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Topic: Anyone For Tennis? (Read 9284 times) |
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SMQ
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Re: Anyone For Tennis?
« Reply #75 on: Feb 3rd, 2009, 3:57pm » |
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(edit: gah, unfortunate page break! This is continued from the last post on page 3...)
Continuing the analysis for completeness:
P(A wins two consecutive points from deuce) = p(1 - q)
P(B wins two consecutive points from deuce) = (1 - p)q
P(A and B each win one point from deuce) = 2pq - p - q + 1
No matter how many times the score returns to deuce the remaining probability is divided proportionately to the above.
P(A finally wins from deuce) = p(1 - q)/[p(1 - q) + (1 - p)q]
P(B finally wins from deuce) = (1 - p)q/[p(1 - q) + (1 - p)q]
Putting it all together:
P(A wins) = P(A wins before deuce) + P(deuce is reached)*P(A finally wins from deuce)
= (210p7q6 - 210p6q7 - 630p7q5 + 630p5q7 + 700p7q4 + 1386p6q5 - 1386p5q6 - 700p4q7 - 350p7q3 - 2240p6q4 + 2240p4q6 + 350p3q7 + 75p7q2 + 1395p6q3 + 2175p5q4 - 2175p4q5 - 1395p3q6 - 75p2q7 - 5p7q - 360p6q2 - 2025p5q3 + 2025p3q5 + 360p2q6 + 5pq7 + 29p6q + 675p5q2 + 1150p4q3 - 1150p3q4 - 675p2q5 - 29pq6 - 69p5q - 600p4q2 + 600p2q4 + 69pq5 + 85p4q + 225p3q2 - 225p2q3 - 85pq4 - 55p3q + 55pq3 + 15p2q - 15pq2 - pq + p)/(-2pq + p + q)
P(B wins) = P(B wins before deuce) + P(deuce is reached)*P(B finally wins from deuce)
= (210p6q7 - 210p7q6 - 630p5q7 + 630p7q5 + 700p4q7 + 1386p5q6 - 1386p6q5 - 700p7q4 - 350p3q7 - 2240p4q6 + 2240p6q4 + 350p7q3 + 75p2q7 + 1395p3q6 + 2175p4q5 - 2175p5q4 - 1395p6q3 - 75p7q2 - 5pq7 - 360p2q6 - 2025p3q5 + 2025p5q3 + 360p6q2 + 5p7q + 29pq6 + 675p2q5 + 1150p3q4 - 1150p4q3 - 675p5q2 - 29p6q - 69pq5 - 600p2q4 + 600p4q2 + 69p5q + 85pq4 + 225p2q3 - 225p3q2 - 85p4q - 55pq3 + 55p3q + 15pq2 - 15p2q - pq + q)/(-2pq + p + q)
And by the way I've arranged the terms it should again be obvious that the equations are symmetrical w/rt p and q. Thus the overall chance of winning depends only on the players respective chances of winning a point and not on which player serves first.
Q.E.D
--SMQ
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| « Last Edit: Feb 3rd, 2009, 4:11pm by SMQ » |
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--SMQ
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ThudnBlunder
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Re: Anyone For Tennis?
« Reply #76 on: Feb 4th, 2009, 5:53am » |
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Wow, une veritable tour de force, SMQ!  Years ago. when this question first occurred to me and I knew no better, I started out trying to calculate those polynomials by hand.  But I only had one ream of paper and soon gave up.
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ThudnBlunder
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Re: Anyone For Tennis?
« Reply #77 on: Feb 17th, 2009, 9:26am » |
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on Sep 10th, 2007, 7:15pm, SWF wrote:Although I didn't check the math, that is an interesting conclusion, ThudanBlunder.
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Actually, I can see this anomaly being hijacked by (m)ad men of the future in the following scenario.
After a fortunate case of mistaken identity, somebody's pleasantly surprised-looking Grandpa, who only nipped out for a packet of cigarettes, shuffles onto the Centre Court to face an ageing Nadal over three sets in the first round of the Seniors' Championship. As the crowd hushes, the new courtside Polynomial Probability Processor (PPP), which concurrently monitors all games and which due to time constraints has been carefully installed directly over the dead bodies of the last Luddite holdouts after Wimbledon's version of the Alamo, shows Grandpa with a chance of winning the first game of between 1 in 3 million and 1 in 4 million, depending on whether he is using his new Zimmer frame. Nadal shapes to serve. On cue, Grandpa turns slowly on his heels and waddles off-court to grab a beer. Unfazed, and still 'digging deep' before each serve, Nadal sends down two consecutive aces. However, becoming slightly complacent, he double-faults on the next point due to the umpire overruling a linesman to call the ball out. Stifling a Spanish version of "You can't be serious!", Nadal demands that the decision be referred to Quarkeye, the quantum successor to the primitive, silicon-based Hawkeye, which quickly decides that the ball was out by roughly 173 Angstroms. Instantaneously, PPP's lights start flashing and popping as it proclaims for all to see that, merely by taking an opportune guzzle of the sponsor's beer, Grandpa has improved his chances of winning the first game to 1 in 2 million, even though he is now behind!
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| « Last Edit: Mar 6th, 2009, 7:07pm by ThudnBlunder » |
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ThudnBlunder
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Re: Anyone For Tennis?
« Reply #78 on: Feb 18th, 2009, 7:39am » |
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In a previous post I mentioned that Federer won more points than Nadal, in spite of losing the match. We could say his ratio of ‘winning’ defeat was 174/173. In a 2n+1-set final (n > 0) what is the limiting maximum ratio of ‘winning’ defeat possible as n ->  ?
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ThudnBlunder
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Re: Anyone For Tennis?
« Reply #79 on: Mar 3rd, 2009, 12:25pm » |
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on Feb 18th, 2009, 7:39am, ThudanBlunder wrote:| In a previous post I mentioned that Federer won more points than Nadal, in spite of losing the match. We could say his ratio of ‘winning’ defeat was 174/173. In a 2n+1-set final (n > 0) what is the limiting maximum ratio of ‘winning’ defeat possible as n -> ? |
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No takers? I worked this out in 5-10 minutes. Still wouldn't mind a second opinion.
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ThudnBlunder
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Re: Anyone For Tennis?
« Reply #80 on: Mar 6th, 2009, 7:01pm » |
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on Feb 18th, 2009, 7:39am, ThudanBlunder wrote:| In a previous post I mentioned that Federer won more points than Nadal, in spite of losing the match. We could say his ratio of ‘winning’ defeat was 174/173. In a 2n+1-set final (n > 0) what is the limiting maximum ratio of ‘winning’ defeat possible as n -> ? |
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I get 65/31. Explanation on request.
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