General Betting
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If you are having a bet on a game of tennis and a person to win the game is 1.25 to 100%, how do you work out the price to win an individual point. 1pt for the correct answer an 1pt extra for showing your working.
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when i say win the game I mean win a service game.
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1.06
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sorry i mislead you. What price is he to win each point? i.e. obviously bigger than 1.25
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1.58 exactly
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50-50, he wins or he doesn't :p
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http://www.betangel.com/tennis/
I always meant to get a hold of this just to see if Britain's phenomenal glut of tennis in-running "experts" (all sleepers pre-betfair) sheepishly follow these odds in their quest to gain pole position in the great rip-off that is betfair shell-and-pea. Thankfully I've got more to do with my time. |
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I wrote a spreadsheet to do that many years ago Feck, can tell you that it's not an effective method of betting on tennis as it's a game where better players can play to the score.
Methods for predicting tennis odds point by point have come a long way from there. |
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Get on Massive is spot on. It's almost exactly 1.58.
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Without doing the stats' math, my first impression of the solution is that to assume a binomial distribution; ans = 80%; known variable = p; => q = 1 - p; set k = 4 [i.e. number of points required to win the game]; n = total number of ways in which the sequence of 4 points can be achieved.
...do I get a mark for my workings? PS Mort Hill assumes a Weibull distribution...two marks to you Sir if you can define it's limitations for I/R tennis. |
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But it's not necssarily (and often isn't) 4 points to win a game, because you can have a very long series of dueces. In this case you need to win 2 more points than your opponent, and this would have to be modelled as a form of sum of an infinite series since it could go on indefinitely.
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In this case you need to win 2 more points than your opponent, and this would have to be modelled as a form of sum of an infinite series since it could go on indefinitely.
No. It's quite simple to deal with the deuce case. If you call the probability of player 1 winning from deuce 'd', then you can create an equation like this: d = (player 1 wins point)^2 + (2 * (player 1 wins point) * (1 - player 1 wins point) * d) and then solve. |
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Lori, "a long way from" where? Are you talking about the betangel software (not that I've ever seen it).
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One point from hell IMHO.
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Listen to some swedish music
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The win from a deuce situation could be
p^2 +2P^3q+4p^4q^2 +etc which is = p^2/[1-2pq] |
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p^2+P^2(2pq)+P^2(2pq)^2+.....
P^2(1+a+a^2+a^3+....) where a =2pq (1-a)^-1 expanded is 1+a+a^2+a^3+.... I think that does the deuce because for each successive deuce situation you have two ways to get to it. |
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A simple model for a game of tennis assuming independence of points is as follows;
P(G)=p^4+4p^4q+10p^4q^2+20p^5q^3/[1-2pq],I think the oddsmakers would have something a little more sophisticated though. |
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Work it out with a dice. 1,2,3,4 means the servers point.
5 and 6 his opponent's point. It works. |