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#1
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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.
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#2
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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
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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
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Re: Likely dumb vig question
Cool, ty for the info guys. Good to know i'm only a partial moron.
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#5
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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
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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
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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
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Re: Likely dumb vig question
Thremp,
If the line is efficient, how would you find the true line at higher prices then? |
#9
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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. |
#10
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Re: Likely dumb vig question
Yeah, I wish I had the time to do research on this biz :/
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