Expected value of winning a dice game I am playing a dice game with another person, the dice are not tweaked in any way. 
The game says that if I throw a bigger or equal number to him I win, otherwise I lose. The game is played by placing a bet of 1 dollar and if I win I receive 2 dollars otherwise the dollar is lost.
How can I calculate the expected value of my winning?
 A: In $6$ cases, the dice roll equally, and you win. In the other $6\times6-6=30$ cases, by symmetry, you’ll win half the time, and lose half the time. Overall, this gives $21$ winning cases, for a probability of $$\frac{21}{6\times6}=\frac{7}{12}$$ of winning, and a probability of $$\frac{15}{6\times6}=\frac{5}{12}$$ of losing. The expected value of the money you’ll win is therefore $$\$\left(\frac{7}{12}-\frac{5}{12}\right)=\boxed{\$\frac{1}{6}.}$$
A: The expected profit from playing the game is
$$P\left(\text{win}\right)\left(\text{net profit on win}\right) + P\left(\text{loss}\right)\left(\text{net profit on loss}\right)\tag{1}$$
There are $6$ possible rolls that your opponent can make, each with probability $1/6$. If he rolls a $1$, you have a $6/6$ probability of winning.  If he rolls a $2$, you have a $5/6$ probability of winning. If he rolls $n$, you have a $(7-n)/6$ chance to win. 
Therefore, your overall probability of winning is 
$$P\left(\text{win}\right) = \frac{1}{6}\left(\frac{6}{6} + \frac{5}{6} +\frac{4}{6} +\frac{3}{6} +\frac{2}{6} +\frac{1}{6}\right) = \frac{21}{36} = \frac{7}{12}$$
The probability of losing is $$P\left(\text{loss}\right) = 1 - P\left(\text{win}\right) = \frac{5}{12}$$

When you win the game, the net profit is \$$1$. When you lose the game, the net profit is $-$\$$1$. The expected profit from playing one round of this game, by $(1)$, is therefore
$$P\left(\text{win}\right)\left(\text{net profit on win}\right) + P\left(\text{loss}\right)\left(\text{net profit on loss}\right) = \frac{7}{12}\left(\$1\right) + \frac{5}{12}\left(-\$ 1\right) = \boxed{\$0.1\overline{6}}$$
