probability question

poker_face

this qn has just been bugging me all day so maybe one of you maths whizz's will let me know how to solve it

There are 2 football teams playing each other. For simplicity sake let's say that they have to either win or lose the game (i.e. a draw isn't a possible outcome).

The probability of the two teams of winning any game is as follows

Team A: P(Win)=.8, P(Lose)=.2
Team B: P(Win)=.4, P(Lose)=.4
The question is, whether there's a way of working out the probability of Team A winning that particular match. A method I've tried which seems to give sensible enough answers is to multiply the probability of team A winning by team B losing. Then I'd multiply the probability of team B winning by team A losing. The sum of these two answers is less than 1, so surely these aren't the true probabilities. I then deduced that the factor of the true probability is to 1, would be the same as the calculated probability (team a winning * team B losing) is to the sum of the two possible outcomes. Hence the true probability of Team A winning would be (team A winning * team B losing) devided by the sum of the two possible outcomes. I hope whoever is reading that could follow that. The problem is I just kind of stumbled across this idea and I can't think of the logic behind. Could someone verify that this is the correct method and if it's not could you show me the correct method.
Cheers

Calculations for the above: Team A win = .8 * .6 =.48
Team B win = .2 * .4 = .08
Sum = .56
True probability of a Win for Team A = (.48/.56)=.85714286

Healio

Kind of relates to bookies overrounds:
This mite help a bit

Pherekydes

poker_face wrote:

There are 2 football teams playing each other. For simplicity sake let's say that they have to either win or lose the game (i.e. a draw isn't a possible outcome).

The probability of the two teams of winning any game is as follows

Team A: P(Win)=.8, P(Lose)=.2
Team B: P(Win)=.4, P(Lose)=.4

This is impossible. The sum of the probabilities must equal 1.

i.e. If P(B wins) = .4, then P(B loses) = .6

The probabilities will change according to their opponent.

So if P(A wins)= .8, then P(B wins) = .2, and so on.

poker_face

This is impossible. The sum of the probabilities must equal 1.
i.e. If P(B wins) = .4, then P(B loses) = .6

Yeah that's what I meant.

P(A Wins)=.8

This is the probability of A winnning any match. Take for example in a league, this value would be got from multiplying the probability for each particular game by each other.

silverside

the way you pose the question doesn't make sense (no offence)

if A has a probability of winning of 0.8, then that holds regardless how good B is. What you are looking for is the probability of winning when A plays B. You can't derive this just from raw statistics, you have to make a subjective guess. If you want, you can do some sort of model, but the answer will still be subjective.