Let's say that Player A is better than Player B, and if they played an infinite number of games would win 60% of the time.
In a best of 1, the probability that Player A loses is
P(0) = 40%.
In a best of 3, the probability that Player A loses is
P(0) + P(1)
= (0.4^3) + (0.4^2 * 0.6 * 3)
= 0.064 + 0.288
= 35.2%
In a best of 5, the probability that Player A loses is
P(0) + P(1) + P(2)
= (0.4^5) + (0.4^4 * 0.6 * 5) + (0.4^3 * 0.6^2 * 10)
= 0.01024 + 0.0768 0.2304
= 31.7%
And the deltas get larger the greater the starting probability you choose. 60% is conservative.
I think there's an assumption behind Player A winning 60% of the time - that is, that Player A never wins via hax. Consider a distribution such as:
Player A 50% to win (normal)
Player A 10% to win (hax)
Player B 30% to win (normal)
Player B 10% to win (hax)
It's totally fine for Player B to win here normally. We're only concerned about the distribution of haxy games. Presumably the odds of a haxy game are equal for both players. So what best-of-five does is weaken the odds of Player B winning
legitimately. Surely making a set "more competitive" reduces the odds of hax occurring
at all - not just for Player B. Player A getting a haxy win is still not ideal, presumably. My argument is that best-of-five increases the number of haxy games - regardless of the winner. Otherwise we're saying that the player statistically likely to win (non-hax) should always be the one to win - that doesn't seem right. Upsets based on play seem perfectly fine.
You might say then "well why not just do best-of-1". And I think it's because of the relative impact. There are advantages, for example, in recognizing player tendencies and decreasing the odds of matchup luck that best-of-1 can't provide, but best-of-3 can. The relative impact of adding more games to the set is not as substantial as best-of-1 to best-of-3 is. The math you provided shows that! Otherwise, if we truly wanted to be competitive, we could just do best-of-69 because that theoretically would increase the odds of Player A winning more. I get that's a slippery slope, but it has the assumption that best-of-5 has some
other advantage over best-of-69, and I'd guess it'd be something like the other 3 points I made that you already acknowledged. But then you might as well just do best-of-three.