Is it possible to calculate the unknowns in these equations?

[jonjacson]

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!

 

[PeroK]

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!

You've almost got the answer yourself, which I've underlined. What you could do is choose a value for $p_a$ and find the value for $p_b$ that makes $PA = 0.8$.

It won't work for every $p_a$, e.g. with $p_a = 0$ you can never get $PA = 0.8$.

You could also try to find the general formula for $p_b$ in terms of $p_a$ and $PA$.

Can you work out how likely it is that the players hold serve? That's an interesting probability problem.

 

[jonjacson]

You've almost got the answer yourself, which I've underlined. What you could do is choose a value for $p_a$ and find the value for $p_b$ that makes $PA = 0.8$.

It won't work for every $p_a$, e.g. with $p_a = 0$ you can never get $PA = 0.8$.

You could also try to find the general formula for $p_b$ in terms of $p_a$ and $PA$.

Can you work out how likely it is that the players hold serve? That's an interesting probability problem.

THanks for your reply.

Well, I can use "trial an error" but I guess I could be in trouble because what I want is calculate pa and pb when the tennis match starts and then project that value for different scores. And I maybe there are two pair of values that work for the start of the match, let's say pa=0,6 and pb=0.55 and pa=0.70 and pb=0.65, but then after some points are played those pairs entail different match winning probabilities.

Well, I could try to do it going to head to head statistics and using those quantities. Let's say that for the last 5 matches one player won 250/400 points on his serve and the other 200/400, I could use that to calculate pa and pb, the problem is that the criteria used to calculate pa and pb is a bit subjective right?

Why do I chose the last 5 matches and not the last 10? And what happens if they have never played agains each other?

 

[PeroK]

THanks for your reply.

Well, I can use "trial an error" but I guess I could be in trouble because what I want is calculate pa and pb when the tennis match starts and then project that value for different scores. And I maybe there are two pair of values that work for the start of the match, let's say pa=0,6 and pb=0.55 and pa=0.70 and pb=0.65, but then after some points are played those pairs entail different match winning probabilities.

Well, I could try to do it going to head to head statistics and using those quantities. Let's say that for the last 5 matches one player won 250/400 points on his serve and the other 200/400, I could use that to calculate pa and pb, the problem is that the criteria used to calculate pa and pb is a bit subjective right?

Why do I chose the last 5 matches and not the last 10? And what happens if they have never played agains each other?

There's a big difference between modelling a single tennis match or tie-break by assuming a fixed probability for each point (based on who's serving) and statisticaly analysing tennis match results. The former is a simple probability question. The latter is full-on statistical analysis of data.

Players play better or worse from one day to the next. They may be in-form or out-of-form or neither and/or recovering from injury. They play on different surfaces etc. A data analysis of tennis results would have to consider at least these factors.

 

[jonjacson]

There's a big difference between modelling a single tennis match or tie-break by assuming a fixed probability for each point (based on who's serving) and statisticaly analysing tennis match results. The former is a simple probability question. The latter is full-on statistical analysis of data.

Players play better or worse from one day to the next. They may be in-form or out-of-form or neither and/or recovering from injury. They play on different surfaces etc. A data analysis of tennis results would have to consider at least these factors.

Yes, I agree.

What I mean is, for sure if you order to 10 PhysicsForums users to perform that analysis I am sure they will get 10 different values for pa and pb because they will interpret and choose differently the data.

 

[PeroK]

Yes, I agree.

What I mean is, for sure if you order to 10 PhysicsForums users to perform that analysis I am sure they will get 10 different values for pa and pb because they will interpret and choose differently the data.

I'd say $p_a$ and $p_b$ don't exist. In the sense that there are no single values that would model all matches beween two players. Any values would, for example, give Nadal the same chance of winning against player X on any surface. You cannot statistically model men's tennis with a single value.

If, however, you are watching a specific match between two players, you might be able to model it based on something as simple as a single $p_a$ and $p_b$.

To illustrate the point, you have already used two values, depending on who's serving. An even simpler model would have a single probability for each point. That might even be valid in some women's tennis, but not men's tennis.

 

[jonjacson]

I'd say $p_a$ and $p_b$ don't exist. In the sense that there are no single values that would model all matches beween two players. Any values would, for example, give Nadal the same chance of winning against player X on any surface. You cannot statistically model men's tennis with a single value.

If, however, you are watching a specific match between two players, you might be able to model it based on something as simple as a single $p_a$ and $p_b$.

To illustrate the point, you have already used two values, depending on who's serving. An even simpler model would have a single probability for each point. That might even be valid in some women's tennis, but not men's tennis.

I agree with you that modelling a tennis match is something very complex and maybe using the model with pa and pb constants is not reallistic. But I was curious about betting odds, I wanted to know if during a real tennis match the odds behave like pa and pb are constant or not. That is my goal, to know if that model works or if the odds behave like they were changing all the time.