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Martingale Betting System- expected value

  1. Jun 12, 2012 #1
    Hi, I have been learning/experimenting with the Martingale betting system recently. I have read a lot about how no "system" works for betting in casinos. However, I want to either prove or disprove the validity of the system by looking at its expected value/payout. I will be using the game of roulette as an example, assuming I have a 50% of getting black or red. Here is how the payout will work-

    If the ball lands on black, I win $5
    If the ball lands on red 7 times in a row, I lose $155

    What is the expected payout of me playing roulette with this system?

    winning payout = $5 * (1/2) ;have a 50% chance of getting black
    losing payout = -$155 * (1/2)^7 ;have a 50% chance of getting red, 7 times

    net payout = winning payout - losing payout
    =$1.29

    So according to this, if I use this system then there will be a positive payout. What do you guys think?

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    If you want to how i got the winning/losing payouts then here is the explanation for that-

    The way the system works is that you first bet your initial, let's say it's $5. If you lose, you double your previous bet, and if you lose again you double up again, etc. So if I start at $5 and keep losing, my bets will go like this - $5,$10,$20,$40,$80....
    Using this system, you will earn $5 every time you win. To see this, assume I lose $5,$10,$20 but on the $40 bet I win. I will have lost $35 from before but gained $40, i.e. $5 profit.

    ***Now, according to this if I have enough money I can eventually earn $5 as long as I have a crap load of money. This is why tables have maximum bets, but many cheap casinos don't.

    Moving on, if i lose 5 times in a row, I will lose 5+10+20+40+80 = $155. Let's say I "restart" after I lose 5 times. Also, because I am so clever, I will bet only until I see two reds. Therefore, I can technically "lose" 7 times in a row and still lose only $155.

    All in all, one can see that this ^ is where the losing payout comes from, and that the winning payout is me simply winning $5 on black.
     
  2. jcsd
  3. Jun 13, 2012 #2

    haruspex

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    Ay, there's the rub. If you assume infinite resources then you may expect paradoxical results. As soon as you plug in a finite limit the paradox disappears.
    I didn't understand the bit about only betting until you see two reds. Please explain.
     
  4. Jun 13, 2012 #3
    There is another problem it is not a 50:50 bet, the table has a white zero wich takes all the stakes and doesn't pay out.
     
  5. Jun 13, 2012 #4

    D H

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    You don't get to set the rules. The house sets the rules. Gambling houses that want to stay in business for any length of time don't set the rules in favor of the bettor. They set the rules in favor of themselves.

    The only way beat a game such as roulette is not to play the game. Any gambling system that does not involve infinite amounts of money is a losing proposition. Roulette is not beatable because the roulette wheel is memoryless and because the payoffs don't reflect the odds. The game is stacked against you.

    Some games such as blackjack are beatable because the payoffs are fixed but the probabilities are not. They instead vary as cards are dealt from the shoe. You can win if you count cards and tailor your betting to the card count, but only if the house doesn't catch you. Card counting is cheating as far as the house is concerned. It violates house rule #1: The house is always supposed to be the winner over the long haul.
     
  6. Jun 13, 2012 #5
    Let us consider a fair game with two equally probable outcomes with probability 1/2 (the roulette is not even this fair, because the probability is 18/37 in Europe, and 18/38 in U.S.). If you win, you get double the amount you bet.

    Suppose you decide to keep increasing your bet every time you lose, so that, if you win in the following turn, you win back all your previous losses, and you make a fixed profit.

    Your bets satisfy the recursion:
    [tex]
    2 x_{k} = S_{k} + a, \ S_{k} = \sum_{j = 1}^{k}{x_{j}}
    [/tex]
    Notice that:
    [tex]
    S_{k+1} - S_{k} = x_{k+1}
    [/tex]
    so, if we subtract the recursion for [itex]k[/itex] from the one for [itex]k + 1[/itex], we get:
    [tex]
    2 x_{k+1} - 2 x_{k} = S_{k+1} - S_{k} = x_{k+1}
    [/tex]
    [tex]
    x_{k + 1} = 2 x_{k}
    [/tex]
    This defines a rising geometric progression with a coefficient 2. In other words, you keep doubling your bets. To find the initial bet, consider the recursion for [itex]k=1[/itex], by noticing that [itex]S_{1} = x_{1}[/itex]:
    [tex]
    2x_{1} = x_{1} + a
    [/tex]
    [tex]
    x_{1} = a
    [/tex]
    The final expression for the martingale is:
    [tex]
    x_{k} = a 2^{k - 1}
    [/tex]
    The total amount deposited up to the kth step is:
    [tex]
    S_{k} = a \left( 2^{k} - 1 \right)
    [/tex]

    The probability to lose in the previous k-1 turns and win in the k-th turn is:
    [tex]
    P_{k} = p \, (1-p)^{k - 1}
    [/tex]
    where p is a little smaller than 1/2.

    Now comes the punchline. You have to quit if you lost N turns. The probability for this is [itex](1-p)^{N}[/itex].

    The mathematical expectation of your win is:
    [tex]
    E[P] = \sum_{k = 1}^{N}{a \, p_{k}} + (-S_{N})(1-p)^{N} = a \, \left[1 - (1 - p)^{N} - (1 - p)^{N} (2^{N} - 1) \right] = a \, \left[ 1 - (2(1-p))^{N} \right]
    [/tex]

    Notice the base of the exponential [itex]2(1-p)[/itex]. I said [itex]p \le 1/2 \Rightarrow 1 - p \ge 1/2, \ 2(1-p) \ge 1[/itex]. This means that only in a fair game your expectation profit is zero. For an unfair game, it is even negative, and rises without bound if you keep increasing the steps.
     
    Last edited: Jun 13, 2012
  7. Jun 13, 2012 #6
    okay i realise some mistakes i have made now. Also, thank you Dickfore for the explanation of the system, however I don't see how that is anything different than what i have done?

    To me it all boils down to one question if anyone can tell me? Here is the situation-

    I want to play roulette and I am betting on black or red with 50:50 odds. When i get to the table I see that two reds have already come up, I then bet on black. However, 5 more reds come up for a total of 7 reds. My question is, is the probability of those 7 reds coming up the same for someone who saw it happen before any red appeared compared to me because I came after there were already 2 reds?
     
  8. Jun 13, 2012 #7

    D H

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    The odds aren't 50:50. The 0 and 00 tilt the odds in favor of the house.

    No different. The roulette wheel, if it's an honest one, is memoryless. There's a small but non-zero probability that red will come up 100 times in a row. The probability that it will be red on the next spin of the wheel is exactly the same as it was at the start of this improbable sequence: 18/37 in Europe, 18/38 in the US.
     
  9. Jun 13, 2012 #8
    I edited my post.
     
  10. Jun 13, 2012 #9
    This is a roulette wheel... there is no deck from which you can take cards to get information about what will come. Red/black is 50/50.

    Martingale betting is just another way to win a little or lose a lot of money.
     
  11. Jun 13, 2012 #10
    Thank you for all of your input! I have figured it out.

    I see that i have a few mistakes in my calculations that basically destroys the system. I was too deep inside the Gambler's Falacy. First the winning probability should be 1-.5^5. This is because I need to get one black in 5 rolls to win $5. Second, the probability of me losing should still be based on losing 5 times in a row, not 7. This is because even if you wait for 2 of the opposite color to come up, the next 5 are independent of the previous 2; it took me a long *** time to completely understand it. After you do both things one can see that the net payout is $0. However, most casinos have a 0 and 00 so the probability of winning decreases and the probability of losing increasing, leading to a negative payout. So in the long run, you lose. =(

    Thanks again!!!
     
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