maths question

"finishing first and last?" - do you mean out of these 3?

If you assumed the events were independent you could work it our easily.... but that's definitely invalid... I'm not sure you could work it out from the info you give here.

I would politely suggest that Dave go back to his workings and try to come up with these chances directly (rather than going through these chances he has worked out)

sure if the events were independent you could work it out easily.

Imagine if the 3 events above were all 50/50 and its 33/33/33 1st 2nd and 3rd.  We know that if Man Utd were to finish ahead of Man City in that situation they'd be 67/33 to have finished ahead of Arsenal.  Therefore I think there is either a definitive answer OR a very good estimate.

I think in that example you have assumed independence.

I'm not completely sure I'm right in my earlier comment... will be interesting to see if you get the answer you are looking for.

I think if the events were truly independent in my example it would be 50/50 rather than 67/33.

It depends on the definition of independence though since every match is assumed independent, yet obviously taking an extreme example at the start of the season had we known Newcastle would be ahead of Chelsea after 33 games the chances that they would be ahead of Sunderland also at that stage given they were ahead of Chelsea would be long odds on rather than 50/50.

i think the probalities listed above are not clear. for instance when u say Man Utd have a 62% chance of finishing ahead of Man City does it mean mancity can be 3rd as well as 2nd

yes so for instance the 62% covers the 3 tricasts..

Man Utd Man City Arsenal
Man Utd Arsenal Man City
Arsenal Man Utd Man City

in that case  i think it should be straight forward calculation. since a team can only be above or below another team the three probalities given above are enough to calculate the probalities of  every possible combination of one team being above or below another

just for city being first
  P(city above utd)* P(city above arsenal)=(1-0.62)*(0.61)
city being second
  P(city above utd)* P(city below arsenal)  +  P(city below utd utd)* P(city above arsenal)= [(1-0.62)*(1-0.61) ]  +  [(0.62)*(0.61)]
similarly city being third would be
    P(city below utd)* P(city below arsenal)

Apr 11, 2012 -- 1:25PM, CLYDEBANK29 wrote:

sure if the events were independent you could work it out easily.Imagine if the 3 events above were all 50/50 and its 33/33/33 1st 2nd and 3rd.  We know that if Man Utd were to finish ahead of Man City in that situation they'd be 67/33 to have finished ahead of Arsenal.  Therefore I think there is either a definitive answer OR a very good estimate.

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

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)  0.61
and when they are not independent we can't multiply unconditional probabilities to get an answer.

I've had a chance to think it over now, and I'm certain we cannot answer the question properly because the events are not independent.

In terms of the definition of independent (from wikipedia): In probability theory, to say that two events are independent intuitively means that the occurrence of one event makes it neither more nor less probable that the other occurs.

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%

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 being above utd donot depend on arsenal or utd being above other.

however as u said, there is a flaw in my logic as well, will look back on it after the evening games.

" 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>ManCity) < 62%

Intuitively knowing Arsenal vs ManUtd tells us something about how ManyUtd have performed.

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 using
Man Utd have a 62% chance of finishing ahead of Man City
Man Utd have a 74% chance of finishing ahead of Arsenal

(as an aside these actually imply Man City have a 63.2% chance of finishing ahead of Arsenal, not 61% as you defined).

The results indicate:

	ManUtd	Man City	Man City
First	52%	31%	17%
Last	17%	30%	53%

Not too far from your guesstimates

Last column should say arsenal (not Man City)

(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.

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

There's something not quite right there.