# Easy Probability Problem

1. Aug 29, 2014

### dgonzo

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

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

3. 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

2. Aug 29, 2014

### PeroK

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

3. Aug 29, 2014

### dgonzo

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

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

4. Aug 29, 2014

### PeroK

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

(Fill in the blanks)

5. Aug 29, 2014

### dgonzo

equally probable?

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

6. Aug 29, 2014

### 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

7. Aug 29, 2014

### PeroK

Have you heard of binomial coefficients?

8. Aug 29, 2014

### 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.

9. Aug 29, 2014

### dgonzo

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

10. Aug 29, 2014

### dgonzo

That's pretty cool. Appreciate the help!

11. Aug 29, 2014

### PeroK

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.

12. Aug 29, 2014

### 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.

13. Aug 29, 2014

### Ray Vickson

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%?

14. Aug 29, 2014

### 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.

15. Aug 29, 2014

### haruspex

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

16. Aug 29, 2014

### 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...

17. Aug 30, 2014

### PeroK

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:

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

18. Aug 30, 2014

### dgonzo

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

19. Aug 30, 2014

### PeroK

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.

20. Aug 30, 2014

### 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))