Clever statistical person test

when i say win the game I mean win a service game.

1.06

sorry i mislead you.  What price is he to win each point?  i.e. obviously bigger than 1.25

1.58 exactly

50-50, he wins or he doesn't :p

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.

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.

Get on Massive is spot on. It's almost exactly 1.58.

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.

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.

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.

Lori, "a long way from" where? Are you talking about the betangel software (not that I've ever seen it).

One point from hell IMHO.

Listen to some swedish music

The win from a deuce situation could be
p^2 +2P^3q+4p^4q^2 +etc
which is = p^2/[1-2pq]

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.

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.

Work it out with a dice. 1,2,3,4 means the servers point.
5 and 6 his opponent's point. It works.