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Is it possible to continuously profit from this? watch

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    Say there's a 50/50 game with 1:1 odds. For example, you place a bet (X tokens) on the side a coin will land on, and if you win, you'll get X tokens and your original back, and if you lose, you lose your bet.

    Is it possible to make a system that continously profits? Would the following not work?

    Place half of your money to one side - we call this the 'bank'. Call the remaining money the 'pot'.

    Bet 10% of the pot and place winnings back into the pot. Continously bet 10% of the pot. As soon as the pot contains over 1.5 times the original value of the pot, collect the bank and pot together and start over.

    Surely from probability, your pot value will vary (averaging around the initial value of course) and so you can simply specify any target (which you eventually must meet), then just bank at that point?

    I can see that it's also possible to continuously lose and hit 1 token, but if our initial pot had millions of tokens, wouldn't this be so unlikely that when it did happen, you'd still profit overall?
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    (Original post by ak9779)
    Say there's a 50/50 game with 1:1 odds. For example, you place a bet (X tokens) on the side a coin will land on, and if you win, you'll get X tokens and your original back, and if you lose, you lose your bet.

    Is it possible to make a system that continously profits? Would the following not work?

    Place half of your money to one side - we call this the 'bank'. Call the remaining money the 'pot'.

    Bet 10% of the pot and place winnings back into the pot. Continously bet 10% of the pot. As soon as the pot contains over 1.5 times the original value of the pot, collect the bank and pot together and start over.

    Surely from probability, your pot value will vary (averaging around the initial value of course) and so you can simply specify any target (which you eventually must meet), then just bank at that point?

    I can see that it's also possible to continuously lose and hit 1 token, but if our initial pot had millions of tokens, wouldn't this be so unlikely that when it did happen, you'd still profit overall?
    Anything could happen. Your expected value is still zero though. And there is variance so it is a bad idea to take this bet - assuming you are not a nutter.
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    (Original post by ak9779)
    so you can simply specify any target (which you eventually must meet), then just bank at that point?
    No.


    You're just as likely to lose as you are to win; that's obvious, so you're not guaranteed anything.
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    I don't think you will get any profit or loss from that.
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    The way to 'beat it' is place X on red.
    Then if you lose, bet 2x on red
    If you lose bet 4x on red.
    Bet 8x on red
    bet 16x on red
    bet 32x on red
    bet 64x on red

    your gonna win 1x!
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    I've been trying to get my head around this..

    Ok so I take half the tokens, put them in my bank, and have the other half in a pot.

    Then I bet 10% of the pot. And suppose it loses.
    Then do I bet 10% again? Or do I bet 10% of what is left in the pot?

    Assuming you mean the latter, I can't see how it will guarantee making money, because, you only have to lose a few times to lose a large chunk of your money. You only have to choose wrong 20 times, and you've lost over 75% of the pot.

    I can't see how it is possible to make money on such a scheme with a finite amount of money. With as much money at your disposal as you require, this method always works;

    Bet one token on Heads.
    If it wins, take the winnings, and bet one token on heads again.
    If it loses, double your stake. Continue to double your stake until you win.
    This way, if you have to bet n tokens before you win, it will always return n+1 tokens.

    Maybe I am not understanding your process correctly. But I believe mine will always work. The problem with mine is that even if you start with 1000 tokens, and you begin by betting 1 token, if you happen to get it wrong 9 times in a row (which isnt inconceivable at around 500/1), then you have lost all your tokens!
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    (Original post by tom29whu)
    I've been trying to get my head around this..

    Ok so I take half the tokens, put them in my bank, and have the other half in a pot.

    Then I bet 10% of the pot. And suppose it loses.
    Then do I bet 10% again? Or do I bet 10% of what is left in the pot?

    Assuming you mean the latter, I can't see how it will guarantee making money, because, you only have to lose a few times to lose a large chunk of your money. You only have to choose wrong 20 times, and you've lost over 75% of the pot.

    I can't see how it is possible to make money on such a scheme with a finite amount of money. With as much money at your disposal as you require, this method always works;

    Bet one token on Heads.
    If it wins, take the winnings, and bet one token on heads again.
    If it loses, double your stake. Continue to double your stake until you win.
    This way, if you have to bet n tokens before you win, it will always return n+1 tokens.

    Maybe I am not understanding your process correctly. But I believe mine will always work. The problem with mine is that even if you start with 1000 tokens, and you begin by betting 1 token, if you happen to get it wrong 9 times in a row (which isnt inconceivable at around 500/1), then you have lost all your tokens!
    Yours is an old technique that mathematically doesn't work.

    I still don't see why my system doesn't work. Imagine a graph where the current value of the pot is varying; occasionally there are peaks up and peaks down, but because you're betting a percentage, you'll never reach 0 tokens. Then when you hit a certain peak, you stop and start over.

    In essence, your bank number never decreases - it only ever increases. Your pot value increases/decreases continously until you a hit a peak, in which case you bank then start again.
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    (Original post by vandub)
    No.


    You're just as likely to lose as you are to win; that's obvious, so you're not guaranteed anything.
    You misunderstand. Your bank money will never decrease - right? Your pot value will increase/decrease dependent on luck, and over an infinite time span will hit all possible values. You simply stop as soon as it hits a certain peak, put that in the bank, then start over.
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    (Original post by Classical Liberal)
    Anything could happen. Your expected value is still zero though. And there is variance so it is a bad idea to take this bet - assuming you are not a nutter.
    How could the money in the bank ever decrease? And surely over an infinite time span, you are guaranteed that your pot value must hit 1.5 times the original at some point? You'll never hit zero if you're betting a percentage of the current pot...
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    (Original post by ak9779)
    Yours is an old technique that mathematically doesn't work.

    I still don't see why my system doesn't work. Imagine a graph where the current value of the pot is varying; occasionally there are peaks up and peaks down, but because you're betting a percentage, you'll never reach 0 tokens. Then when you hit a certain peak, you stop and start over.

    In essence, your bank number never decreases - it only ever increases. Your pot value increases/decreases continously until you a hit a peak, in which case you bank then start again.
    Are you assuming you can have a fractional number of tokens? I think the amount in the pot will tend towards zero. If you lose, you have to win more than once to make up for it. If you win, a loss on the next go would more than undo that winning.

    That's my intuition anyway.

    Edit. Actually... my intuition may be off a little...
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    (Original post by ak9779)
    You misunderstand. Your bank money will never decrease - right? Your pot value will increase/decrease dependent on luck, and over an infinite time span will hit all possible values. You simply stop as soon as it hits a certain peak, put that in the bank, then start over.
    Including zero, right? Doesn't matter that for your 'first' pot, you may be more likely to hit your target than to hit zero, because, on average, you'll lose as much money as you make (say your target is 5 wins in a row: you're as likely to get five losses in a row) . What happens if you lose all of the first pot? Tidy little profit there.
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    (Original post by Woodworth)
    The way to 'beat it' is place X on red.
    Then if you lose, bet 2x on red
    If you lose bet 4x on red.
    Bet 8x on red
    bet 16x on red
    bet 32x on red
    bet 64x on red

    your gonna win 1x!
    So I lose 1 pound on red,
    6 more bets down the line - 127 quid gone and still only a 50% chance of winning?

    You're a genius, absolute genius.
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    (Original post by vandub)
    Including zero, right? Doesn't matter that for your 'first' pot, you may be more likely to hit your target than to hit zero, because, on average, you'll lose as much money as you make (say your target is 5 wins in a row: you're as likely to get five losses in a row) . What happens if you lose all of the first pot? Tidy little profit there.
    It's actually impossible to hit 0 by simply betting 10% each time, but let's say that going below 10 tokens counts as a failure (since you can't do 10% anymore).

    If you fail, you still have half of your money in the bank. Thus hitting <10 tokens (unlikely) halves your money.

    The target isn't a series of wins - it's a value of the pot. Say the target is 1.5 * the value of the original pot.

    If you manage to hit the target (after 5 games, 50 games or 500 million games) your overall bank will increase by (x/2) + (x*1.5) where x was the value of the bank before.

    So it will take 4 'successes' to more than double your money.

    Summary:

    Every 4 'successes' (hitting the target amount) doubles your money.
    Every 'failure' (reaching <10 tokens) halves your money.

    So can you not choose your initial pot amount to be large enough such that the chance of hitting <10 tokens is small enough so this gives a profit?
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    (Original post by Woodworth)
    The way to 'beat it' is place X on red.
    Then if you lose, bet 2x on red
    If you lose bet 4x on red.
    Bet 8x on red
    bet 16x on red
    bet 32x on red
    bet 64x on red

    your gonna win 1x!
    This only works if you have an infinate amount of money. It is possible to go on a streak of 20 losses and if you do this a considerable amount of times that sort of thing can happen which blows the bank.
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    (Original post by gummers)
    So I lose 1 pound on red,
    6 more bets down the line - 127 quid gone and still only a 50% chance of winning?

    You're a genius, absolute genius.
    ur bound to win a quid unless ur not rich enough (hope that doesn't sound snobbish) and well unlucky.
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    Relevant:
    http://en.wikipedia.org/wiki/Marting...ette_system%29
    http://en.wikipedia.org/wiki/St._Petersburg_paradox
 
 
 
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