Friday, December 29, 2006

If they Giants win on Saturday, just assume they are in the playoffs

If the Giants win their game on Saturday, they have a tremendously high chance of getting into the playoffs as a Wildcard. Here is the playoff scenario:

New York Giants can clinch playoff berth with:
1) NYG win + NYG clinch strength of victory tiebreaker over GB**, OR
2) NYG win + GB loss or tie, OR
3) NYG tie + GB loss or tie + STL loss or tie + ATL loss or tie + CAR loss or tie, OR
4) GB loss + STL loss + ATL loss + CAR loss

**N.Y. Giants clinch strength of victory tiebreaker over Green Bay if:

a) DET loses OR
b) MIN loses OR
c) any two of the following results occur: ARI loss, MIA loss, SF loss, CAR win, HOU win, TB win.

Here are all the ways the Giants make it into the playoffs if they do win:
- GB loses: at best, they have about a 59% chance of losing.
- if GB wins:
- DET lose: 87%
- or MIN lose: 55%
- the odds of both DET and MIN winning is (13% x 45%) = 6%. So there is a 94% chance that at least one of them will lose.

If these were the only scenarios, then:

41% (GB win) x 6% (DET and MIN both win) = 2.5%

Now, we have to multiply that by the chances that any two of the following results don't occur (ARI loss, MIA loss, SF loss, CAR win, HOU win, TB win). My head is already spinning by thinking how to calculate those chances. I'm sure your head is spinning by reading this far already too! In cases like this, I don't think it is necessarily to look too deep into the math. Just round it off. If you are betting on the Giants to get into the playoffs, be a bit conservative and assume if they win their game, they have a 98% chance of getting in. You won't be hurt if you make that assumption from an EV point of view. Taking the time to calculate the other numbers just isn't worth it for the extra value it could bring. What is the difference between 98% and 99% or 99.5% when it comes to making a bet? I can't think of any bet where that would make a difference in whether I make a bet or not.

No comments: