# League tables

• Jul 20th 2011, 11:37 AM
RichardJBentley
League tables
3 teams (A,B,C) play in a league of 20. Using historical data for the mean and SD of points totals one can compute P(A finishes season higher than B) "P(A>B)" etc.

Q1. Given just P(A>B) and P(B>C) can one calculate P(A>C)?

Q2 Given the above how do you then calc. P(A finishes higher than BOTH B & C)?
• Jul 20th 2011, 02:01 PM
TKHunny
Re: League tables
Didn't we already do this? If you have 20 years of data, why not just take it from the data?

There remain only six outcomes:

ABC
ACB
BAC
BCA
CAB
CBA

Event(A>B) = U(ABC,ACB,CAB)
Event(B>C) = U(ABC,BAC,BCA)
Event(A>C) = U(ABC,ACB,BAC)
Event(A>Both) = U(ABC,ACB)

Off the top of this third rumination, I'd still say you need more information.
• Jul 21st 2011, 01:03 PM
RichardJBentley
Re: League tables
I re-posted because I was still stuck without a formula relating P(A>B), P(B>C),
P(A>C) and P(A>B&C). And still am! I understand your expressions but don't know how to calculate.

I don't have 20 years of data - it's a league of 20 teams. But regardless of the data, I'm just interest in the theory and annoyed with myself that I cannot find a formula.

Lets put some numbers in and you'll see my dilemma.

1. If P(A>B) = P (B>C) = 0.5 then (intuitively) P(A>C) = 0.5 and P(A>B&C) = 0.3333

In this scenario the three teams have equal strength.

2. If P(A>B) = 1 and P(B>C) = 0.5 then answers are also intuitive, being 1 and 1.

3. But say P(A>B) = 0.6 and P(B>C) = 0.55 then what formulae do I plug 0.6 and 0.55 into to get results for P(A>C) and P(A>both)?

That's where I'm stuck.

This formula must work for any scenario of course
• Jul 21st 2011, 04:05 PM
TKHunny
Re: League tables
Quote:

Originally Posted by RichardJBentley
I don't have 20 years of data - it's a league of 20 teams.

My mistake. If you have historical data, why not just take it from the data?

Quote:

annoyed with myself that I cannot find a formula.
Perhaps it would be wiser to be annoyed becaue you do not recognize there is insufficient information.

Quote:

In this scenario the three teams have equal strength.
If they have equal strength, each of the six is eqully likely.

ABC
ACB
BAC
BCA
CAB
CBA

P(A>B) = 3/6
P(B>C) = 3/6
P(A>C) = 3/6
P(A>B&C) = 2/6

Where's the tricky part?
• Jul 22nd 2011, 12:32 AM
RichardJBentley
Re: League tables
OK, let's consider a subset of all possible team strengths. In this subset teams B & C have equal strength [P(B>C) = P(C>B) = 0.5] and A may vary. Since B & C are equal then P(A>B) = P(A>C) = P1 for present purposes. Some trial & error work gives the following definite results:

If P1 = 0 then P(A>both B&C) = 0
If P1 = 1/3 then P(A>both B&C) = 1/6
If P1 = 1/2 then P(A>both B&C) = 1/3
If P1 = 2/3 then P(A>both B&C) = 1/2
If P1 = 1 then P(A>both B&C) = 1

What I want is an equation which relates P1 to P(A > both B&C), so that I can plug in any value for P1 and get the result. That equation must deliver the above results and give the result for any value of P1 in between the above easy cases.

Then I want to be able generalise that equation to the situation where B & C are not equal strength teams as in the subset above.

There cannot be any information missing. There are no other relevant variables as far as I can see. I have first class degree in maths from a good english unversity but I'm just rusty after 25 years!!! I get the feeling there are equations for surfaces needed here somewhere and perhaps the equation of the line that solves the above subset is a line on the general surface?