I’ve seen it floating around that the Bucs are somehow at a disadvantage tomorrow because they’ve beaten the Saints twice already and it’s hard to go 3-0.

A short clarification thread...

From a probability perspective, don’t ask, “How likely is it that the Bucs go 3-0 against the Saints?” but rather, “How likely is it that the Bucs win the 3rd game given that they’ve gone 2-0 in the first two?”

The answers to those questions are *very* different - the second is necessarily larger than the first. Let me show you with an example...

(Note: | means “given” and P denotes “probability of”)

Let’s play out two scenarios, one where game outcomes are independent, and one where probability of winning the next game increases if you win the previous one because you’ve shown you have some advantage.

Case 1)

Each game has 50% win probability and is independent (unlikely but fine for us right now). Then the initial probability (before any games have been played) of three wins is 0.5*0.5*0.5 = 12.5%.

Now, we want to know P(win 3rd | won 2nd & 3rd), which is just 50% because the probability of each outcome is independent, and 50% > 12.5%.

Case 2)

Assume we’re given:
P(win 1st) = 0.5
P(win 2nd | won 1st) = 0.55
P(win all 3) = 0.165

This makes plausible sense - the odds of winning the next game rise as you learn that Tampa has had an advantage in the past.

The missing probability is P(win 3rd | won 1st & 2nd), which, by definition of probability, is just P(win all 3) / P(win 1st & 2nd).

We were given the first term (0.165). The second term = P(win 1st)*P(win 2nd | won 1st), both of which were given, and the product is 0.275.

Thus, P(win 3rd | won 1st & 2nd) = 0.165 / 0.275, or 60%, and 60% > 16.5%.

Of course, these particular probabilities are hypothetical and designed under the assumption that winning a previous game gives some info that you have a competitive advantage.

Even if you reverse that assumption and allow the probability of winning to fall given previous victories, P(win 3rd | won 1st & 2nd) *must* be greater than P(win all 3) because the former equals P(win all 3) / P(win 1st & 2nd), and the denominator must realistically be <1.

It’s easy to assume that P(win all 3) is low, so P(win tomorrow) must be low, but we have to remember that the first two wins are now taken as given.

So, in short (haha), even if you think the games are dependent such that winning twice makes them less likely to win a third time (not sure why that would be), that probability is still higher than the unconditional probability of going 3-0.