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 07-19-2012, 07:33 PM #1 Pooh-Bah     Join Date: Oct 2005 Location: UYD Nation Posts: 4,432 Pretty simple math question for you guys. Hello all. We were discussing a 7 game series of Starcraft 2 that is coming up in the appropriate subforums of 2+2 and according to pinnacle the player known as "MC" is a 1.63 : 1 favorite over "Seed." We were then wondering how we get from that information (62% vs 38% to win the 7 game series) to what the odds are for each individual game in the series. Pinny does have a line up for just the 1st game that would indicate the answer is ~57.12%.
 07-19-2012, 07:54 PM #2 Pooh-Bah     Join Date: Sep 2004 Posts: 4,329 Re: Pretty simple math question for you guys. You can't unless you assume that all of the games are independent (would you set the line for the 4th game the same as the first if the score were 3-0 or 0-3?), or you specify all the correlations. If you do assume they're independent, then MC wins the match 61.97% of the time, then just plugging it in excel =binomdist(3,7,x,true) (the odds somebody wins 3 or fewer out of 7 if they're x to win each individual game) and iterating x to give .6197 gives a 1-game probability of .5554. There isn't any quick formula to do that conversion for you.
 07-19-2012, 08:45 PM #3 veteran     Join Date: Aug 2009 Location: Stanford, CA USA Posts: 3,331 Re: Pretty simple math question for you guys. You need to win 4 games first and as TomCowley said you cannot do it unless you assume each game doesnt have a stress of its own coming from already created score and is independent. So you are looking at how often something that wins a game with chance p and loses 1-p will arrive first at 4 in a simplified independent model. So its possible to win from 3-0 or 3-1 or 3-2 or 3-3. So Pwin = P(3-0)*p+P(3-1)*p+P(3-2)*p+P(3-3)*p Now P(3-0)=p^3 and P(3-1)=4!/3!/1!*p^3*(1-p) and P(3-2)=5!/3!/2!*p^3*(1-p)^2 and P(3,3)=6!/3!/3!*p^3*(1-p)^3 The chances for score n-m with n,m<4 are easy to find using the number of ways you can distribute n+m games in n wins m losses. Basically we realize that you can only win in 1 game if you have already 3-0,3-1,3-2,3-3 scores so all we need is the chance of each one of these scores to appear. (create the tree diagram of all possible ways=branches the series can end to see why) So basically Pwin=p^4*(1+4*(1-p)+10*(1-p)^2+20*(1-p)^3) Pwin=35p^4-84p^5+70p^6-20p^7 So you solve 0.61977=35p^4-84p^5+70p^6-20p^7 and Mathematica gives you as only acceptable sub 1 positive solution p=0.555428 verifying above excel work.
 07-20-2012, 03:09 AM #4 Pooh-Bah     Join Date: Dec 2005 Location: Chicago Posts: 5,670 Re: Pretty simple math question for you guys. interesting, so in something like a "best of 99" series, there's no quick formula? it's just simply taking that Pwin formula and adding a bunch more scenarios?
07-20-2012, 04:10 AM   #5
veteran

Join Date: Aug 2009
Location: Stanford, CA USA
Posts: 3,331
Re: Pretty simple math question for you guys.

Quote:
 Originally Posted by GBP04 interesting, so in something like a "best of 99" series, there's no quick formula? it's just simply taking that Pwin formula and adding a bunch more scenarios?
So you mean who gets first to 50 wins in your example?

A general formula would be ; (m=number of games needed to be won-1,p=probability to win a game]

Pwin[m,p]=(1/(m! (1 + m)!))(m! (1 + m)! - (-(-1 + p) p)^(1 + m) *(1 + 2 m)! Hypergeometric2F1[2 + 2 m, 1, 2 + m, 1 - p])

So for m=49 (50-1) its;

Pwin=p (p^49 + 50 (1 - p) p^49 + 1275 (1 - p)^2 p^49 +
22100 (1 - p)^3 p^49 + 292825 (1 - p)^4 p^49 +
3162510 (1 - p)^5 p^49 + 28989675 (1 - p)^6 p^49 +
231917400 (1 - p)^7 p^49 + 1652411475 (1 - p)^8 p^49 +
10648873950 (1 - p)^9 p^49 + 62828356305 (1 - p)^10 p^49 +
342700125300 (1 - p)^11 p^49 + 1742058970275 (1 - p)^12 p^49 +
8308281242850 (1 - p)^13 p^49 + 37387265592825 (1 - p)^14 p^49 +
159518999862720 (1 - p)^15 p^49 +
648045936942300 (1 - p)^16 p^49 +
2515943049305400 (1 - p)^17 p^49 +
9364899127970100 (1 - p)^18 p^49 +
33516481089577200 (1 - p)^19 p^49 +
115631859759041340 (1 - p)^20 p^49 +
385439532530137800 (1 - p)^21 p^49 +
1243918491347262900 (1 - p)^22 p^49 +
3894005712043605600 (1 - p)^23 p^49 +
11844267374132633700 (1 - p)^24 p^49 +
35059031427432595752 (1 - p)^25 p^49 +
101131821425286333900 (1 - p)^26 p^49 +
284667349197102273200 (1 - p)^27 p^49 +
782835210292031251300 (1 - p)^28 p^49 +
2105556772509601296600 (1 - p)^29 p^49 +
5544632834275283414380 (1 - p)^30 p^49 +
14308729894903957198400 (1 - p)^31 p^49 +
36218972546475641658450 (1 - p)^32 p^49 +
89998659054878867151300 (1 - p)^33 p^49 +
219702608869263116869350 (1 - p)^34 p^49 +
527286261286231480486440 (1 - p)^35 p^49 +
1244981450259157662259650 (1 - p)^36 p^49 +
2893740668169934025792700 (1 - p)^37 p^49 +
6625143108704848953788550 (1 - p)^38 p^49 +
14949040860667351485471600 (1 - p)^39 p^49 +
33261615914984857055174310 (1 - p)^40 p^49 +
73013303228015539877211900 (1 - p)^41 p^49 +
158195490327367003067292450 (1 - p)^42 p^49 +
338464770002738704236997800 (1 - p)^43 p^49 +
715391445687606806682745350 (1 - p)^44 p^49 +
1494373242103000885070623620 (1 - p)^45 p^49 +
3086205608690980088732809650 (1 - p)^46 p^49 +
6303739115624129542943611200 (1 - p)^47 p^49 +
12738806129490428451365214300 (1 - p)^48 p^49 +
25477612258980856902730428600 (1 - p)^49 p^49)

Or for the example above with p=0.555428

Pwin=P(win 50 first)=0.866331

I am thinking there may be an asymptotic expansion limit expression for large N (like whoever gets first to N with N very big) but i am still thinking about how to do it. If i get any ideas i may post it.

07-20-2012, 02:54 PM   #6
Pooh-Bah

Join Date: Sep 2004
Posts: 4,329
Re: Pretty simple math question for you guys.

Quote:
 Originally Posted by masque de Z I am thinking there may be an asymptotic expansion limit expression for large N (like whoever gets first to N with N very big) but i am still thinking about how to do it. If i get any ideas i may post it.
http://en.wikipedia.org/wiki/De_Moiv...aplace_theorem

 07-20-2012, 06:47 PM #7 Pooh-Bah     Join Date: Sep 2004 Posts: 4,329 Re: Pretty simple math question for you guys. If you're in the realm of competitiveness, then just take the cdf of the approximate normal. For more tail-y situations, it seems like it needs a 3rd generation convergence theorem. jason posted something about those, in his thread I think, basically 1/LLN tells you about the mean. 2/CLT tells you about the distribution. 3/? tells you about tail convergence properties. I don't know anything about them beyond that though.
 07-20-2012, 06:53 PM #8 veteran     Join Date: Aug 2009 Location: Stanford, CA USA Posts: 3,331 Re: Pretty simple math question for you guys. Yes i removed what i typed to rethink what i wrote it (because it may turn out its a simple binomial to normal without major error for usual p~around 1/2 and n>>1). I simply would like the thread to have an expression for large n just for the fun of it although most athletic events etc are usually only best of 7 , 9 etc.
07-20-2012, 07:00 PM   #9
Pooh-Bah

Join Date: Sep 2004
Posts: 4,329
Re: Pretty simple math question for you guys.

Quote:
 Originally Posted by masque de Z Yes i removed what i typed to rethink what i wrote it (because it may turn out its a simple binomial to normal without major error for usual p~around 1/2 and n>>1. I simply would like the thread to have an expression for large n just for the fun of it although most athletic events etc are usually only best of 7 , 9 etc.
No you have a real point- even for the best of 99 you posted about, which is nominally competitive at 55-45 games and 87-13 series wins, the normal approximation sucks.

 07-21-2012, 07:28 PM #10 Pooh-Bah     Join Date: Oct 2005 Location: UYD Nation Posts: 4,432 Re: Pretty simple math question for you guys. Thanks guys
07-22-2012, 07:25 PM   #11

Join Date: Apr 2006
Location: Swimming with sharks
Posts: 865
Re: Pretty simple math question for you guys.

Quote:
 Originally Posted by TomCowley If you're in the realm of competitiveness, then just take the cdf of the approximate normal. For more tail-y situations, it seems like it needs a 3rd generation convergence theorem. jason posted something about those, in his thread I think, basically 1/LLN tells you about the mean. 2/CLT tells you about the distribution. 3/? tells you about tail convergence properties. I don't know anything about them beyond that though.
IIRC, the jason thing was large deviations. (I may have mis-rembered, since it was a while ago.)

Here's a paper:
http://arxiv.org/PS_cache/arxiv/pdf/...804.2330v1.pdf

It has an approximation for P(n,k) which may be useful in this instance, using Kullback/Liebler entropy.

 07-23-2012, 04:40 AM #12 adept     Join Date: Apr 2006 Location: Swimming with sharks Posts: 865 Re: Pretty simple math question for you guys. To add to my earlier post, here's Jason's original post on large deviations: http://forumserver.twoplustwo.com/sh...&postcount=516 It seems to be what Tom referred to.
07-23-2012, 07:36 AM   #13
Carpal \'Tunnel

Join Date: Sep 2002
Posts: 8,907
Re: Pretty simple math question for you guys.

Quote:
 Originally Posted by TomCowley No you have a real point- even for the best of 99 you posted about, which is nominally competitive at 55-45 games and 87-13 series wins, the normal approximation sucks.
It doesn't suck for 55-45 and P(win) = 0.5554 if you use the proper correction for continuity.

1 - BINOMDIST(54,99,0.5554,TRUE)

=~ 54.04%

1 - NORMDIST(54.5,0.5554*99,SQRT(99*0.5554*0.4446),TRU E)

=~ 53.90%

If you used 55 instead of 54.5, you get 49.9%, and if you used 54 you get 57.9%. For 87-13 the normal approximation is off by a factor of 85 because the answer is so tiny, only 1 in 250 billion.

 07-23-2012, 08:03 AM #14 veteran     Join Date: Aug 2009 Location: Stanford, CA USA Posts: 3,331 Re: Pretty simple math question for you guys. But shouldnt the answer for 99 be p(win 50 first)=0.86633 for p=0.555428? PS. river_tilt thanks for the references.
07-23-2012, 08:14 AM   #15
Carpal \'Tunnel

Join Date: Sep 2002
Posts: 8,907
Re: Pretty simple math question for you guys.

Quote:
 Originally Posted by masque de Z But shouldnt the answer for 99 be p(win 50 first)=0.86633 for p=0.555428?
Yes, and the normal approximation is very accurate for that too.

1-BINOMDIST(49.5,99,0.555428,TRUE)

=~ 86.633%

1-NORMDIST(49.5,0.555428*99,SQRT(99*0.555428*(1-0.555428)),TRUE)

=~ 86.647%

If you want to find p for P(win 50 first) = 87%, the exact binomial gives 55.623%, and the normal gives 55.620%. I see that this is what Tom actually meant, not what I posted before, but it is accurate.

Last edited by BruceZ; 07-23-2012 at 08:42 AM.

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