New Posts For Thread: Clever statistical person testhttps://community.betfair.com/general_betting/go/thread/view/94082/29099361/clever-statistical-person-testIf 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.Work it out with a dice. 1,2,3,4 means the servers point.5 and 6 his opponent's point. It works.https://community.betfair.com/general_betting/go/thread/view/94082/29099361/clever-statistical-person-test?post_id=519402889#519402889Work it out with a dice. 1,2,3,4 means the servers point.5 and 6 his opponent's point. It works.Sun, 06 May 2012 16:14:04 -0500A 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.https://community.betfair.com/general_betting/go/thread/view/94082/29099361/clever-statistical-person-test?post_id=519393903#519393903A 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.Sun, 06 May 2012 12:15:08 -0500p^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.https://community.betfair.com/general_betting/go/thread/view/94082/29099361/clever-statistical-person-test?post_id=519393327#519393327p^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.Sun, 06 May 2012 11:55:49 -0500The win from a deuce situation could be p^2 +2P^3q+4p^4q^2 +etcwhich is = p^2/[1-2pq]https://community.betfair.com/general_betting/go/thread/view/94082/29099361/clever-statistical-person-test?post_id=519393119#519393119The win from a deuce situation could be p^2 +2P^3q+4p^4q^2 +etcwhich is = p^2/[1-2pq]Sun, 06 May 2012 11:49:08 -0500Listen to some swedish musichttps://community.betfair.com/general_betting/go/thread/view/94082/29099361/clever-statistical-person-test?post_id=519374553#519374553Listen to some swedish musicSat, 05 May 2012 19:48:31 -0500One point from hell IMHO.https://community.betfair.com/general_betting/go/thread/view/94082/29099361/clever-statistical-person-test?post_id=519373393#519373393One point from hell IMHO.Sat, 05 May 2012 18:55:10 -0500Lori, "a long way from" where? Are you talking about the betangel software (not that I've ever seen it).https://community.betfair.com/general_betting/go/thread/view/94082/29099361/clever-statistical-person-test?post_id=519370691#519370691Lori, "a long way from" where? Are you talking about the betangel software (not that I've ever seen it).Sat, 05 May 2012 17:07:42 -0500In 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 probabilityhttps://community.betfair.com/general_betting/go/thread/view/94082/29099361/clever-statistical-person-test?post_id=519366873#519366873In 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 probabilitySat, 05 May 2012 15:07:44 -0500But 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 serhttps://community.betfair.com/general_betting/go/thread/view/94082/29099361/clever-statistical-person-test?post_id=519366453#519366453But 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 serSat, 05 May 2012 14:56:09 -0500Without 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 thttps://community.betfair.com/general_betting/go/thread/view/94082/29099361/clever-statistical-person-test?post_id=519364917#519364917Without 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 tSat, 05 May 2012 14:17:53 -0500