clear would be my advice.


  • Options
    #3
    cmssjone
    Registered Users Posts: 355 ✭✭cmssjone


    Yakuza wrote: »
    The first thing I'd say is that he could be waiting for those 6 reds to appear. In a formula I found online, the expected number of trials for 6 successes (where P(success) = 18/37 (assuming the wheel is 18 red, 18 black and one green)) is 144 trials. If a spin (and all the attendant moving of chips, paying winners etc) takes (conservatively) 30 seconds, then he could be waiting over an hour. On time value alone, he'd probably be better off playing blackjack!

    (1-p^r) / ((p^r) * (1-p) gives the expected number of trials needed for r successes with probability p

    See here for the formula, it looks plausible! :
    https://stats.stackexchange.com/questions/91518/waiting-time-for-successive-occurrences-of-a-result-when-rolling-a-die

    As you correctly say, the fact that there have been 6 reds makes no difference to how the 7th trial turns out (the "memoryless" property) but your uncle could be forgiven for using "surely a black is due up now" logic but it has no mathematical basis. In the long run (on an unbiased wheel and ball), you would expect the distribution of results to be in proportion to the relative frequency of the colours but for any given individual trial, the probability of a red / black / green is unchanged.

    By the way, it's the green slot that on average gives the house its edge - people tend to bet on black or red (and only get double their stake back even though their chances are slightly less than evens). Also, the 35 to 1 payout is less than the number of slots (37 (or 38 in the US with both 0 and 00)) so your long run expected earnings are negative. Steer clear would be my advice.

    Thanks Yakuza,

    I was pretty sure that it was pure luck or exaggeration. I'll check out the link you provided and try to explain in layman's terms why it doesn't work to my father.


  • Options
    #4
    mikeymouse
    Closed Accounts Posts: 568 ✭✭✭mikeymouse


    cmssjone wrote: »
    Hi guys,

    My father's told me about his brother who claims that he has system (yeah I know) that is making him money. I am absolutely sure it has no mathematical validity to it but thought I'd throw it out to you guys for verification. So his system is as follows:

    1) Wait for 6 blacks or 6 reds to come up consecutively
    2) Example: If 6 reds come up then place a bet on black. If bet wins, then restart system. If red, then bet on black a second time, placing enough money to cover the first bet and come out with the same winnings as if he would have won on the first spin.

    So if at 2:1 he wants to win 100, he waits until 6 of any colour (say red) is spun consecutively. He then bets 100 on black.

    If he wins then he stops with 100 profit
    If he loses he then has to bet 150 200 to cover the previous bet and still walk away with 100 profit.
    If he loses then he stops and he has made a 250 loss. 300 loss

    Mathematically I can't see how this system can work and said as much to my father. As this is a Bernoulli trial and each spin is independent of each other I can't see how my father's brother is making a profit using this system other than pure dumb luck. Any insights to where my logic is failing (if it is) would be highly appreciated.

    At odds of evens he has to double his stake each turn to recoup his losses and make his profit.
    A better system would be to wait for a run of say 20 before commiting.
    That way he'll probably never have a bet, which is the only way to not lose at roulette


  • Options
    #5
    i71jskz5xu42pb
    Closed Accounts Posts: 1,806 ✭✭✭i71jskz5xu42pb


    cmssjone wrote: »
    each spin is independent of each other

    That about sums it up, no?


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