Maths question.

On June 04 2015 10:29 Romm3l wrote: Show nested quote + On June 03 2015 21:46 Floofy wrote: Here is my take on this. Me and baal are both right 75% of the time. If we tend to have the exact same bets 100% of the time because we think alike, then the odds will simply remain 75% when we agree. If we tend to disagree a lot, this means that when we don't agree together, the odds of being right are lowered, and therefore when we agree, odds should be greater than 75%. This would mean that, if you can find 2 winning sport betters, who tend to disagree together a lot, if both of them shares the same bet, it should be a great bet. correct? this is right, well done. if we can model each of your thought process as a random variable which will tell you which side to bet on, and gets the side correct with a known 75% probability, then getting two independent signals in the same direction suggests that direction is now correct with 93.75% probability. there's a pretty cool model for rational herding behaviour in the stock markets based on these kinds of ideas. If you think others make trading decisions based on their 'private signals' which are just as likely to be correct as your private signal (>50% probability of being correct) and you observe two people in front of you betting on A but your private signal says B, you will rationally decide to ignore your own private signal and bet on A as well, since two people are more likely to be right than one (you). But the person behind you will observe three people betting on A and will be even more likely to bet A as well, and so on in a line, causing an information cascade.

Yep Tom is incorrect, I find that an easy way to see these kind of problems is to try a different scenario with more extreme odds.

Lets say a guy guesses the correct answer 99% of the time, another one guesses INCORRECTLY 99% of the time.

When we have this information the odds of any of them being wrong isnt 1% or 99% but 50%

On June 04 2015 22:23 Floofy wrote: In a more concrete approach, lets say we have 2 LPers with 60% win rate at sport betting 60=xy+0.5 *(100-x) This gives us: y(x) = 10/x + 0.5 So essentially, if odds both guys agree together = 60% of the time, they will win 66.6666% of the time when they do agree together. Doesn't seem like a super huge gain in this case. But moral of the story is, if a lot of pro gamblers think a line is really good, its probably not a good idea to bet against them (unless that line changes a lot obviously).

I dont get how did you assume that formula, the combined probability they agree and win plus half the probability they disagree? wut

btw dont use percentages in formulas especially if you are not using the symbol, if you were to substitute x or y for integers their multiplication would result in an extra 0, use decimal form.

surely it depends on what information you use to get your answer. if 10 people use the same information to think a team is going to win with 75% certainty then it will probably be close to 75%, if you're using completely independent information to come to your conclusion, ie 1 person uses past statistical data only to make their decisions, another person uses only player form to make their decisions etc etc then theoretically 2 people would be closer to the 93.75%. Obviously in reality people use similar but not identical information but there will be some differences so the more "75% winning sports bettors" agree on a bet the higher the chance of winning will be but it wont be going from 75% to 93.75% when two players agree on a bet, more realistically 75 to something like 78 or 82 depending on how much information is independent.

Sorry for not answering this patricularly mathematically but I believe in real world application this makes the most sense.

However, paragraph one first example assumes there is no overlap between the variables, so past statistica data would have to have no overlap with player form which is clearly unlikely to be the case, so in reality two sports bettors agreeing on a bet would never really increase winning chances to 93.75%

In reality I believe if 10 sports bettors agree on a bet it is still unlikely to be > 90% to win (or certainly not close to 98% or anything like that!!!!), for a lot of quite obvious reasons touched on here. I think I can prove this mathematically if anyone disagrees

edit: Ah, I didn't read responses, I feel like everyone said what I said much clearer. Also interesting point floofy with overlapping bets with sports bettors you disagree with regularly.

someone take me to this magical land of floofy where 10 verifiably profitable sportsbettors ever share their picks about anything, let alone the same event, and 75% is an attainable winning percentage (assuming adjusted for odds, otherwise it's meaningless as anyone can win 75% of the time betting -400 favorites and still be a losing bettor).

On June 05 2015 15:03 bigredhoss wrote: someone take me to this magical land of floofy where 10 verifiably profitable sportsbettors ever share their picks about anything, let alone the same event, and 75% is an attainable winning percentage (assuming adjusted for odds, otherwise it's meaningless as anyone can win 75% of the time betting -400 favorites and still be a losing bettor).

You are right 75% is most likely not possible to achieve. This is only theory.

As i proved, 2 winners of just 55%, actually does not gain much from sharing.
But what if 10 winners of 55% shared? that might actually give a good number.

On June 05 2015 15:03 bigredhoss wrote: someone take me to this magical land of floofy where 10 verifiably profitable sportsbettors ever share their picks about anything, let alone the same event, and 75% is an attainable winning percentage (assuming adjusted for odds, otherwise it's meaningless as anyone can win 75% of the time betting -400 favorites and still be a losing bettor).

it's called a hypothetical situation

and the world in which the hypothetical takes place sounds nice so i would like someone to take me there -.-

edit: no actually i need to think this through again

Oops - We've all done something very silly in accepting this 93.75% number in op.

If the probability a bettor gets the wrong signal is 25%, the probability of two independent bettors getting the wrong signal is indeed 25%^2. But doing [1 - 25%^2] (=93.75%) just gives us the probability both bettors getting the wrong signal doesn't happen (which includes outcomes where bettors don't agree on which side to bet). What we're trying to get is the probability a certain direction is right given both have agreed on it, and we haven't done that.

P(both get it right) = 75%^2 = 0.5625
P(both get it wrong) = 25%^2 = 0.0625
P(bettors disagree) = 2*75%*25% = 0.375

P(both right given both agree) = 0.5625 / (0.5625+0.0625) = 0.9


Another way to get the same result is using Bayes formula, starting with prior probability of a bet being correct being 0.5 (based on those close odds), then updating that probability with the information from your signal (to get 0.75) then updating it again with the new information from your bettor friend's independent signal (to get to our 0.9 result)

On June 06 2015 11:49 Romm3l wrote: Oops - We've all done something very silly in accepting this 93.75% number in op. If the probability a bettor gets the wrong signal is 25%, the probability of two independent bettors getting the wrong signal is indeed 25%^2. But doing [1 - 25%^2] (=93.75%) just gives us the probability both bettors getting the wrong signal doesn't happen (which includes outcomes where bettors don't agree on which side to bet). What we're trying to get is the probability a certain direction is right given both have agreed on it, and we haven't done that. P(both get it right) = 75%^2 = 0.5625 P(both get it wrong) = 25%^2 = 0.0625 P(bettors disagree) = 2*75%*25% = 0.375 P(both right given both agree) = 0.5625 / (0.5625+0.0625) = 0.9 Another way to get the same result is using Bayes formula, starting with prior probability of a bet being correct being 0.5 (based on those close odds), then updating that probability with the information from your signal (to get 0.75) then updating it again with the new information from your bettor friend's independent signal (to get to our 0.9 result)

None of this make any sense

You are assuming both betters are completly independant, a bit like TalentedTom.
In reality, you can't know how often they will both get it right, because it depends on how often they agree.

They could agree 100% of the time, 90% of the time, 80% of the time, etc. There is no way to know this.
I also think this would depend on the actual fights happening.

yes you're right i am assuming the necessary assumptions to arrive at the 93.75% number you got in op, where we have independence: P(player1 gets it wrong) = P(player1 gets it wrong given player2 also gets it wrong) = P(player1 gets it wrong given player2 gets it right).

93.75% = 1-P(both wrong) = P(both right) + P(both disagree)

The only point of my post is that your 93.75% number is wrong given this assumption

On June 06 2015 12:03 Romm3l wrote: yes you're right i am assuming the necessary assumptions to arrive at the 93.75% number you got in op, where we have independence: P(player1 gets it wrong) = P(player1 gets it wrong given player2 also gets it wrong) = P(player1 gets it wrong given player2 gets it right). 93.75% = 1-P(both wrong) = P(both right) + P(both disagree) The only point of my post is that your 93.75% number is wrong given this assumption

oh ok i understand now.

My 93.75% number wasn't a serious number tho.

On June 05 2015 05:42 Baalim wrote: Show nested quote + On June 04 2015 22:23 Floofy wrote: In a more concrete approach, lets say we have 2 LPers with 60% win rate at sport betting 60=xy+0.5 *(100-x) This gives us: y(x) = 10/x + 0.5 So essentially, if odds both guys agree together = 60% of the time, they will win 66.6666% of the time when they do agree together. Doesn't seem like a super huge gain in this case. But moral of the story is, if a lot of pro gamblers think a line is really good, its probably not a good idea to bet against them (unless that line changes a lot obviously). I dont get how did you assume that formula, the combined probability they agree and win plus half the probability they disagree? wut btw dont use percentages in formulas especially if you are not using the symbol, if you were to substitute x or y for integers their multiplication would result in an extra 0, use decimal form.

ok i will try to explain the formula.

60 (percentage of being right for both guys) = x(% of the time both guys agrees) * y (% of the time they win when they agree)+ 0.5 (if they disagree, then their combined chance to be right is 50%) * (100 -x) (this is % of the time both guys disagree)

Ok i have a second question

let's assume a marginal winning sport better, at 55%, named BAAL
He has a friend who's a big winner, at 75%

Now, lets say the friend is nice and shares all his picks with Baal.

Obviously, Baal's optimal strategy is probably to copy everything and get a 75% win rate.

But what if he decides to only copy the picks he agrees with. What will his new win % be?

78.57% assuming independence again

On June 06 2015 22:08 Romm3l wrote: 78.57% assuming independence again

There's no independence thought...

On June 07 2015 03:56 Floofy wrote: Show nested quote + On June 06 2015 22:08 Romm3l wrote: 78.57% assuming independence again There's no independence thought...

if they hold the same info then the winrate can never go over 75%

On June 06 2015 12:38 Floofy wrote: Show nested quote + On June 05 2015 05:42 Baalim wrote:   On June 04 2015 22:23 Floofy wrote: In a more concrete approach, lets say we have 2 LPers with 60% win rate at sport betting 60=xy+0.5 *(100-x) This gives us: y(x) = 10/x + 0.5 So essentially, if odds both guys agree together = 60% of the time, they will win 66.6666% of the time when they do agree together. Doesn't seem like a super huge gain in this case. But moral of the story is, if a lot of pro gamblers think a line is really good, its probably not a good idea to bet against them (unless that line changes a lot obviously). I dont get how did you assume that formula, the combined probability they agree and win plus half the probability they disagree? wut btw dont use percentages in formulas especially if you are not using the symbol, if you were to substitute x or y for integers their multiplication would result in an extra 0, use decimal form. ok i will try to explain the formula. 60 (percentage of being right for both guys) = x(% of the time both guys agrees) * y (% of the time they win when they agree)+ 0.5 (if they disagree, then their combined chance to be right is 50%) * (100 -x) (this is % of the time both guys disagree)

Oh that was my question I didnt know where the 0.5 came from it makes it confusing when you mix percentages and decimals

On June 07 2015 03:56 Floofy wrote: Show nested quote + On June 06 2015 22:08 Romm3l wrote: 78.57% assuming independence again There's no independence thought...

yeah sure, it's just a useful benchmark to show the best-case scenario. With this result we know the true answer is likely to lie in the interval [75%, 78.57%] and the extra winrate the weak bettor gets probably isn't enough to justify having to miss out on a lot of good bets because his 55% signal didn't agree with the stronger bettor's 75% signal (which happens over half the time under the benchmark independence case).

Ofcourse if they agree only 30% of the time with these winrates then they win 100% when they agree, as in your example before. I might have struggled to see how disagreeing more than the amount predicted by independence while maintaining the same winrates would actually be possible in the real world, but i think your example of each player making non-overlapping mistakes due to systematic biases is a great one to highlight that. However I'm skeptical about the degree to which this is useful in the real world. true winrates are neither observable nor constant, and it could be a 'winning' bettor you frequently disagree with has just been running good (or you have been)

But I doubt that its a bad idea to follow it even if the edge gain in average is minimal