AO 2011 odds -
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post #1 of 3 (permalink) Old 01-24-2011, 11:48 PM Thread Starter
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AO 2011 odds

Since it's possible to calculate the odds of winning a match given the odds of winning a point ( ), I thought I'd do just that given the fact that we now know the QFs. (It would be a nearly impossible task at the R128!) I used data from H2Hs with regards to points (and a little bit of guessing where there was no H2H).

Then, doing a Monte Carlo simulation (i.e. running simulations using the probability of each match pairing), it's possible to have 'simulated AOs' given these probabilities. Obviously, this is based just on past matches, and not on any of the intangibles (like who is better at slams, or clutch moments, etc.). Nonetheless, I thought it might be interesting to some of you. (I ran 500 simulations.)

Probability of winning:
Federer 32.4%
Nadal 31.6%
Murray 16.8%
Djokovic 13.8%
Berdych 3.2%
Ferrer 1.4%
Wawrinka 0.8%
Dolgopolov 0.0%

10 most likely final results:

Nadal d. Federer 17.0%
Federer d. Nadal 14.2%
Federer d. Murray 13.2%
Murray d. Federer 11.4%
Nadal d. Djokovic 9.2%
Djokovic d. Murray 7.0%
Djokovic d. Nadal 5.6%
Federer d. Ferrer 4.4%
Murray d. Djokovic 4.0%
Nadal d. Berdych 2.8%

Semi-final (top half):

Nadal d. Murray 46.0%
Murray d. Nadal 29.8%
Murray d. Ferrer 9.8%
Nadal d. Dolgopolov 6.4%
Ferrer d. Murray 5.6%
Ferrer d. Dolgopolov 1.6%
Dolgopolov d. Nadal 0.6%
Dolgopolov d. Ferrer 0.2%

Semi-final (bottom half):

Federer d. Djokovic 40.4%
Federer d. Berdych 20.6%
Djokovic d. Federer 19.0%
Djokovic d. Wawrinka 8.4%
Berdych d. Federer 4.2%
Wawrinka d. Berdych 2.8%
Wawrinka d. Djokovic 2.4%
Berdych d. Wawrinka 2.2%

If there's any interest, I can update as each result comes in.

(Also, I'm not sure if this is better for 'Statistics' or one of the main boards, but I figured that here was fine.)
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post #2 of 3 (permalink) Old 01-25-2011, 12:42 PM
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Re: AO 2011 odds

Pretty cool!
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post #3 of 3 (permalink) Old 01-26-2011, 07:41 PM
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Re: AO 2011 odds

Great idea! In case the website is yours, the calculations can be simpler. Instead of

P(6 to 2) = R(5,6)zw + R(4,6)w2 = 21w6z2
P(6 to 3) = R(5,6)z2w + R(4,6)[wz + zw]w + R(3,6)w3 = 56w6z3
P(6 to 4) = R(5,6)z3w + R(4,6)[3w2z2] + R(3,6)[3w3z] + R(2,6)w4 = 126w6z4
P(5 to 5) = R(5,6)z4 + R(4,6)[4wz3] + R(3,6)[6w2z2] + R(2,6)[4w3z] + R(1,6)w4 = 252w5z5

one can do

P(6 to 2) = R(5,7)w = 21w6z2
P(6 to 3) = R(5,8)w = 56w6z3
P(6 to 4) = R(5,9)w = 126w6z4
P(5 to 5) = R(5,10) = 252w5z5

The results stay the same.

Also, did you assume best-of-5? And point H2H for only hardcourt of all surfaces?

Interesting results though, with some refinements this could be useful to bookmakers.
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