Imagine we played a game using a dice. I start, then you get a turn, then me again and so on. The first person to get a 6 wins. What is my probability of winning if I start?
Now consider we did the same but with a coin. Without calculation, would this probability increase or decrease?
Here's what I was thinking:
On average, you get 6 every $6$ rolls. So on average, the $6$th person to roll will win. So you going first doesn't change the probability (three rolls each), which must therefore be $\frac 12$.
Using this logic for a two-sided die, we'd get the same probability: $\frac 12$.
These results are extremely unintuitive to me and and almost certainly incorrect. I'd like to know (1) the flaw in the reasoning, and (2) how to get the correct solution.
Best Answer
Let $P$ be the probability that the person about to roll wins the game.
Let $p$ be the probability that a particular roll is a winning result.
Therefore $P= p+ (1-p)(1-P)$, because either the person about to roll immediately obtains a winning result, or they don't and then their opponent is about to roll next. With a little arithmentic rearrangement :
$$P= \frac{1}{2-p}$$
This is also the probability that whoever starts the game wins the game.
Sure, the average count for rolls until winning is $6$, but this sais nothing about the probability for who wins. The probability for obtaining this expected value is only about $0.067$, and it is but one outcome in the event for you not winning.
You want the probability that the count for rolls until winning is an odd number.$$P~{=\sum_{k=0}^\infty \mathsf P(X=2k+1)\\=\sum_{k=0}^\infty (1-p)^{2k}p\\ = p+(1-p)^2p+(1-p)^4p+\cdots}\qquad\text{Where }X\sim\mathcal {Geo}_1(p)$$