Hi Geno

It is nice to see you.

That is what I thought but it could be that the 6 was meant to be a 7 in which case it would still be half.

or that we were suppose to include the possibility of a tie, whch would make the problem a tad harder.

**Say we were meant to include the possibility of a tie.**

Then the outcomes culd be 2:4, 4:2, or 3:3

I do not think it could be 1:5 or 0:6 because then the series would have finished earlier.

The number of ways a team can win 2 games in 6 is 6C2 = 15

The number of ways a team can win 3 games in 6 is 6C3 = 20

The number of ways a team can win 4 games in 6 is 6C4 = 15

So the probability that they score 3 games each is 20/50 = 4/10

The remaining possibilities are split equally between A and B wining 0.5(1-0.4) = 0.3

So if they play 6 games the probabilities are

P(tie) = 0.4

P(A wins) = P(Bwins) = 0.3