I would like to bet on NHL games, but first I want to find out the
odds. If there are 15 hockey games in one night, and you have to pick
a winner from each game, what are the odds that you will get 1/15
games right, 2/15 games right,... 15/15 games right. I believe the
total number of possibilities is 2 to the power of 15, but I can't
figure out how many combinations there are for each.

If you could reply in list form, that would be great.

IE:

1/15 = 15/32,768
2/15 = x/32,768
...
14/15 = x/32,768
15/15 = 1/32,768

Please explain how you got your answer.

Thanks for your time.

Hi egerbeaver,

Presuming an equal chance that you will win or lose each of your bets,
the odds would be:

0/15 = 1/32768
1/15 = 15/32768
2/15 = 105/32768
3/15 = 455/32768
4/15 = 1365/32768
5/15 = 3003/32768
6/15 = 5005/32768
7/15 = 6435/32768
8/15 = 6435/32768
9/15 = 5005/32768
10/15 = 3003/32768
11/15 = 1365/32768
12/15 = 455/32768
13/15 = 105/32768
14/15 = 15/32768
15/15 = 1/32768

"The formula for finding the number of combinations of k objects you
can choose from a set of n objects is:

            n!
n_C_k = ---------- 
        k!(n - k)!

"Permutations and Combinations"
The Math Forum @ Drexel
http://mathforum.org/dr.math/faq/faq.comb.perm.html

In essence, your question asks how many combinations of k objects you
can choose from n objects (and what the ratio of that number of
combinations is to the total number of combinations).  There are 15
objects here -- the 15 hockey games.  You "choose" 0 games if you win
none of the bets; you "choose" 1 game if you win 1 bet; etc.

There is only 1 way to choose 0 games -- by losing every bet.  [
15_C_0 = 15! / 0! * (15-0)! = 1 ] There is likewise only 1 way to
choose 15 games -- by winning every bet.  [ 15_C_15 = 15! / 15! *
(15-15)! = 1]

There are 15 ways to choose 1 game -- you could win only game 1, only
game 2, and so on up to only game 15.  [ 15_C_1 = 15! / 1! * (15-1)! =
15! / 14! = 15 ]  There are likewise 15 ways to choose 14 games -- by
losing only game 1, losing only game 2, and so on up to losing only
game 15.  [ 15_C_14 = 15! / 14! * (15-14)! = 15! / 14! = 15 ]

You will find that the number of combinations for choosing 0, 1, and
up to 15 games are arranged according to the 15th row of Pascal's
Triangle.

"Pascal's Triangle to Row 19"
The Math Forum @ Drexel
http://mathforum.org/dr.cgi/pascal.html

"Pascal's Triangle"
The Math Forum @ Drexel
http://mathforum.org/dr.math/faq/faq.pascal.triangle.html

The total number of combinations -- winning 0, 1, and up to 15 games
-- is indeed 2 to the 15th power, or 32,768.  This is equal to the
numbers in the 15th row of Pascal's Triangle.  The odds given above
have the numbers from that row (i.e., the numbers calculated by the
formula for n_C_k) as numerators, and the total number of combinations
as the denominator.

See also:

"Introduction to Probability"
The Math Forum @ Drexel
http://mathforum.org/dr.math/faq/faq.prob.intro.html

Please let me know if you need any clarification.

- justaskscott


Search strategy --

Searched on Google for:

combinations permutations

Awesome! Exactly what I was looking for. Thank you very much.

That is a great answer if you are just tipping blindly, but anyone
following the league should be able to improve his chances by
considering the realistic chances of top teams beating bottom teams.

i don't really follow hockey (heresy in the northeast u.s., so don't
tell anyone), so i've never bet on it, but i assume like football or
baseball, you have to either give odds, or bet on a point spread, so
unless you know the league better than the market does, 50% odds are
probably a reasonable guess.

-cab

I don't follow hockey either but is there not a chance of a draw. If
so there are 3 possible outcomes.

Thanks for the comments. You are right that you can improve your
chances by knowing which teams are better than others. I just wanted
an approximation of how well I was doing. There was a girl who won
$444,000 on a $5 bet by picking all 13 football games for the week
right. I figured I would give it a shot.

There are a few different ways to bet on hockey. You can bet on
individual games (minimum 3) (win, lose, or tie), and you can bet
whether the goal total will be over or under the average (5.5).

On the weekends, they have have a pool where you pay $5 and pick the
winner out of 15 games. Who ever gets the most right, wins the pot. If
multiple people get the same amount of games right, they share the
pot. There are no ties in this game because if the game goes past
overtime, they have a shootout to determine the winner.