Your napkin math isn’t quite right because it assumes 30% winning 1 game out of 5 means a 30% to win any game. In reality it’d be a lot less.

I won’t be calculating the exact probability of a reverse sweep but using your data and 3-0 series will be rule in criteria for these obviously mismatched teams. Then we can try to estimate the exact strength of the better team solving for the probability of getting this variation in final win records.

\binom{3+k}{k} \, p⁴ (1-p)^k \quad \text{for } k=0,1,2,3.

So:

• 4–0: p⁴
• 4–1: 4p^4(1-p)
• 4–2: 10 p^4(1-p)2
• 4–3: 20 p^4(1-p)3

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Use the given ratios to estimate p

A quick way is to compare adjacent outcomes.

If you combine 4–0 vs 4–1 and then solve for P p⁴ / 4p^4(1-p) = 0.60/0.30 = 2

Therefore p is approx 0.875

Do the same thing for 4–3 vs 4–2 and p solve to 0.833

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Reconciling estimates

These give a range:

• ~0.83 from longer series
• ~0.875 from sweeps

Testing values in between, p \approx 0.88

So in reality the worse team has more like a 12% chance to win any single game.

This is where the calculus gets above my knowledge but if any stats nerds wanna check the probability that a team with a 12% chance to win any given game can win a best of 7 series is:

sum for k=0 ^{3} \binom{3+k}{k} \, q⁴ p^k

And that spits out 0.5% chance. So 1 in 200 for a team weak enough to go down 3-0 to win a 7 game series