Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    both produce a win. But
    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].
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook