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50 heads in a row

  1. Jan 22, 2014 #1
    If your betting on a coin flip but person a is the banker and person b can quit at any time, does person b have an advantage.

    The martingale theory of doubling a losing bet doesn't work if there is a maximum bet. Start at 5 and going to the max of 1000 a person can bet 5,10,20,40,80,160,320,640.

    So they can bet 8 times. The odds of losing 8 in a row are 1 in 128.

    If there is no maximum bet, and the person can bet 50 times, what are the odds of losing 50 coin flips in a row.

    I know it's something like (1/2) to the 50th power. But how does this translate into more usable numbers, such as 1 in whatever?

    Would you feel comfortable placing a bet in this scenario?
  2. jcsd
  3. Jan 22, 2014 #2


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    How did you get 128 for 8 tosses? Do the same thing with 50 to get the results for 50 tosses.

    Oh, and by the way, you have the wrong answer for 8 tosses. It's 1 out of 256
  4. Jan 22, 2014 #3
    Double post
  5. Jan 22, 2014 #4
    My math is wrong the first time, so do the same thing and get another wrong answer?

    You see the irony here right
  6. Jan 23, 2014 #5


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    The probability of 8 consecutive coin flips with the same outcome is 2^n - 1, where n is the number of flips. The first iteration is free, that outcome is guaranteed. After that, it's a simple probability of 50-50 repeating the first outcome with each flip. The odds of 8 consecutive outcomes of the same result is 2^7, or 128 to one. If you insist the first outcome be 'heads', the probability becomes 256 to 1. If you wish to weird up the probabilities, you can always toss in the odds of the coin resting on its edge. That would constitute house odds for a coin flip casino game - sort of like green on the roulette wheel, but, a lower payout.
  7. Jan 24, 2014 #6
    2 to the tenth to the fifth power is equal to 2 to the 50th power. 2 to the tenth is about one thousand. Three zeros. So 2 to the 50th is about fifteen zeros, which is a quadrillion.
  8. Jan 26, 2014 #7


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    A player, b, who can quit when he is ahead has a distinct advantage over a player, a, who must keep playing. This is an example of a 1-dimensional random walk, which is guaranteed to eventually allow player b to be ahead and quit. Don't confuse this with the real gambling situation, where player b is forced to quit when he goes broke and is behind.

    The odds of 50 heads in a row is one in 1.126 x 10^15.
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