# Ladder game based on coin flips

1. Feb 19, 2009

### Keiser

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. Feb 19, 2009

### mXSCNT

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
$$\sum_{i=10}^\infty \frac{(2/3)^i}{3} \\ = \frac{(2/3)^{10}}{3}\sum_{i=0}^\infty (2/3)^i \\ = (2/3)^{10}$$.

3. Feb 27, 2009

### bpet

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
$$\sum_{k=0}^\infty (1/4)^k*(1/2)=(1/2)/(1-1/4)=2/3$$
and thus to win 10 steps the probability is $$(2/3)^{10}$$.