New Posts For Thread: maths question

There's something not quite right there.

Wed, 11 Apr 2012 20:45:32 -0500

There's something not quite right there.

FIRST LASTMan U 57.34% 12.42%Man C 29.53% 30.39%Arsenal 13.14% 57.19%

Wed, 11 Apr 2012 20:35:03 -0500

FIRST LASTMan U 57.34% 12.42%Man C 29.53% 30.39%Arsenal 13.14% 57.19%

(as an aside these actually imply Man City have a 63.2% chance of finishing ahead of Arsenal, not 61% as you defined).That would figure as the 62% was actually meant to be 60%. I just remembered it incorrectly when I typed the original question.

Wed, 11 Apr 2012 17:51:23 -0500

(as an aside these actually imply Man City have a 63.2% chance of finishing ahead of Arsenal, not 61% as you defined).That would figure as the 62% was actually meant to be 60%. I just remembered it incorrectly when I typed the original question.

Last column should say arsenal (not Man City)

Wed, 11 Apr 2012 16:56:31 -0500

Last column should say arsenal (not Man City)

I've backed out an answer (estimate):I assumed all teams have the same variance of normally distributed league points and backed out a mean (average) from just usingMan Utd have a 62% chance of finishing ahead of Man CityMan Utd have a 74% chance of

Wed, 11 Apr 2012 16:52:39 -0500

I've backed out an answer (estimate):I assumed all teams have the same variance of normally distributed league points and backed out a mean (average) from just usingMan Utd have a 62% chance of finishing ahead of Man CityMan Utd have a 74% chance of

" the probality of mancity being above utd donot depend on arsenal or utd being above other."I think they do.If I know nothing then P(ManUtd>ManCity) = 62%If I know Arsenal < ManUtd then P(ManUtd>ManCity) > 62%If I know Arsenal > ManUtd then P(ManUtd

Wed, 11 Apr 2012 16:37:55 -0500

" the probality of mancity being above utd donot depend on arsenal or utd being above other."I think they do.If I know nothing then P(ManUtd>ManCity) = 62%If I know Arsenal < ManUtd then P(ManUtd>ManCity) > 62%If I know Arsenal > ManUtd then P(ManUtd

if we define the event as one team being above the other team in the ranking of 3 teams, the probality of one event as clarrified by clydebank before 3 posts doesnt depend on the occcurance of another.that is, for example, the probality of mancity be

Wed, 11 Apr 2012 12:28:16 -0500

if we define the event as one team being above the other team in the ranking of 3 teams, the probality of one event as clarrified by clydebank before 3 posts doesnt depend on the occcurance of another.that is, for example, the probality of mancity be

I instinctively feel there is a method to estimate it with a very high degree of accuracy. My guesstimate given those figures would be that the chances of each team finishing 3rd would be around 53% 34% and 13%

Wed, 11 Apr 2012 12:26:27 -0500

I instinctively feel there is a method to estimate it with a very high degree of accuracy. My guesstimate given those figures would be that the chances of each team finishing 3rd would be around 53% 34% and 13%

But Henok, you did:just for city being first: P(city above utd)* P(city above arsenal)=(1-0.62)*(0.61)But when these things are not independent (as here):P(city above arsenal | city above utd) 0.61and when they are not independent we can't multiply

Wed, 11 Apr 2012 11:46:41 -0500

But Henok, you did:just for city being first: P(city above utd)* P(city above arsenal)=(1-0.62)*(0.61)But when these things are not independent (as here):P(city above arsenal | city above utd) 0.61and when they are not independent we can't multiply

in this case i would expect a team to have a liklihhod of 1st,2nd and 3rd as 0.25, 0.5, 0.25

Wed, 11 Apr 2012 10:23:15 -0500

in this case i would expect a team to have a liklihhod of 1st,2nd and 3rd as 0.25, 0.5, 0.25