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Thread: odds of this happening

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    odds of this happening

    fully aware martingale is a bad idea. ive tried it and lost 600+ fairly quick a few months back
    at my local cardroom that i go to regularly i dont think ive ever lost 7 hands in a row. not like that matters but just throwing it out there. that being said, its possible to do that next time im in there without question.
    ive lost 300 the past 3 times ive been in there (just throwing out big bets at the wrong time..) so its not the end of the world if i waste 600+ if the chances of loosing all that are slimmer than just shooting fat bets out like ive been.
    $5 min $500 max.
    5-10-20-40-80-160-320
    buy in for 640. what are the odds of walking away just 100 or 200 ahead? at 50/50 odds an estimate would be in 40 hands you would be around that $100 mark IF you dont hit that 7 loosing streak obviously..
    (it would be more hands taking in consideration pushes and 3+ loosing streaks but that can go both ways i suppose..) what are the odds of hitting that 7 loose streak in that 40 hands? how about in 80 hands?

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    right..BUT was just wondering what the actual odds were of the situation

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    What situation?

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    (1-p)^x=y

    p = prob of win
    x = consecutive trials to lose
    y = chance of losing x in a row

    Chance of NOT losing x in a row is (1-y)

    So, chance of NOT losing x in a row over z trials is (1-y)^z

    For your example of p = .5 (not realistic), and losing 7 in a row, we get:

    (1-.5)^7 = 0.0078125

    Chance of not losing 7 in a row is 1-0.0078125=0.9921875

    Chance of winning 40 units before losing 7 in a row:

    .9921875^40 = .7307

    80 units before 7 losses in a row:

    .9921875^80 = .5339

    At 89, your chances are better of losing 7 in a row than not.

    Of course this doesnt take into account the fact that the chance of you winning a hand of blackjack (negating pushes) is more like 47 to 48% (assuming no AP techniques), and this is assuming no doubles, splits, etc.

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    The odds of losing a hand once pushes are negated in blackjack is not 50% but 52.51% on average. You always win about the same percentage but as the count increases you push more hands and lose less. You are in all practical terms never at as high as 50% to win the next hand if you negate pushes. The odds of losing 7 in a row is (.5251)^7 = .011007638 or about 1 in 91 times. So you win $5 90 times and lose $635 once for every 91 progressions you start. That is a net of -$235 for 91 progressions. Now you need to factor in the blackjack bonus. Each blackjack will give either $2.50, $5, $10, $20, $40, $80 or $160. In a six deck game the odds of getting a blackjack are about 4.75%.If we make your assumption that each lose is 50.50 to make this calculation easy it will be generous on the amount. In that case the 90 winning progressions each have a weighted average of a bonus of $2.50 as each amount is doubled and half as likely as the previous step in the martingale.
    $2.50*90*.0475 = $10.6875 or about $10.69
    If we add this to the previous total you lose $224.31 for every 91 progressions started.

    For a nice round number you expect to lose $2.50 for every progression you start.

    The question you ask for a 50/50 game the answer is .5^7 = .0078125 or 1 every 128 times. As you can see it is way off the odds of losing 7 hands of blackjack in a row which was 1 in 91 times.

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    Senior Member Tom's Avatar
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    Firstly, the odds of winning a hand are never 50/50 due to the house edge and possibility of pushes. Under AC rules the probability of hitting a 7 hand loosing streak at any point is close to 2%. Forty hands, eighty hands, whatever.....doesn't matter. It could happen on your first shoe, or it might not happen for a hundred shoes. The reason it doesn't matter is because probability doesn't change without new information (counting cards), playing Blackjack is always an independent event.

    The bottom line is you WILL hit that loosing streak, eventually, and paired with table limits, that's why martingale always fail. Think about it, you only hit that streak 2% of the time.....but when you do, you erase any "gains" from the other 98% of the time. You will even have a 9 hand losing streak almost 1% of the time, and based on your table max you wouldn't be able to martingale that. Also, the probability of having a X hand losing streak is actually never zero, no matter than # of hands. It might be very very tiny but it will never be zero. Another reason martingale always fail. You would have to have infinite capital and infinite table max in order for it to work.

    Edit: Ah, Tthree beat me to it
    ~Tom (aka "PT")

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    Quote Originally Posted by zengrifter View Post
    What you ask makes pseudo-sense. Martingale WILL win in short-run.
    ZG, you know I love you, but this is obviously wrong. Martingale will fail on the short run, long run, mid run
    Chance favors the prepared mind

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    I assume by "win", ZG means that OP will show a profit more often than he will show a loss, in the short run, which is clearly true.

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    Senior Member Frostbyte's Avatar
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    Quote Originally Posted by zengrifter View Post
    The love is mutual, mon ami ... but why does martingale work in tourneys then?
    The standard assumptions accompanying the concepts of ruin and profit don't apply the same way in a tournament setting.
    "Wait a minute. How do you beat someone to death with their own skull? That doesn't seem physically possible." "That's what Jimmy kept screaming: 'This doesn't seem physically possible!'"

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