TITLE: Maximizing probability of picking the dime
QUESTION [6 upvotes]: A money pouch contains a certain number of cents and only one dime. You and your friend are playing a game: They alternate turns and pick one coin at a time, which they put in their pockets. Whoever picks the dime wins. You are trying to decide whether it is better to play first or second: Initially you think that if you pick the dime in your first draw, you immediately win, while if you opt to play second, your probability to pick the dime increases. Which of the two options is more favorable for you? We assume that you can’t see the coins before taking them out of the pouch and also that you can’t tell from the size which coin is which.
My thought: Let's say we have $n$ coins in total. If I play first, I have probability of picking the dime: $\frac {1}{n}+\frac {1}{n-2}+\frac {1}{n-4}...$ while if I play second, I have $\frac {1}{n-1}+\frac {1}{n-3}+\frac {1}{n-5}...$. Clearly the second option gives a larger number.
Is this correct?
Thank you.

REPLY [4 votes]: The first essential point is that the dime has no preference for position, thus its Pr of being at any position is $\frac{1}{n}$
The second essential point is that we don't know whether $n$ is odd or even, and if it is odd, we get one extra try if we go first
Thus it is always preferable to go first