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I'll extend this contest another 24 hours (to 48 hours from when I first posted the question).Comment -
I think 6x^5-15x^4+10x^3-p=0 is the equation to solve...
Fitting the curve with the sigmoidal function I get this:
p=p(win)>=0.5 ... p(score first)=[-ln(1/p-1)/12]^0.825 - 0.5
p=p(win)<0.5 ... p(score first)=0.5-[ln(1/p-1)/12]^0.825
The results break down towards extremes 0 an 1 but are fairly accurate in the range p(win)[0.15, 0.85]Comment -
100%Comment -
this one may seem very unscientific, but it's a good approximation...
Because the function is symmetric around 0.5, 0.5 we can fit the 5th order polynom by two second order polynomials, first on range [0, 0.5] and second on (0.5, 1].
I get these coefficients for A, B and C
p=p(win) < 0.5 ... A=2.37388, B=-0.17162, C=0.00104
p=p(win)>=0.5 ... A=-2.37388, B=4.57615, C=-1.20330
The solutions for each of the ranges are
p(score first)=[-B+sqrt(B^2-4×A×(C-p(win))]/2A
...just plug in the corresponding coefficients.
RobkorComment -
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Is the real world inspiration for this question about deriving game moneylines from series' moneylines? Are you thinking a series price would maybe be more accurate than a game price? For advanced betting, anyway?
Or are you researching "Team A to win Series in X games" props?
I'm just wondering what your real world application is.Comment -
Grand slam tennis mach case!
If you know prob that one player will won match (take that from pinnacle odds) how calculate prob that he will won frst set?
Spectre, If best of five is too complex for us how about only best of 3 case!
Me seem that your numerical solution from post #24 is correct!Last edited by SBRAdmin3; 07-14-14, 02:30 PM.Comment -
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So to make things clear you're assuming probabilities of points scored are independent? If you're doing this for the relevant baseball prop you won't even get close. If you're doing it for soccer then independence might be okay but I imagine the prop would only be offered as a three way line, in which case your assumption of binary outcomes is violated.Comment -
Also TomG already answered this. If you're too lazy to work it out and plug it into a math program that will simplify the algebra as much as possible, which can then easily be put into excel, then you don't really deserve an answer.Comment -
Unfortunately I don't think I'm going to have time to figure this out before the deadline I just have too much going on before I leave for my trip, but I'll throw my thoughts as they are right now and hopefully it leads someone else to the full solution.
If you read through the posts so far you get a pretty good about how relate p to x, and I agree with what has been proposed as far as that's concerned. As Justin said, he doesn't want a look-up chart or answers for given p values. He would like a formula expressed in terms of the x for what many have already shown is:
p = 6x^5 - 15x^4 + 10x^3
He also stated that the chance to get a point remains constant throughout the match, so I don't think we need to over-complicate this. We would just like a general algebraic solution.
Now, as someone already pointed out, quintic functions provide no general algebraic solution (Abel–Ruffini theorem).
However, (and Justin has re-iterated this several times) on the range 0 < p < 1 the function loses the other local maxima/minima and inflection points (see below) and in actuality looks more like a cubic or logarithmic function (or at least could be expressed as such). Now outside this range the general solution would of course fall apart, which I think he already knows but is not concerned about.
Essentially we are looking for a piecewise solution for p on the interval (0,1) and nothing more.
To help with what I've written above, here is a graph of the quintic:
It is trivial to just use numerical methods or online calculators to get an actual solution, but to express what we care about in the general case will take some more figuring out and by looking at this graph I tend to agree with Justin that I believe it is possible for that given range.
The logic being applied in Robkor's posts above is what needs to be done (I have not verified his formulas or calculations however). But I believe if someone spent some time to get the exact formula required and expressed it as he has, that would be exactly what Justin is looking for.
I really wish I could spend some more time and research on this, but need to let it rest. Good luck guys.
PS-- love this kind of stuff Justin, please do another in 2 weeks!
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it's just the standard uniform distribution which has EV (a+b)/2 in this case where a=0, b =1 and variance (b-a)^2/12
this is how computers generate random numbers.
when you use this in your next book will i get co-author creditComment -
btw if you want real answer have to include things like correlation between 1st set/full game etc
from info can only look at the case where they're independent so prob(winning full game) = prob(winning a set)Comment -
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I disagree that you will find a piecewise cubic function to fit this fifth order equation exactly between 0 and 1.Unfortunately I don't think I'm going to have time to figure this out before the deadline I just have too much going on before I leave for my trip, but I'll throw my thoughts as they are right now and hopefully it leads someone else to the full solution. If you read through the posts so far you get a pretty good about how relate p to x, and I agree with what has been proposed as far as that's concerned. As Justin said, he doesn't want a look-up chart or answers for given p values. He would like a formula expressed in terms of the x for what many have already shown is: p = 6x^5 - 15x^4 + 10x^3 He also stated that the chance to get a point remains constant throughout the match, so I don't think we need to over-complicate this. We would just like a general algebraic solution. Now, as someone already pointed out, quintic functions provide no general algebraic solution (Abel–Ruffini theorem). However, (and Justin has re-iterated this several times) on the range 0 < p < 1 the function loses the other local maxima/minima and inflection points (see below) and in actuality looks more like a cubic or logarithmic function (or at least could be expressed as such). Now outside this range the general solution would of course fall apart, which I think he already knows but is not concerned about. Essentially we are looking for a piecewise solution for p on the interval (0,1) and nothing more. To help with what I've written above, here is a graph of the quintic: It is trivial to just use numerical methods or online calculators to get an actual solution, but to express what we care about in the general case will take some more figuring out and by looking at this graph I tend to agree with Justin that I believe it is possible for that given range. The logic being applied in Robkor's posts above is what needs to be done (I have not verified his formulas or calculations however). But I believe if someone spent some time to get the exact formula required and expressed it as he has, that would be exactly what Justin is looking for. I really wish I could spend some more time and research on this, but need to let it rest. Good luck guys. PS-- love this kind of stuff Justin, please do another in 2 weeks!
SpektreComment -
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I'm starting to think this is true. We can definitely get some real nice approximations and for the purposes of betting and odds would more than suffice. But an exact, general algebraic solution may not exist even on a closed interval of the quintic. I'm curious if we get an answer later on.Comment -
also btw,
how can something take any value in the open interval, which doesn't include the endpoints, (0,1) but not include 0 and 1?
in these cases
0<p<1 is the same thing as 0<=p<=1 so you can't exclude 0,1
unless you want to violate the assumption that P and Q are real numbers between 0 and 1Comment -
How exactly?
Here is the information given
a) 3 points
b) Prob winning entire game = p, losing = 1-p
c) P is contained in (0,1) for all reals
d) P can't be 0 or 1 which is wrong if c) is true
Nothing about a team that wins the 1st set is 90% to win the entire game, etc. You can't tell anything about set 1 from the game results because you don't know anything about how set 1 affects the final game. For all you know winning set 1 could make you more likely to lose the entire game than win.
The only valid answer without that information is to assume they are independent, in which case you just use the standard uniform distribution which implies their chance to win the entire game is equal to their chance to win the first set.Comment -
There's not going to be a closed-form solution that works for the set of all possible quintics, but solving it numerically or via excel solver is trivial.Comment -
How exactly? I don't think Justin7 intended this to be, but this thread is turning out to be a great example between the differences of probability and statistics.
Lets say we wanted to look at the the % of times a 1H under hits when the full game under hits (which is basically this problem). We would need to make some "statistical inference" based on data collected to determine what is the relationship between the 1H total and the full game total. We could do this by looking at the 1H scores based on the full game total and model the relationship between them. Then we can use that data to determine the relationship between them.
In this example there's no data or implied relationship between them so we can't just assume one, therefore the way it is stated it is not calculable. You can't even go the route to assume every condition because the chances of all the conditions being equally likely is essentially 0.
If he gave some fixed probabilities or data about the relationship between 1st set and the total game then sure you can calculate it, but without that or data to determine a relationship between them you can't calculate anything unless you assume independence.Comment
