- #1
jonjacson
- 453
- 38
Well folks,
I am a bit confused with this apparently "simple" system of equations.
pa=Probability player A wins a point when he is on serve
qa=1-pa=Probability player A loses a point when he is on serve
pb, and qb are the same for player B when he is on his own serve
I have calculated the probability of winning a Tie Break when they are tied at 6,6:
Probability player A wins a Tie Break from score PA(6,6)=(pa*qb)/(1-pa*pb-qa*qb) equation 1
Probability player B wins a Tie Break from score PB(6,6)=(pb*qa)/(1-pa*pb-qa*qb) equation 2
PA(6,6)+PB(6,6)=1 equation 3
If I know pa and pb I can easily calculate PA and PB, let's say pa=0.6 and pb=0.6, then we get:
PA(6,6)= 0.6*0.4/(1-0.6*0.6-0.4*0.4)=0.5
PB(6,6)=0.6*0.4/0.48=0.5
For pa=0.9 and pb=0.5 I get PA=0.9 and PB=0.1 so player A will win the Tie Break 90% of the time and player B only 10%.
Now the problem is the input data is PA and PB, so let's say player A wins 80% of the Tie Breaks and player B wins only 20%. Can I calculate probabilities pa and pb?
Apparently I have only two unkowns pa, and pb, and I have enough equations to solve it. But when I try to do it I find an absurd result.
I guess the equations are not independent, so they mean the same and it is not possible to calculate pa and pb.
Because for every value of pa, I can choose a pb that satisfies the values of PA and PB.
Am I correct? Is it impossible to calculate pa and pb? Or is there any way to get those values?
Thanks for your reply!
I am a bit confused with this apparently "simple" system of equations.
pa=Probability player A wins a point when he is on serve
qa=1-pa=Probability player A loses a point when he is on serve
pb, and qb are the same for player B when he is on his own serve
I have calculated the probability of winning a Tie Break when they are tied at 6,6:
Probability player A wins a Tie Break from score PA(6,6)=(pa*qb)/(1-pa*pb-qa*qb) equation 1
Probability player B wins a Tie Break from score PB(6,6)=(pb*qa)/(1-pa*pb-qa*qb) equation 2
PA(6,6)+PB(6,6)=1 equation 3
If I know pa and pb I can easily calculate PA and PB, let's say pa=0.6 and pb=0.6, then we get:
PA(6,6)= 0.6*0.4/(1-0.6*0.6-0.4*0.4)=0.5
PB(6,6)=0.6*0.4/0.48=0.5
For pa=0.9 and pb=0.5 I get PA=0.9 and PB=0.1 so player A will win the Tie Break 90% of the time and player B only 10%.
Now the problem is the input data is PA and PB, so let's say player A wins 80% of the Tie Breaks and player B wins only 20%. Can I calculate probabilities pa and pb?
Apparently I have only two unkowns pa, and pb, and I have enough equations to solve it. But when I try to do it I find an absurd result.
I guess the equations are not independent, so they mean the same and it is not possible to calculate pa and pb.
Because for every value of pa, I can choose a pb that satisfies the values of PA and PB.
Am I correct? Is it impossible to calculate pa and pb? Or is there any way to get those values?
Thanks for your reply!