wilmott.com

There is a soccer team "T"There are two players "A" and "B" in the team.In last 10 year history::Whenever "A" played for "T", 40% of the time "T" won, 60% lost, no draw.Whenever "A" did not play for "T", 50% of the time "T" won, 50% lost, no draw.Whenever "B" played for "T", 70% of the time "T" won, 30% lost, no draw.Whenever "B" did not play for "T", 45% of the time "T" won, 55% lost, no draw.We have 4 quadrants herei) Neither played for "T"ii) "A" played, "B" didn'tiii) "B" played, "A" didn'tiv) Both playedQuestion1) Can you divide the percentage of games played by "T" in last 10 years into the above 4 quadrants?Question2) Do we have multiple solutions? no solutions or unique solutions?Question3) If we have multiple solution, which additional condition can make it give unique solution?

We need to know the probability of A and B playing:P(both play), P(only A plays), P(only B plays), P(neither plays)Of course, the sum of these four should be one. So three additional degrees of freedom.

ymous, the degree of freedom is 2, can you tell me how?

since the number of games won (when worked out for player A and B seperatley) have to be the same, if you know the nu,ber of games played by one you know for the other thats one equation.then the 4 probs have to add to one thats another equation. 4 unknowns and 2 equations is 2 degrees of freedom

Denote:W = event of T winningA = event of A played on TB = event of B played on T~A = event of A did not play on T~B = event of B did not play on TThen the given info says:.4 = P(W|A).5 = P(W|~A).7 = P(W|B).45 = P(W|~B)The 4 "quadrants" correspond to:i) P(~A~B) = P(W~A~B) + P(~W~A~B)ii) P(A~B) = P(WA~B) + P(~WA~B)iii) P(~AB) = P(W~AB) + P(~W~AB)iv) P(AB) = P(WAB) + P(~WAB)There are 8 unknowns (atoms) in these equations: P(WAB), P(WA~B), P(W~AB), P(W~A~B), P(~WAB), P(~WA~B), P(~W~AB), P(~W~A~B).Expanding the given information in terms of the atoms, we have.4 = P(W|A) = P(WA)/P(A) = [P(WAB) + P(WA~B)] / [P(WAB) + P(WA~B) + P(~WAB) + P(~WA~B)].5 = P(W|~A) = P(W~A)/P(~A) = [P(W~AB) + P(W~A~B)] / [P(W~AB) + P(W~A~B) + P(~W~AB) + P(~W~A~B)].7 = P(W|B) = P(WB)/P(B) = [P(WAB) + P(W~AB)] / [P(WAB) + P(W~AB) + P(~WAB) + P(~W~AB)].45 = P(W|~B) = P(W~B)/P(~B) = [P(WA~B) + P(W~A~B)] / [P(WA~B) + P(W~A~B) + P(~WA~B) + P(~W~A~B)]Another equation is that the 8 atoms add to 1.So there are 3 degrees of freedom in determining the full probability space. The (simplified?) matrix equations are Gv = (0,0,0,0,1)' (v being the vector with coefficients equal to the probabilities of the atoms as listed above) where G is:( 3 3 0 0 -2 -2 0 0 )( 0 0 1 1 0 0 -1 -1 )( 3 0 3 0 -7 0 -7 0 )( 0 11 0 11 0 -9 0 -9 )( 1 1 1 1 1 1 1 1 )However, we're only looking for the probabilities of the 4 quadrants; these make up the range of Mv where v satisfies the above equation and M is:( 1 0 0 0 1 0 0 0 )( 0 1 0 0 0 1 0 0 )( 0 0 1 0 0 0 1 0 )( 0 0 0 1 0 0 0 1 )We can change coordinates via C (w = Cv) where C is:( 1 0 0 0 0 0 0 0 )( 0 1 0 0 0 0 0 0 )( 0 0 1 0 0 0 0 0 )( 0 0 0 1 0 0 0 0 )( 1 0 0 0 1 0 0 0 )( 0 1 0 0 0 1 0 0 )( 0 0 1 0 0 0 1 0 )( 0 0 0 1 0 0 0 1 )with inverse (v = C^{-1}w) given by C^{-1} equal to:( 1 0 0 0 0 0 0 0 )( 0 1 0 0 0 0 0 0 )( 0 0 1 0 0 0 0 0 )( 0 0 0 1 0 0 0 0 )( -1 0 0 0 1 0 0 0 )( 0 -1 0 0 0 1 0 0 )( 0 0 -1 0 0 0 1 0 )( 0 0 0 -1 0 0 0 1 )Here, w = (P(WAB), P(WA~B), P(W~AB), P(W~A~B), P(AB), P(A~B), P(~AB), P(~A~B))', so the last 4 coordinates correspond to our target variables.Then we reduce to solving for the degrees of freedom in the last 4 coordinates of w in the system:G*C^{-1} w = (0,0,0,0,1)'Gaussian elimination reduces G*C^{-1} to (up to row permutation)( 1 0 0 -1 0 0 .22 .22 | -.14 )( 0 0 1 1 0 0 -.5 -.5 | 0 )( 0 1 0 1 0 0 -.22 -.22 | .54 )( 0 0 0 0 0 1 .4 1.4 | 1.2 )( 0 0 0 0 1 0 .6 -.4 | 1 )Only the bottom 2 rows have zero coefficients in the first 4 coordinates, so there are only 2 constraint equations in P(AB), P(A~B), P(~AB), P(~A~B):P(A~B) + .4*P(~AB) + 1.4*P(~A~B) = 1.2P(AB) + .6*P(~AB) - .4*P(~A~B) = 12 equations in 4 unknowns is 2 degrees of freedom. If we knew P(A) and P(B), then we'd have the two final equations that would be enough to ensure a unique solution; the final constraint equations being:P(AB) + P(A~B) = P(A)P(AB) + P(~AB) = P(B)

The solution I had in mind is the following. This should be similar or identical to what EscapeArtist999 mentionedLet Q1 = P(~A~B)Q2 = P(A~B)Q3 = P(~AB)Q4 = P(AB)Then we have two equations as follows, and 4 unknowns, hence 2 degrees of freedom.Q1 + Q2 + Q3 + Q4 = 1 <The total probability = 1>0.4*(Q2 + Q4) + 0.5*(Q1 + Q3) = 0.7*(Q3 + Q4) + 0.45*(Q1 + Q2) <The wins when calculated from both team's side are same>

OK with you but even with 2 degrees of freedom, I can't find a solution.Will try again soon

There are 4 more constraints which says none of this can be negativeBut still we have many solutionsExample:Q1=0.54Q2=0.44Q3=0.01Q4=0.01

You have more constraints.Q must be <1 and proba of winning for each Q must be in [0;1].I find this (first number is the number of games second percentage of games and third percentage of winning):Q1: 750 76.92% 44.92%Q2: 100 10.26% 46.25%Q3: 75 7.69% 100%Q4: 50 5.13% 26.25%

I missed to mention tht each Q<1 constraint, even though I took care of it in my solution.As far as the winning % =[0,1] constraint is concerned, I think that is not required since the other constraints mentioned so far will automatically ensure it.

I just noticed, even Q<=1 constraint it not required since other constraints will take care of it.

For the winning %, try to find a solution with Q1=85%; Q2=0; Q3=2.5%; Q4 =12.5%The solution implies that winning% of Q3 (noted a3)= 220% because a1=0.45 and a4=0.4.

Yeah, ur second constraint(winning %) is correct, thanks for pointing that out. Q3 has 2.5% games and 5.5% wins ==> 220% Now we have 6 equations for 8 unknowns{Q1, Q2, Q3, Q4, WQ1, WQ2, WQ3, WQ4}Where WQi is the % of wins in ith quadrant w.r.t total number of games played in all 4 quadrantsWe also have restrictions on Qis and WQis, with some trial and error we might be abe to get a solution.

I wrote a software to solve for {Q1, Q2, Q3, Q4, WQ1, WQ2, WQ3, WQ4}, pls find a few below.During this, I realized we have 3 degrees of freedom, and not 2. Can anyone tell me why?Solutions::solution1 = (0.7743, 0.0829, 0.0829,0.0600,0.3786,0.0071,0.0500,0.0500)solution2 = (0.7757,0.0671,0.1171,0.0400,0.3564,0.0229,0.0900,0.0200)solution3 = (0.7957,0.0471,0.0971,0.0600,0.3664,0.0129,0.0800,0.0300)solution4 = (0.8157,0.0271,0.0771,0.0800,0.3764,0.0029,0.0700,0.0400)solution5 = (0.680,0.220,0.070,0.030,0.320,0.085,0.055,0.015)solution6 = (0.690,0.210,0.060,0.040,0.325,0.080,0.050,0.020)solution7 = (0.700,0.200,0.050,0.050,0.330,0.075,0.045,0.025)solution8 = (0.710,0.190,0.040,0.060,0.335,0.070,0.040,0.030)

here is one approach that seems leading to somewhere.changing the notation a bit, consider the number of matches played and the winning percentage within 4 categories: N0 and p0 in (i), NA and pA in (ii), NB and pB in (iii) as well as N2 and p2 in (iv). sop2*N2+pA*NA=0.4*(N2+NA),pB*NB+p0*N0=0.5*(NB+N0),p2*N2+pB*NB=0.7*(N2+NB),pA*NA+p0*N0=0.45*(NA+N0),which can be written in the following form:[$]\left(\begin{array}{cccc}p_2-0.4&p_A-0.4&0&0\\0&p_A-0.45&p_0-0.45&0\\0&0&p_0-0.5&p_B-0.5\\p_2-0.7&0&0&p_B-0.7\end{array}\right)\left(\begin{array}{c}N_2\\N_A\\N_0\\N_B\end{array}\right)=\left(\begin{array}{c}0\\0\\0\\0\end{array}\right).[$]besides the constraints 0<=p<=1 and the two elements of each row need to be opposite signs (also need to be rational since N's are integers, but this can be ignored for now), one gets the constraint det=0.consider pA>0.45 (the other branch is hairy), one gets p2<0.4, pB>0.7, p0<0.45, so[$]\displaystyle\frac{0.4-p_2}{0.7-p_2}\frac{p_A-0.45}{p_A-0.4}\frac{p_B-0.7}{p_B-0.5}=\frac{0.45-p_0}{0.5-p_0}[$],one can quickly check that 0<r.h.s.=l.h.s<(4/7)*(11/12)*(3/5)=11/35 (max with p2=0, pA=1, pB=1), i.e., one can get a class of solution by choose41/96<p0<0.45,then accordingly pick0<p2<(96p0-41)/(90p0-35),0.45<pA<0.45+0.05/(0.6*(0.5-p0)/(0.45-p0)*(0.4-p2)/(0.7-p2)-1),then pB is inferred from the equation above and would satisfy all the constraints.