Two Plus Two Newer Archives Likely dumb vig question
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#1
11-30-2007, 05:47 PM
 Daliman Senior Member Join Date: Sep 2002 Location: Wongs are two things, (at least). Posts: 10,376
Likely dumb vig question

Let's say I'm making a bet with a friend, and we obv split the vig. If the game is -110 both sides, pretty obv an even \$\$\$ bet, but if it is a higher number like say -300/=240, is it as simple as going to the middle? It's seems I have heard something somwhere where going to the middle actually infers a slight edge to one side, ( I think the negative?) as the higher negative must be offset by a higher spread or some crap. EIther way, was just wondering, flame if needed.
#2
11-30-2007, 05:59 PM
 centris Senior Member Join Date: Jul 2006 Posts: 232
Re: Likely dumb vig question

Since the relationship between the price and its associated probability is not linear the mean of the two prices is not the mean of their associated probabilities. So no you cannot just take the average of the two.

For each price calculate the associated probability and then take the mean of those two probabilities and translate this back to a line.

Probability asssocaited with -300= 300/400=0.750

Probability associated with -240= 240/340=0.706

(.75+.706)/2=.728

Line= 100*.728/(.728-1)=-268 (versus -260 by just taking the average)
#3
11-30-2007, 06:03 PM
 Thremp Senior Member Join Date: Nov 2005 Location: Free Kyleb Posts: 10,163
Re: Likely dumb vig question

http://forum.sbrforum.com/players-ta...ical-hold.html

No-vig line isn't the true line at higher prices.
#4
11-30-2007, 07:28 PM
 Daliman Senior Member Join Date: Sep 2002 Location: Wongs are two things, (at least). Posts: 10,376
Re: Likely dumb vig question

Cool, ty for the info guys. Good to know i'm only a partial moron.
#5
11-30-2007, 07:30 PM
 Daliman Senior Member Join Date: Sep 2002 Location: Wongs are two things, (at least). Posts: 10,376
Re: Likely dumb vig question

[ QUOTE ]
Since the relationship between the price and its associated probability is not linear the mean of the two prices is not the mean of their associated probabilities. So no you cannot just take the average of the two.

For each price calculate the associated probability and then take the mean of those two probabilities and translate this back to a line.

Probability asssocaited with -300= 300/400=0.750

Probability associated with -240= 240/340=0.706

(.75+.706)/2=.728

Line= 100*.728/(.728-1)=-268 (versus -260 by just taking the average)

[/ QUOTE ]
Ty for the work, but wouldn't the "average" be -270?

Can I also infer that the neg side will ALWAYS need to be a bit lower?
#6
11-30-2007, 08:06 PM
 Performify Senior Member Join Date: Aug 2004 Location: Sports Betting forum Posts: 3,847
Re: Likely dumb vig question

Yes, the average would be -270. Probably just a typo.

For those following along at home, you can do all the math above with a Moneyline Converter, for example http://www.smartcapper.com/tool_mone...converter.html

Just find the two percentages for the two lines and take the average of those (add them, divide by two, ldo).

-P
#7
11-30-2007, 08:28 PM
 Thremp Senior Member Join Date: Nov 2005 Location: Free Kyleb Posts: 10,163
Re: Likely dumb vig question

http://forum.sbrforum.com/players-ta...tml#post112449

This is what I was looking for.

While the above posts are a step in the correct direction, they aren't correct.

Buchdahl's book touches on this subject in the opening paragraph.
#8
12-01-2007, 12:13 AM
 hedgie43 Senior Member Join Date: Sep 2006 Location: Rugby Heaven Posts: 382
Re: Likely dumb vig question

Thremp,

If the line is efficient, how would you find the true line at higher prices then?
#9
12-01-2007, 12:16 AM
 Thremp Senior Member Join Date: Nov 2005 Location: Free Kyleb Posts: 10,163
Re: Likely dumb vig question

[ QUOTE ]
Thremp,

If the line is efficient, how would you find the true line at higher prices then?

[/ QUOTE ]

Guess? I have not really gotten that far. Though from the research and economic arguments I've seen (And anecdotal evidence), it pretty clearly points to large MLs not being accurately represented by the no vig line. Which ironically may actually make a bastardized version like the one presented above or a true average more useful.
#10
12-01-2007, 11:58 AM
 centris Senior Member Join Date: Jul 2006 Posts: 232
Re: Likely dumb vig question

[ QUOTE ]
Since the relationship between the price and its associated probability is not linear the mean of the two prices is not the mean of their associated probabilities. So no you cannot just take the average of the two.

For each price calculate the associated probability and then take the mean of those two probabilities and translate this back to a line.

Probability asssocaited with -300= 300/400=0.750

Probability associated with -240= 240/340=0.706

(.75+.706)/2=.728

Line= 100*.728/(.728-1)=-268 (versus -260 by just taking the average)

[/ QUOTE ]

Oh yeah, gosh that is embarrassing I made two mistakes here. If you take the average it should just be -270. And then as Thremp wrote this is a totally bastardized calculation. Just go to the linked page and see the correct calculation. But for the sake of comparison:

Probability asssocaited with -300= 300/400=0.750

Probability associated with +240 = 100/340=0.294

Probability of fav winning = .75/(.75+.294)=0.718

Vig-free line = -255

So there is actually a fairly large difference:
Average -270
My bastardized calculation -268
Correct vig free calc -254

As to which one actually works the best I have no clue.

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