What is the probability distribution of the net payoff of my sports bets? I am placing 6 sportsbets this weekend each with a probability of 50% of winning and 50% of losing.
I would like to know how to figure out the probability of the following outcomes:
a) winning 0 bets
b) winning 1 bet
c) winning 2 bets
d) winning 3 bets
e) winning 4 bets
f) winning 5 bets
g) winning 6 bets
I can figure out that 0 and 6 are both the same at 1/64 or 1.56% and that winning 3 is 50%.  Is this right and how do I figure out the others?  
 A: There are exactly
$$\binom 6n = \frac{6!}{n!(6 - n)!}$$
combinations of $n$ elements chosen from the $6$ total elements. As you already noted, there are in total $2^6 = 64$ combinations.
Assuming that each bet is independent from each other, it follows that the probability of winning $n$ bets is
$$p(n) = \binom 6n \left(\frac12\right)^6.$$
So we have that
$$\begin{align}
p(0) = p(6) = \frac1{64}\\[0.6em]
p(1) = p(5) = \frac3{32}\\[0.6em]
p(2) = p(4) = \frac{15}{64}\\[0.6em]
p(3) = \frac{20}{64} = \frac5{16}.
\end{align}$$
A: Check out the $6$th row in Pascal Triangle:
$$\binom60,\binom61,\binom62,\binom63,\binom64,\binom65,\binom66$$
Or equivalently:
$$1,6,15,20,15,6,1$$
Your probabilities are:
$$\frac{\binom60}{2^6},\frac{\binom61}{2^6},\frac{\binom62}{2^6},\frac{\binom63}{2^6},\frac{\binom64}{2^6},\frac{\binom65}{2^6},\frac{\binom66}{2^6}$$
Or equivalently:
$$\frac{1}{64},\frac{6}{64},\frac{15}{64},\frac{20}{64},\frac{15}{64},\frac{6}{64},\frac{1}{64}$$
A: The probability of winning exactly three bets is $${6\choose 3}\left(\frac{1}{2}\right)^6=\frac{20}{64}=31.25\%$$
In this expression ${6\choose 3}$ is a binomial coefficient that counts how many ways there are to choose which three of the six bets you win.
