I'm just curious how many parlay bets can be made with 16 games. A parlay bet is just betting on multiple games as 1 bet. So if there are only 3 games (2 teams), then there can only be 3 parlay bets like the following:
Games:
A v B
C v D
E v F
So the parlays are:
1. (A, C, E)
2. (A, C, F)
3. (A, D, E)
4. (A, D, F)
5. (B, C, E)
6. (B, C, F)
7. (B, D, E)
8. (B, D, F)
9. (A, C)
10. (A, D)
11. (A, E)
12. (A, F)
13. (B, C)
14. (B, D)
15. (B, E)
16. (B, F)
17. (C, E)
18. (C, F)
19. (D, E)
20. (D, F)
I haven't done this kind of math in years so I'm not sure if this correct. Is my approach correct? Here is my thought process:
SUM((2 ^ # of games in parlay) * (games not in parlay))
So the example above holds since (2^3 * 3 choose 0) + (2^2 * 3 choose 1) = (8 * 1) + (4 * 3) = 20
So for 16 games it would be:
(2 ^ 16) * (16 choose 0) +
(2 ^ 15) * (16 choose 1) +
(2 ^ 14) * (16 choose 2) +
.
.
.
Is this correct?
Yes, you're right, but your formula can be simplified considerably.
$$\begin{align} \sum_{k=2}^n\binom{n}{n-k}2^k&=\sum_{k=0}^n\binom{n}{n-k}2^k-2\binom{n}{n-1}-1\\ &=(1+2)^n-2n-1\\ &=3^n-(2n+1) \end{align}$$ where we have used the binomial theorem to simplify the sum.
When $n=3$, this gives $3^3-7=20$, in conformity with your example.
EDIT
Since you say you're out of practice, here's another way to look at it. How many ways are there to bet on the games. For each game, we have $3$ choice: bet on the home team, bet on the visiting team, or don't bet on the game. That gives $3^n$ choices for all the games, but we want to exclude the cases where we bet on fewer than two games. There's only one way to bet on none of the games (don't place a bet) and there are $2n$ ways to bet on one game ($n$ ways to pick the game and $2$ ways to pick the team). That make $3^n-(2n+1)$ as above.