Calculate ev spread question

**Pancho sanza** · 02-11-11, 03:12 PM

Calculate the win probability using pinny's line, -121/110

Works out to .5348

So at -110 your edge would then be +2.1 %

**Winner_13** · 02-11-11, 03:31 PM

im ok at math ...but this is making me feel dumb

u divided negative121/110?

i tried 121/110/
121/210
121/221

all didnt get that answer

**bztips** · 02-11-11, 04:19 PM

I got a different answer than Pancho...

Assuming Pinny is the right line, prob of winning at -121 = -121/(-121 - 100) = .5475

which yields an edge of about 4.5% at -110

**Winner_13** · 02-11-11, 04:31 PM

yes...thats the answer i got.

I didn't think it was that simple.

The other side is +110??
It's more difficult because pinny isn't using a 20 cent line.

**Winner_13** · 02-11-11, 04:32 PM

so wouldn't it be different cause its not a standard 20 cent line?

**bztips** · 02-11-11, 04:36 PM

oops, my calculation is with no vig; Pancho is correct once you account for the vig
No vig prob at -121 = .5475
No vig prob at +110 = .4762
Total = 1.0237

Then .5475/1.0237 = .5348 = Pancho's answer

**Inspirited** · 02-11-11, 04:44 PM

I'll try to help. Hope this is the idea. If not someone will crush me and I will immediately surrender

.

For favorites -f/(-f+100) and for underdogs 100/(u+100)

Thus you get 121/(100+121) and 100/(110+100). Once you have these probabilities you need to add them together. You'll notice that they are great than 100%. This is the vig. You need to remove the vig.

The results were. 0.5475 and 0.4762. This sums up to 1.0237. Divide both by this and you get 0.5348 and 0.4652. Now you have the true probabilities of each team winning, and you can do a simple EV calculation.

.5348(100)-110(.4652) = 53.48- 51.17 = +2.31

For every $100 you bet you're expected to make $2.31. I suppose this is your caculated edge -> 2.31% ???

**Winner_13** · 02-11-11, 04:48 PM

thanks

i knew how to do it b4 with moneylines.

I got confused with the point spread.

**That Foreign Guy** · 02-12-11, 05:38 AM

So glad I use decimal odds.