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

[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.