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Ladder game based on coin flips

  1. Feb 19, 2009 #1
    Hi there, found a situation where I'm trying to find the probability of winning a ladder type game.

    There are 10 steps and the goal is to win all 10 steps without busting (i.e. a streak, given one exception below in event B). On any given step three different events can occur:
    Event A: 50% chance to win, move up to next step.
    Event B: 25% chance to lose but not bust, stay at same step. A "life saver" if you will.
    Event C: 25% chance to lose and bust. Start over.

    Think of it like flipping a coin, but with a double elimination. So on a heads you move up a step. On a tails you flip again, a heads after the tails and you reset to this level, but if you get tails again, you lose. The best way I could think of it mathematically is what is the probability that you will have 10 more heads than tails and never have two tails in a row.

    For instance
    HHHHHHHHHH
    HHHthHHthHHHthHthH
    both produce a win. But
    HHHthHHtt
    loses after winning 5 steps.

    Any ideas? Been a few years since I've done any higher math. The 10 more heads than tails portion sounds like standard deviation, but I never used that in any class I took.
     
  2. jcsd
  3. Feb 19, 2009 #2
    Well, you can ignore the B's, and suppose it is A with 2/3 chance and C with 1/3 chance. On any given "run" (that is, starting from 0) the probability that you get n A's followed by a C is (1/3) * (2/3)^n. The probability that you win on a given run is therefore
    [tex]\sum_{i=10}^\infty \frac{(2/3)^i}{3} \\
    = \frac{(2/3)^{10}}{3}\sum_{i=0}^\infty (2/3)^i \\
    = (2/3)^{10}[/tex].
     
  4. Feb 27, 2009 #3
    It wasn't obvious to me that A has 2/3 chance, but I'm guessing you worked it out this way: To win a single step needs one of H, thH, ththH,... which has total probability
    [tex]\sum_{k=0}^\infty (1/4)^k*(1/2)=(1/2)/(1-1/4)=2/3[/tex]
    and thus to win 10 steps the probability is [tex](2/3)^{10}[/tex].
     
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