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Possbility (Gambling Question)

  1. Jun 3, 2013 #1
    1. The problem statement, all variables and given/known data

    I am just wondering, this is a problem solving question that I had thought of. If a Gambler was to gamble and the odds were always 50%, and he loses, if he had given his next bet (2X the amount of his previous bet he lost [Martingale] ) to another gambler and the second gambler gambled,

    My question now is, does doing this minimize your chances of losing? And also does it maximise your chances of winning? As in will you have a lower chance of losing + will you have higher chance of winning if you kept doing this in a cycle with 4 gamblers? Any maths to this? Sorry I'm just looking for other peoples opinions
     
  2. jcsd
  3. Jun 3, 2013 #2

    Simon Bridge

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    What you have described is an old gambling "system" ... it goes like this:

    On a fair 50:50 split, even money, each time you lose, double your next bet.
    Each time you win, just bet the base amount.

    The trouble is that your winnings increase linearly but your losses increase exponentially - the system breaks down the first time you lose 1/3 or more of your money... which happens quickly because wins and losses come in runs and losses build exponentially.

    YAdd to this that it relies on the game being fair. In a casino - there is always an edge for the house.
     
  4. Jun 4, 2013 #3
    Okay I think I know what you mean. But what if the 1st gambler lost $5, and gives $10 (2X the bet) to the 2nd gambler to gamble the $10, will it mean 75% Win, 25% Loss? Because 1st Gambler has 50% chance and 2nd gambler has 50% chance
     
  5. Jun 4, 2013 #4

    Simon Bridge

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    Each time the odds are still 50:50.
    The second gambler is no more likely to win than the first one.

    Why bother with a long chain of gamblers? Why not just have one gambler who can bet many consecutive times?

    After two turns, the combinations are:
    win win 25% +5+5=+10
    lose win 25% -5+10=+5
    win lose 25% +5-5=0
    lose lose 25% -5-10=-15

    expectation: (10+5+0-15)/4 = 0

    Repeat for the long term - the expectation is that you break even.
    However, this requires that your initial stake holds out.
    Give the gambler an initial $100 and run the simulation... look at the expectation over many turns.
    The more turns you run the simulation for the more likely the gambler has to stop playing because he's run out of money.

    You should realize that most everyone thinks of this system sometime - nobody uses it for long.
    It's part of a class of systems called Martingale Systems

    Also see:
    http://wizardofodds.com/gambling/betting-systems/
     
    Last edited: Jun 4, 2013
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