Dice Roll Probabilities I'm trying to figure out the probabilities for the following casino game:
You and the dealer each roll a pair of dice and the person with the highest individual die roll wins. If its a tie, you win. 
First, what is the probability you win?
Second, given that you've won, whats the probability that the game resulted in a tie?
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Here are my thoughts:
For the first - the expected value of your die roll is 4.25; which means you will win unless the dealer rolls a 5 or 6, giving you a 2/3 chance of winning.
For the second I'm not sure how to think about it. 

Denote $a,b$ are the results of my dices with $6 \ge a>b \ge 1$ We consider $6$ cases:


*

*If the largest result of my dice is $1$, or $a=1$ then $b=1$. Hence, the dealer's results must less than or equal to $1$. We obtain $1$ way to determine the dealer's results such that the condition is satisfied. Thus, there are $1 \times 1$ ways that I won when $1$ is my largest result.

*If $a=2$ then $b=1$ or $b=2$. Now we need to determine the dealer's results so that I could win. This means, the largest result of the dealer is $2$ or $1$. There are $2$ ways that the dealer gets the same result in two dices ($11$ and $22$). We consider the case that the result in two dices is different. We get $\frac{(2-1) \times 2}{2}=1$ ways. This means the dealer has $1+2=3$ ways to get the result. Therefore, there are $2 \times 3$ ways so that I won when my largest result is $2$

*If $a=3$ then $b=1,2$ or $3$. Similarly, there are $3$ ways so that the dealer can get the same result in 2 dices ($11,22$ and $33$) and $\frac{(3-1) \times 3}{2}=3$ ways in order to get the different results in 2 dices. This means the dealer has $3+3=6$ ways to get the result. Hence, there are $6 \times 3$ ways so that I won when my largest result is $3$.

*We obtain the formula $n \times \left(n+ \frac{(n-1)n}{2} \right)=\frac{n^2(n+1)}{2}$, which is the number of ways that I won when my largest result is $n$.

*In conclusion, the probability to win is $$\dfrac{\sum_{n=1}^6 \frac{n^2(n+1)}{2}}{\left[ 6+ \frac{(6-1)6}{2} \right]^2}= \dfrac{\sum_{n=1}^6 \frac{n^2(n+1)}{2}}{21^2}$$

