Probability. Fair dice with $n>6$ sides. Who wins? I am fairly new to statistics so bear with me.
Two players are playing a game with a fair dice that has $n>6$ sides. The sides are labeled $1,2,3,..,n$.
The one who throws a larger number wins but if the throws are equal then player one wins. What are the chances that player two wins? (the probability is a function of $n$)
I'm completely lost as to how to start. If anyone could give me a hand it would be greatly appreciated.
 A: We have two players. They have been given the labels one and two. I would rather call them Alicia and Beti.
Alicia wins if she gets a greater number than Beti, or if she and Beti get the same number. We want to find the probability that Beti wins.  In order for Beti to win she must get a number larger than Alicia's.
We may suppose that Alicia tosses first (it doesn't matter). Whatever she gets, the probability Beti matches it is $\frac{1}{n}$.
So the probability the dice don't match is $1-\frac{1}{n}$, or equivalently $\frac{n-1}{n}$. By symmetry, the probability the dice don't match and Beti has the larger number is $\frac{n-1}{2n}$.  
Alicia has an edge in this game, since she also wins if the dice match. Her probability of winning is $\frac{n+1}{2n}$.
A: Hint: If the two throws are unequal, then the probability of player one winning is $1/2$ (Why?). The only other way player one can win is if the two throws are equal. Can you calculate that probability?
A: If the throws are different, then I claim that the chances of player $2$ winning is $1/2$. What's more, $P(\text{throws are different}) + P(\text{throws are the same}) = 1$. So, if we can work out $P(\text{throws are the same})$ then we can work out your answer. So, how many different ways can the throws be the same and what is the probability of each of these different outcomes?
Well, if player $1$ throws $i$ and player $2$ throws $i$ then they are the same, and this happens with probability $1/n \times 1/n = 1/n^2$. What's more, there are $n$ possible vales that $i$ can take, that is, $i\in\{1, 2, 3, \dots, n\}$. So, $$P(\text{throws are the same}) = n \times \frac1{n^2} = \frac1n\quad \text{ and } \quad P(\text{throws are different}) = \frac{n-1}n$$ and finally $$P(\text{player 2 wins}) = \frac{n-1}{2n}.$$
