A bet is decided between 2 people by flipping a coin until someone gets heads. Is this fair and is there a more fair way to do this? I can't tell how to decide this because it feels as though the first person has an advantage as they have a 1/2 chance of success and the second has a 1/2 X 1/2 = 1/4 chance but I don't know how to state this rule well or what a fairer way would be.
 A: The first player has a 1/2 + 1/8 + 1/32 + 1/128 + .... = 2/3 chance of winning and the second player has a 1/4/ + 1/16 + 1/64 + ... = 1/3 chance of winning.
This is kind of interesting in that it means if the two players played a game where they flip a coin until it was heads; the first player wins if the total number of flips is odd and the second player wins if the total number of flips is even; this game is not fair even though intuitively it is.
For a fair game, flip the coin once, of course.
A: Let $p$ be the probability that the first player wins. You can calculate $p$ without having to deal with an infinite sequence or series as follows. 
First, $p$ is clearly the probability that the first player wins on the first toss plus the probability that he wins after first getting a tail. 


*

*The probability that he wins on the first toss is $\frac12$. 


Suppose that he gets a tail on the first toss; then in effect the game starts over with the second player as first player, and at that point the probability that the original second player wins is $p$, so the probabillity that he loses is $1-p$.


*

*Thus, the probability that the first player initially gets a tail and the second player then loses is $\frac12(1-p)$. 


Putting the pieces together, we see that
$$p=\frac12+\frac12(1-p)\;,$$
and solving for $p$ yields $p=\dfrac23$.
A: The probability that the first person wins at the ith trial is
$i=1: \frac12$
$i=3: \frac12^3$
$i=5: \frac12^5$
$i=n: \frac12^n$
The sum is $\sum_{k=0}^{n} \frac12^{2k+1}$
$\frac12^{2k+1}=\frac14^k\cdot \frac12$
$\sum_{k=0}^{n} \frac14^k\cdot \frac12=\frac12\sum_{k=0}^{n} \frac14^k$
The partial sum of $\sum_{k=0}^{n} \frac14^k$ is $\frac{1-\frac14^n}{1-\frac14}=\frac43\cdot (1-\frac14^n)$
Let $n \to \infty$
$$\lim_{n \to \infty}\frac12\cdot \frac43\cdot \left(1-\frac14^n \right)=\frac23$$
