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cormack15.
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February 13, 2012 at 22:53 #20980
As part of a failed and absent-minded lifelong attempt to get to grips with probability theory I continue to read on the subject.
Here’s an interesting exercise I came across tonight.
If my expectation of winning a tennis <i>point </i>against you are 1 in 3 (or 2/1 against in racing odds parlance) what are my chances of winning any individual <i>game</i> (not a set or match – a game)?
All other things which normally affect UK tennis players such as choking being equal of course.
February 13, 2012 at 22:59 #391213Errm…is the answer Geoff Hurst in the 1966 World Cup Final?
Mike
February 13, 2012 at 23:28 #391215Are you British or a dour Scot?
February 13, 2012 at 23:30 #3912161.01
February 14, 2012 at 01:29 #391229I put this to my Maths genius friend and occasional TRF poster –
becher’s chair
.
After thinking about it, he said
"I believe it would be impossible to get a full exact answer because in a game of tennis you could keep returning to deuce infinitely, so you can’t get one answer."
February 14, 2012 at 02:32 #391232Well it’s 80/1 to win to love, 134/1 to 15, 224/1 to 30 and 1124/1 on the 1st deuce.
February 14, 2012 at 08:20 #391244I put this to my
Maths
genius friend and occasional TRF poster –
becher’s chair
.
After thinking about it, he said
"I believe it would be impossible to get a full exact answer because in a game of tennis you could keep returning to deuce infinitely, so you can’t get one answer."
Statisticians and mathematicians are different. A mathematician (male of course!) was taken into a room where there was a very attractive woman standing with her back against the far wall. The mathematician was told he could advance toward her, but in each move he could only halve the distance between them. He said it would be a waste of time since he would never reach her, and he left the room. A statistician (female this time) was taken into a room where there was a very attractive man standing with his back against the far wall. She was told she could advance toward him, but in each move she could only halve the distance between them. She said: "Great, for all practical purposes, I will eventually get near enough."
For all practical purposes, the odds will limit out at approximately 35.5 to 1.
February 14, 2012 at 08:51 #391247I’d play this particular game in Monte Carlo
February 14, 2012 at 09:05 #391249This is fairly routine stuff.You can get simple freeware that will tell you probabilities over one game up to the whole match over 5 sets. See below. Betfair odds tend to move on a point by point basis pretty much in line with these calcs.
http://www.downloadplex.com/Windows/Edu … 17243.html
Incidentally if you do any each way on horses, you could do worse than putting them in as doubles with Roger Federer for the French open. He is 12/1 at WH , about right for the win (depends on Nadal being injured and clay not being Djoko’s forte), but I think his odds of reaching the final are about 3/1 (could happen if he faces Djoko in the semis rather than Nadal) so half odds is good value.
February 14, 2012 at 09:11 #391252VG
Marginal Value and Indocine
Halving the distance is a concept akin to drawing the diagonal (root 2) of the Unit Square.
Root 2 (1.414…) is an irrational number and therefore has no ‘end’ :
ergo
it cannot be represented as a finite line
It’s all Greek to me
February 14, 2012 at 09:50 #391256Is this game being played by A Murray in a grand slam final?
February 14, 2012 at 14:11 #391304Incidentally if you do any each way on horses, you could do worse than putting them in as doubles with Roger Federer for the French open. He is 12/1 at WH , about right for the win (depends on Nadal being injured and clay not being Djoko’s forte), but I think his odds of reaching the final are about 3/1 (could happen if he faces Djoko in the semis rather than Nadal) so half odds is good value.
Madness. Doesn’t matter what Djokovic’s forte is, the man is a machine. He was thrashing Nadal on clay in the ATP events before the French last year. I do agree that only Federer can beat Djokovic though.
February 14, 2012 at 15:09 #391315I make it around 6/1 to win the game.
I’m particularly tempted by the 1124/1 quoted on winning on the first deuce.
I make it 22% chance of getting to deuce:
(6!/(3!*3!))*(power(0.333,3))*(power(0.66667,3))
From there you’re a 20% chance of winning, so a 22/1 poke.
February 14, 2012 at 15:34 #391321Incidentally if you do any each way on horses, you could do worse than putting them in as doubles with Roger Federer for the French open. He is 12/1 at WH , about right for the win (depends on Nadal being injured and clay not being Djoko’s forte), but I think his odds of reaching the final are about 3/1 (could happen if he faces Djoko in the semis rather than Nadal) so half odds is good value.
Madness. Doesn’t matter what Djokovic’s forte is, the man is a machine. He was thrashing Nadal on clay in the ATP events before the French last year. I do agree that only Federer can beat Djokovic though.
If Nadal is injured Feds would be seeded 2 and avoid Djoko until the final. If the top 4 all play Feds would in theory meet Djoko in the semis and would lose IMO
February 14, 2012 at 17:11 #391328Nadal near enough won the Aussie with an injured shoulder. No way will he skip the French.
February 14, 2012 at 17:15 #391329I make it around 6/1 to win the game.
I’m particularly tempted by the 1124/1 quoted on winning on the first deuce.
I make it 22% chance of getting to deuce:
(6!/(3!*3!))*(power(0.333,3))*(power(0.66667,3))
From there you’re a 20% chance of winning, so a 22/1 poke.
I’ll lay you 2p at 1124/1 and the rest at evens.
I don’t know what that equation means. I just know it’s 4/6 for him to lose the point, 2/1 to win it. 1.66667 * 1.66667 * 1.66667 (3.629/1) takes him to 0-40, leaving him to win 5 points to win the game from there. So 4.629*3*3*3*3*3 = 1125.
February 14, 2012 at 17:19 #391330The Janet & John way.
http://codepad.org/SbfDaQZj
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