Sports question. Team Race to 6 points chances

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themaestro
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Was hoping someone might be able to assist in this problem
Scenario is have a American football game with two team A and team B.

Am trying to work out the probability of each team getting to 6 points first, from the following infomation:

Let A be the event team A scores equal to or more than 6 points
Let B be the event team B scores equal to or more than 6 points
Assuming P(A) and P(B) are independent so P(A and B)=P(A)P(B)

Also making the assumption that Team A has possession and have a 0.62 chance of scoring a touchdown (worth 6 points) on this possession
Let this be labelled as P(Apo)

let Af be the event team A gets to 6 points first
let Bf be the event team B gets to 6 points first

Let P(A)=0.7, P(B)=0.7
P(Af given (A and B))=0.75
P(Bf given (A and B))=0.25

so question is what are P(Af) and P(Bf)?
when I try to calculate I get P(Af)=0.5775 which clearly can't be right as need
P(Af)>=P(Apo).
Is there any problems with the assumptions?
 
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here is details of my attempt.


P(Af)=P(Af| (A and B)) *P(A and B) + P(Af| not(A and B))*P(not(A and B)) (1)

since P(Af| not(A and B)) can only occur if Team A reaches 6 points and B does not then this is equal to

(P(A)- P(A and B)) / (P(not(A and B))


sub into (1) and simplfying gives

P(Af)=P(Af| (A and B)) *P(A and B) +P(A)- P(A and B)


sub in values for these and I get P(Af)=0.557.


can't see where I have made a mistake here. If it is correct then it would imply I am not free to set P(Apo)>0.557 but I am not sure why this would be true. Surley as long as P(Apo)<P(A) i am free to set whatever value I want for this?
 
Just wondering if anyone has any ideas about where I have gone wrong? i reckon there must be some other relationship between P(Af) and P(Apo) that I have not taken account of.