What probability p corresponds to an expected number of 10 turns The problem below is from a problem set for a Game Theory course. We never really touched on much probability, probability distributions, etc so I was surprised when I saw this question...
"In probability theory, if an event occurs with fixed probability $p$ and ends with probability $(1-p)$, then the expected number $v$ of event turns is defined by:
$v=(1-p)\sum_{n=1}^\infty np^{(n-1)}=\frac{1}{1-p}$
What probability $p$ corresponds to an expected number of 10 turns?"
Okay, so, I have no idea what I am being asked here. All I can think of doing is plugging $v=10$ to get $p=0.9$, but that seems all too easy for a question for a graduate level course.
Any help?


Okay, so, I have no idea what I am being asked here. All I can think of doing is plugging $v=10$ to get $p=0.9$, but that seems all too easy for a question for a graduate level course.

Yes, that is exactly what is needed.   You were being asked: "Given that $v = \frac 1{1-p}$, when $v=10$, what does $p$ equal?"   Everything else was just bloated text.   The question was testing your ability to not be distracted by that flood of irrelevant information.
You passed.
