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%.

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.

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.

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.

http://en.wikipedia.org/wiki/De_Moiv...aplace_theorem

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.

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.

Thanks guys

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.

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.

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.

But shouldnt the answer for 99 be p(win 50 first)=0.86633 for p=0.555428?

PS. river_tilt thanks for the references.

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.