Maths question. - Page 2

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