How can I convert game winning percentage to set winning percentage?

[dgonzo]

Homework Statement

I've been struggling all morning with seems like a very easy problem. But I'm having difficulty wrapping my head around it.

In the sport of tennis, I am trying to convert game winning percentage to set winning percentage. Is there a short-cut equation, short of mapping out every possible game by game scenario?

Homework Equations

If it helps we can give Player A a game winning percentage of .67

The Attempt at a Solution

My best attempt at a solution (in excel) looks like this:

A 6-0 0.67 0.67 0.67 0.67 0.67 0.67 0.090458382
0.090458382

0.67 0.67 0.67 0.67 0.67 0.33 0.67 0.029851266
0.67 0.67 0.67 0.67 0.33 0.67 0.67 0.029851266
0.67 0.67 0.67 0.33 0.67 0.67 0.67 0.029851266
0.67 0.67 0.33 0.67 0.67 0.67 0.67 0.029851266
0.67 0.33 0.67 0.67 0.67 0.67 0.67 0.029851266
A 6-1 0.33 0.67 0.67 0.67 0.67 0.67 0.67 0.029851266
0.179107597

and so on and so on

 

[PeroK]

My best attempt at a solution (in excel) looks like this:

A 6-0 0.67 0.67 0.67 0.67 0.67 0.67 0.090458382
0.090458382

0.67 0.67 0.67 0.67 0.67 0.33 0.67 0.029851266
0.67 0.67 0.67 0.67 0.33 0.67 0.67 0.029851266
0.67 0.67 0.67 0.33 0.67 0.67 0.67 0.029851266
0.67 0.67 0.33 0.67 0.67 0.67 0.67 0.029851266
0.67 0.33 0.67 0.67 0.67 0.67 0.67 0.029851266
A 6-1 0.33 0.67 0.67 0.67 0.67 0.67 0.67 0.029851266
0.179107597

and so on and so on

Do you notice anything in common between all those 6-1 sets?

 

[dgonzo]

well, .67 must occur 6 times and .33 once.

(6)*(.67^6*.33^1) seems to produce our answer

 

[PeroK]

well, .67 must occur 6 times and .33 once.

(6)*(.67^6*.33^1) seems to produce our answer

Which means that all 6-1 sets are ------- ------

(Fill in the blanks)

 

[dgonzo]

equally probable?

I guess getting to the number of combinations is the difficult part.

 

[dgonzo]

I'm just thinking aloud... need an equation for how often .33 occurs

0.67 0.67 0.67 0.67 0.67 0.33 0.33 0.67 0.009850918
0.67 0.67 0.67 0.67 0.33 0.67 0.33 0.67 0.009850918
0.67 0.67 0.67 0.33 0.67 0.67 0.33 0.67 0.009850918
0.67 0.67 0.33 0.67 0.67 0.67 0.33 0.67 0.009850918
0.67 0.33 0.67 0.67 0.67 0.67 0.33 0.67 0.009850918
0.33 0.67 0.67 0.67 0.67 0.67 0.33 0.67 0.009850918
0.67 0.67 0.67 0.67 0.33 0.33 0.67 0.67 0.009850918
0.67 0.67 0.67 0.33 0.67 0.33 0.67 0.67 0.009850918
0.67 0.67 0.33 0.67 0.67 0.33 0.67 0.67 0.009850918
0.67 0.33 0.67 0.67 0.67 0.33 0.67 0.67 0.009850918
0.33 0.67 0.67 0.67 0.67 0.33 0.67 0.67 0.009850918
0.67 0.67 0.67 0.33 0.33 0.67 0.67 0.67 0.009850918
0.67 0.67 0.33 0.67 0.33 0.67 0.67 0.67 0.009850918
0.67 0.33 0.67 0.67 0.33 0.67 0.67 0.67 0.009850918
0.33 0.67 0.67 0.67 0.33 0.67 0.67 0.67 0.009850918
0.67 0.67 0.33 0.33 0.67 0.67 0.67 0.67 0.009850918
0.67 0.33 0.67 0.33 0.67 0.67 0.67 0.67 0.009850918
0.33 0.67 0.67 0.33 0.67 0.67 0.67 0.67 0.009850918
0.67 0.33 0.33 0.67 0.67 0.67 0.67 0.67 0.009850918
0.33 0.67 0.33 0.67 0.67 0.67 0.67 0.67 0.009850918
0.33 0.33 0.67 0.67 0.67 0.67 0.67 0.67 0.009850918
A 6-2 0.206869274

 

[PeroK]

I'm just thinking aloud... need an equation for how often .33 occurs

Have you heard of binomial coefficients?

 

[dgonzo]

6-2: in 7 slots, how many combinations can .33 occur exactly twice. 6-3: in 8 slots how many times can .33 occur exactly 3 times.

I am still perplexed, but seems to take a pattern:
6-1: 6/1= 6
6-2: (7*6)/(2*1)= 21
6-3: (8*7*6)/(3*2*1)= 56

Anyone want to connect the dots for me? I don't understand why that is the solution.

 

[dgonzo]

Oh, no I haven't. I'll read up...

 

[dgonzo]

That's pretty cool. Appreciate the help!

 

[PeroK]

6-2: in 7 slots, how many combinations can .33 occur exactly twice. 6-3: in 8 slots how many times can .33 occur exactly 3 times.

I am still perplexed, but seems to take a pattern:
6-1: 6/1= 6
6-2: (7*6)/(2*1)= 21
6-3: (8*7*6)/(3*2*1)= 56

Anyone want to connect the dots for me? I don't understand why that is the solution.

You are working out the binomial coefficients for yourself there:

$$\binom{8}{3} = \frac{8!}{(8-3)!3!} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1}$$

This is the number of ways that a set can get to 5-3, equal to the number of different ways that one player can win 3 games out of 8. Why 5-3 and not 6-3? Why not:

$$\binom{9}{3} = \frac{9!}{(9-3)!3!} = \frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1}$$

For 7-5 and 7-6, you have to be extra careful.

 

[dgonzo]

Ah, I see it!

The 6-3 set requires the assumption that player A win the 9th game. So the solution is (.67)*8!/(8−3)!3!
ie. Player A's final game W%(same as any other game) * the odds Player B wins exactly 3 of 8 games

If I used 9 and 3 it may include a combination where B wins the final game, which would of course not be possible.

I might run 7-5 and 7-6 by you as well.

 

[Ray Vickson]

Homework Statement

I've been struggling all morning with seems like a very easy problem. But I'm having difficulty wrapping my head around it.

In the sport of tennis, I am trying to convert game winning percentage to set winning percentage. Is there a short-cut equation, short of mapping out every possible game by game scenario?

Homework Equations

If it helps we can give Player A a game winning percentage of .67

The Attempt at a Solution

My best attempt at a solution (in excel) looks like this:

A 6-0 0.67 0.67 0.67 0.67 0.67 0.67 0.090458382
0.090458382

0.67 0.67 0.67 0.67 0.67 0.33 0.67 0.029851266
0.67 0.67 0.67 0.67 0.33 0.67 0.67 0.029851266
0.67 0.67 0.67 0.33 0.67 0.67 0.67 0.029851266
0.67 0.67 0.33 0.67 0.67 0.67 0.67 0.029851266
0.67 0.33 0.67 0.67 0.67 0.67 0.67 0.029851266
A 6-1 0.33 0.67 0.67 0.67 0.67 0.67 0.67 0.029851266
0.179107597

and so on and so on

Do you really mean 0.67 percent? That is, you are saying that A wins the fraction 0.67/100 of the games, or wins 67 in 10,000 games. Do you really mean that, or do you mean 67 out of 100 = 67%?

 

[dgonzo]

So, I mean that A wins game 1 67% of the time. Same with game 2, 3, 4, 5, 6, and so on. I was just multiplying those rows.

I think I've got everything squared away. But thanks again for all the help! 7-5 and 7-6 weren't too bad.

 

[haruspex]

The 6-3 set requires the assumption that player A win the 9th game. So the solution is (.67)*8!/(8−3)!3!

You mean 0.67⁶*0.33³*8!/(8−3)!3!, right?

7-5 and 7-6 weren't too bad.

It would be more realistic to start with the probability of winning a point, and that depending who serves. From that you can derive the probabilities of A winning: a service game, a receiving game, a tie breaker.

 

[dgonzo]

I do mean 0.67^6*0.33^3*8!/(8−3)!3!. Looking back, what I wrote was very unclear.

I agree that a single point probability would be the purest input for forecasting a match, and that each point would have different probabilities, but I won't be able to acquire that data.

And I am stuck somewhat on the 7-6 matches just because I have to make a probability for the tie-breaker. It seems it would be possible to set up an equation to solve for point probability based off of game probability. Then use that as an input in another binomial coefficient problem to solve for the tie-break probability. I might leave that for another day though...

 

[PeroK]

It would be more realistic to start with the probability of winning a point, and that depending who serves. From that you can derive the probabilities of A winning: a service game, a receiving game, a tie breaker.

If the probability of A winning a service point is p, then what is the probability of holding serve? Let's call that P. That's one problem.

If the probability of A winning a receiving point is q, then you can apply the above to get Q, the probability of A breaking serve.

The second problem is: given P and Q, what is the probability of A winning a set?

(There's also the tie-break problem to solve in this case.)

I wonder if that would give a realistic simulation of tennis scores?

Anyway, I make it that:

$$P = p^4(\frac{15-34p+28p^2-8p^3}{1-2p+2p^2})$$

 

[dgonzo]

$$P = p^4(\frac{15-34p+28p^2-8p^3}{1-2p+2p^2})$$

Could you explain this more? What steps led you to that equation?

 

[PeroK]

Could you explain this more? What steps led you to that equation?

I thought you might like to try to derive the formula for yourself. It's much the same as what you've already done, although you'll need to be able to sum an infinite geometric series.

 

[dgonzo]

I would like to try it. It seems like we're under the assumption we're playing out the Ad like the pros do.

Question, as a component I need to the odds of winning two points in a row. Of course this should equal p^2, but when I use binomial coefficients it comes to 1; ie. (2!)/(2!*(2-2)!) = 1

Also I don't understand why 0! = 1

Otherwise the problem is broken into:
the odds of winning 4/4 points + 3/4 points and one more + 3/5 points and one more + the ad probability

the ad = the odds of 3/6 points in any order * ((2 points in a row) + (1/2 points * 2 points in a row) + (1/2 points * 1/2 points * 2 points in a row) + (1/2 points ^3... etc. * 2 points in a row))

 

[PeroK]

I would like to try it. It seems like we're under the assumption we're playing out the Ad like the pros do.

Question, as a component I need to the odds of winning two points in a row. Of course this should equal p^2, but when I use binomial coefficients it comes to 1; ie. (2!)/(2!*(2-2)!) = 1

Also I don't understand why 0! = 1

Otherwise the problem is broken into:
the odds of winning 4/4 points + 3/4 points and one more + 3/5 points and one more + the ad probability

the ad = the odds of 3/6 points in any order * ((2 points in a row) + (1/2 points * 2 points in a row) + (1/2 points * 1/2 points * 2 points in a row) + (1/2 points ^3... etc. * 2 points in a row))

Yes, that's the way to do it. Re binomials:

The odds of winning two points are p^2. The binomial coefficient tells you that there is only 1 way to win both points.

0! = 1 by convention. If you left it undefined, then you'd need to say that:

$$\binom{n}{k} = \frac{n!}{(n-k)!k!} \ (k ≠ 0, n) \ \ and \ \ \binom{n}{0} = \binom{n}{n} = 1$$

It's easier to allow 0! =1.

 

[dgonzo]

I think I'm getting something different. For the various points did you get:

W at 40-0: p^4
W at 40-15: 4p^4(1-p)
W at 40-30: 10p^4(1-p)^2
W at 40-40: 20*(p^5*(1-p)^3+(p^5*(1-p)^3)/(1-(p*(p-1)))

and P = sum of the above

 

[haruspex]

I think I'm getting something different. For the various points did you get:

W at 40-0: p^4
W at 40-15: 4p^4(1-p)
W at 40-30: 10p^4(1-p)^2
W at 40-40: 20*(p^5*(1-p)^3+(p^5*(1-p)^3)/(1-(p*(p-1)))

and P = sum of the above

Try a sanity check: put p = 0.5.

 

[dgonzo]

1st 3 sound about right: .0625, .125, .15625

40-40 produces an error

 

[haruspex]

1st 3 sound about right: .0625, .125, .15625

40-40 produces an error

Quite so.
Having arrived at deuce, how did you calculate the prob that A wins from there?

 

[dgonzo]

6!/3!*(6-3)!=20 combinations
prob of getting to 40-40= 20*p^3*(1-p)^3
prob of winning once score 40-40= P^2 + P^2*(((p)*(p-1)+((p)*(p-1))^2+((p)*(p-1))^3 ... ^ infiniti)
r=((p)*(p-1))
so reduce to p^2 + p^2/(1-((p)*(p-1)))

finally multiply together: 20*p^3*(1-p)^3 * p^2 + p^2/(1-((p)*(p-1)))

I've got to think its the geometric series that I am doing wrong, but it reads right to me

 

[haruspex]

prob of winning once score 40-40= P^2 + P^2*(((p)*(p-1)+((p)*(p-1))^2+((p)*(p-1))^3 ... ^ infiniti)

A couple of problems there: you don't mean p-1; each deuce to deuce transition can occur in either of two orders.

 

[dgonzo]

Right: (1-p) is how that should read. And I included an extra p^2

So, starting over:
6!/3!*(6-3)!=20 combinations
prob of getting to 40-40= 20*p^3*(1-p)^3
prob of winning once score 40-40:
r=((p)*(1-p))
so prob = p^2/(1-((p)*(1-p)))

so all together: 20* p^2/(1-((p)*(1-p))) * p^3*(1-p)^3

reduces to: 20p^5*(1-p)^3/(1-((p)*(1-p)))

I'm still doing something wrong though, in excel the odds of winning the game only add to 44%

 

[dgonzo]

Ohh, sorry I see it. Cool, thanks

 

[PeroK]

Right: (1-p) is how that should read. And I included an extra p^2

So, starting over:
6!/3!*(6-3)!=20 combinations
prob of getting to 40-40= 20*p^3*(1-p)^3
prob of winning once score 40-40:
r=((p)*(1-p))
so prob = p^2/(1-((p)*(1-p)))

so all together: 20* p^2/(1-((p)*(1-p))) * p^3*(1-p)^3

reduces to: 20p^5*(1-p)^3/(1-((p)*(1-p)))

I'm still doing something wrong though, in excel the odds of winning the game only add to 44%

You're still going wrong with the deuce transitions. You need:

P(winning after 1st deuce) = $p^2$

P(getting to second deuce) = ?

P(winning after 2nd deuce) = $p^2$

P(getting to third deuce) = ??

Then, the overall probability of winning from deuce is the sum of an infinite series with $p^2$ as the first term, and r = ?.

 

[dgonzo]

r=(2)*(p)*(1-p)
and that seemed to fix it. ie. odds of winning once at deuce is (2)*(p)*(1-p) because the two events can occur in either order.

 

[PeroK]

Can you get to this?

$$P = p^4(\frac{15-34p+28p^2-8p^3}{1-2p+2p^2})$$
For the total probability of winning the game.

 

[dgonzo]

ahh, I finally did. Took me a while to get that denominator throughout the fraction.

Thanks again all, very cool thread.

 

[haruspex]

I investigated tie-breaker games. These are tricky because the service keeps changing, so there are two parameters for the probability of winning a point.
Seems to me the easiest approach is to pretend that an even number of points is always played. This can result in scores like 8-2 instead of 7-2, but it doesn't matter. I used three transition probabilities
p = A wins both
q = one each
r = B wins both
p+q+r=1.
Having obtained a formula in p, q and r, I then eliminated q (this was the hard part).
Not counting the case of reaching 6-6, the expression has 21 terms!

The same approach (combining results in consecutive pairs) can be used to get from game probabilities to set probabilities.

 

[PeroK]

I investigated tie-breaker games. These are tricky because the service keeps changing, so there are two parameters for the probability of winning a point.
Seems to me the easiest approach is to pretend that an even number of points is always played. This can result in scores like 8-2 instead of 7-2, but it doesn't matter. I used three transition probabilities
p = A wins both
q = one each
r = B wins both
p+q+r=1.
Having obtained a formula in p, q and r, I then eliminated q (this was the hard part).
Not counting the case of reaching 6-6, the expression has 21 terms!

The same approach (combining results in consecutive pairs) can be used to get from game probabilities to set probabilities.

For a set, I did it the same way:

P = probability of holding serve, Q = probability of breaking serve;

A = PQ (probability of winning two games)
B = P(1-Q) + Q(1-P) (winning one game)
C = (1-P)(1-Q) (losing both games)

Then, for winning a non-tie-break set, I got:

$P(Set \ without \ TB) = A^3 + 3A^3B + 6A^2B^2 + 3A^3C + 12A^3BC + 12A^2B^2C + 4A^2B^3 + 4AB^4$
$+ 6A^3C^2 + 12A^2B^2C + AB^4 + 30A^3BC^2 + 20A^2B^3C + AB^5$

$P(Tie-break) = 30A^2B^2C^2 + 20AB^4C + B^6$

The set formula would need to be extended (with p, q, a, b and c) to get the odds of winning the tie-break 7-5 or better. The tie-break formula is the same to get to 6-6 in the tie-break (with a, b, c); from where the odds of winning the tie-break are a/(1-b).

And, I think that's everything! Hopefully there are no typos.