Sports Gambling odds question?

[Phishy McFish]

With the 17 week schedule in the NFL.....and assuming the individual has a 50/50 chance of picking a game right against the spread.....what are the odds they will be able to do this?

Pick one 3 team wager correctly
Pick one 2 team wager correctly
& Pick one single game wager correctly

They would only be allowed to make one of the attempts each week (so they couldn't try for the 3-teamer the same week they were trying for the 2-teamer).

If they fail in a week, they can still try the next week until the regular season is over.....they must complete all 3 by end fo week 17 to succeed.

[Phishy McFish]

[img]/images/graemlins/smirk.gif[/img]

[geek]

Just a swing at this...

To simplify, I'll assume you have to win the three game ticket first, followed by the two game ticket, then the single game.
The possible ways to lose for the season are to:
(1) Never make the 3 game ticket.
(2) Make the 3 game ticket, but fail to make the 2 game ticket.
(3) Make the 3 and 2 game tickets, but fail to make the individual game.
Sum the (1), (2) and (3) together for your probability of failure.

(1) NEVER MAKE THE THREE GAME TICKET:

0.875^17 = 0.1033

(2) MAKE THE THREE GAME TICKET BUT FAIL TO MAKE THE TWO:

SUM(from n = 1 to 17): 0.875^(n-1) * 0.125 * 0.75^(17-n)

RESULT = 0.0958
[note: n represents the week in which the 3 game ticket was made]

(3) MAKE THE THREE AND TWO, BUT FAIL ON THE SINGLE GAME:

(a) make 3 game ticket on week 1:

0.125 * 0.875^0 * SUM(from n = 2 to 17): 0.75^(n-2) * 0.25 * 0.5^(17-n)

[note: here n represents the week in which the 2 game ticket was made]

(b) make 3 game ticket on week 2:

0.125 * 0.875^1 * SUM(from n = 3 to 17): 0.75^(n-3) * 0.25 * 0.5^(17-n)

.
.
.

(p) make 3 game ticket on week 16:

0.125 * 0.875^15 * SUM(from n = 17 to 17): 0.75^(n-17) * 0.25 * 0.5^(17-n)

I don't have a chance to solve for (3) right now [lunch over], but it shouldn't take too long. I'll probably have a little time to complete this afternoon.

[aloiz]

Don't know a real good way of figuring this out without going through each possibility, but my estimate is around 74%.

aloiz

[geek]

I get a probability for case(3) of 0.0614.
Hopefully I caught all the logical branches and (more importantly) didn't make typo errors.

I get:
ProbSuccess = 1 - 0.1033 - 0.0958 - 0.0614
(or 74%)

How did you do that Aloiz?

[Phishy McFish]

.....but since you are calculating over my head, this does take into account that a 3-teamer "ticket" needs to be right on all 3 games....I'm figuring the chance of winning just a 3-teamer ticket in a season would be 17 times the odds of flipping a coin and it being HEADS all 3 times.

[geek]

Just to make sure I'm on the same page as you, are you saying there is a 12.5% chance that the three team parlay will be successful on any given week? That's the way I see it.

[pudley4]

[ QUOTE ]
I'm figuring the chance of winning just a 3-teamer ticket in a season would be 17 times the odds of flipping a coin and it being HEADS all 3 times.

[/ QUOTE ]

This is obviously not correct. Here's why:

The odds of getting 3 heads in a row is 1/2 * 1/2 * 1/2 = 1/8. The probability of hitting at least once in 17 tries is not 17 * 1/8 = 2.125 (The probability of anything happening can never be greater than 1)

[Shawsy]

Okay, now I think I get the question.

If you have a 50/50 chance of picking any one game correctly against the spread, then the chances of:
Winning a 3-game ticket is (0.5)^3=0.125 (1 in 8)
Winning a 2-game ticket is 0.5)^2=.25 (1 in 4)
Winning any one game bet is 0.5 (1 in 2)

If n1 is the number of weeks it takes to win on the 1-game ticket, n2 is the number of weeks it takes to win on the 2-game ticket, and n3 is the number of weeks to win on the 3-game ticket, then you want the probability that n1+n2+n3 is 17 or less.
I can work the probability that any one of these tasks is completed in x weeks, but I have a tough time trying to add them together. Individually they follow geometric distributions,
E[n1]=2 with a variance of 2.
E[n2]=4 with a variance of 12.
E[n3]=8 with a variance of 56.
So you could expect to accomplish all three tasks in 14 weeks, but I don't know how to figure the probability that all 3 would be completed in 17 weeks.
(Actually, I could work it out, but there must be a better way than the one I am thinking of.)

[Phishy McFish]

That sure is rather obvious.....and I work with statistics in my job every day.....sigh.

So I'll need to open my old text books.....