It’s around P = 0.71 for team A to win both because you have to multiply the probabilities together to get an answer for P>0.5 for both trials. I say .71 because .7*.7 = .49.

My intuition has my brain balk at the problem as soon as it detects there is no real (non-gambling, non-math-problem) reason to care what the answer is.

But, forcing myself to think about it, it seems to me that any value other than p = 0.5 would have a double win or loss be more likely than not, because it’s most likely that the better side will win, and then it’s more likely that that same side will win again.

I’ll wait for someone to confirm whether that’s correct or not. I’m not sure without working it out in a confirming way.

Your intuition is correct, though this is not as obvious as it may seem. Suppose team A has a 60% chance of winning so, p=.6 while team B’s chance is 1—.6=.4. The probability of team A sweeping is .6x.6=.36, less than half. However, if we add in B’s chance of sweeping, .4x.4=.16, we get .36+.16 = .52, which is greater than a half.

Here is the algebra. The chances of a sweep are p^{2 + (1-p)}2. The chances of a split are p(1-p) + (1-p)p = 2p(1-p). If we subtract the probability of a split from the probability of a sweep, we get p^{2 + (1-p)}2 – 2p(1-p) = (p-(1-p))^{2 = (1–2p)}2 >= 0.

If p=.5, then there is an equal chance of a sweep or a split. Otherwise the probability is always greater for having a sweep.