What is the probability your friend wins on his n'th toss? You and your friend roll a sixth-sided die. If he lands even number then he wins, and if you land odds then you win. He goes first, if he wins on his first roll then the game is over. If he doesn't win on his first roll, then he pass the die to you. repeat until someone wins.
What is the probability your friend wins on his $n^{th}$ toss? : $(\frac{1}{2} )^{2n-1}$
I want to understand why this is the answer and how do I derive this answers? I asked my professor but he said that there isn't any formula to base this off of.
 A: Both of you have the same probability of winning and losing on a single roll, $\frac{1}{2}$
The probability that he wins in his first toss is $\frac{1}{2}$.
He wins in his second toss when he and you both don't win your first tosses and he wins in his second toss. The probability that this happens is $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}$.
He wins in his third toss when he and you both don't win your first two tosses and he wins in his third toss. The probability that this happens is $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}$.
He wins in his $n$th toss when he and you both don't win your first $n-1$ tosses and he rolls a win in his $n$th toss. The probability that this happens is $\frac{1}{2} ^{n-1} \frac{1}{2}^{n-1} \cdot \frac{1}{2} $ Which simplifies to your desired answer.
A: The probability of an odd number on a single toss is $\frac{3}{6}$, that is, $\frac{1}{2}$, so it's just like tossing a fair coin. 
Your friend wins on his $n$-th toss if the $2n-2$ tosses before that, ($n-1$ by her and $n-1$ by you) were tails, and he finally gets a head on her $n$-th toss. This has probability $\left(\frac{1}{2}\right)^{2n-2}\cdot \frac{1}{2}$.
