# Ladder game based on coin flips

• Keiser
In summary, the conversation discusses a ladder type game with 10 steps and three possible events on each step. The goal is to reach the top without busting, with one event acting as a "life saver." The probability of winning the game is equivalent to the probability of getting 10 more heads than tails and never having two tails in a row, which can be calculated using a mathematical formula. The conversation also mentions a method for calculating the probability of winning a single step and extrapolating it to 10 steps.
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.

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

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

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 $$(2/3)^{10}$$.

## 1. How does the ladder game based on coin flips work?

The ladder game based on coin flips is a game where two players take turns flipping a coin. Each time the coin is flipped, the player moves up or down the ladder based on the outcome of the coin flip. If the coin lands on heads, the player moves up one rung on the ladder. If the coin lands on tails, the player moves down one rung. The first player to reach the top of the ladder wins the game.

## 2. Is there a strategy to winning the ladder game based on coin flips?

Since the outcome of each coin flip is random, there is no specific strategy that guarantees a win in the ladder game based on coin flips. However, some players may choose to use a specific pattern or strategy to try and increase their chances of winning.

## 3. Can the ladder game based on coin flips be played with more than two players?

Yes, the ladder game based on coin flips can be played with any number of players. However, as the number of players increases, the game may become more complex and take longer to complete.

## 4. What happens if the coin lands on its edge in the ladder game based on coin flips?

In the unlikely event that the coin lands on its edge, the flip is considered invalid and must be redone. The player must flip the coin again until it lands on either heads or tails.

## 5. Is the ladder game based on coin flips purely a game of chance?

Yes, the ladder game based on coin flips is a game of chance as the outcome of each coin flip is random. However, some players may use strategies or patterns to try and increase their chances of winning.

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