This question is inspired by: A longer series is better for a better team: Can you see this at a glance?
Also obviously inspired by the NBA playoffs happening right now. :)
Suppose two teams are playing a series of $2k-1$ games, and the first team to win $k$ games wins the overall series. (No game can end in a tie.) Moreover, team A is "better" than team B.
Let $A_i$ denote the event that team A wins game $i$.
Let $A_{series}$ denote the event that team A wins the series, i.e., A wins $k$ or more games.
If the game results are i.i.d., and $P(A_i) = p > 1/2$ for any game $i$, then a longer series (larger $k$) increases A's chance of winning the series, i.e. $P(A_{series})$ is an increasing function in $k$. This is intuitively obvious, and a proof can be found in the link above (although that post asks an excellent question re: why such an "obvious" result requires an algebraically convoluted proof).
I wanna know under what conditions, i.e. under what probability model, would a longer match be BAD for the better team A. Perhaps some dependence that encodes "reversion to the mean" and/or (opposite of) "momentum"?
What I tried
My $0$-th attempt: If we allow "fatigue" in the form of decreasing $P(A_i),$ then a longer series can be bad for A, even when all $P(A_i) > 1/2.$ A simple 3-game example: if $P(A_1) = 1, P(A_2) = P(A_3) = 0.51$, then A always wins in a 1-game "series" ($k=1$) but B has a chance ($0.49^2$) in a 3-game series ($k=2$). @Henry in his answer gave an infinite length example. So I'm looking for something where the marginal probabilities $P(A_i)$ are constant (i.e. no fatigue).
My 1st attempt: let $P(A_1) = p > 1/2$, and thereafter every game $i$ result = game 1 result. This means: (1) $P(A_i) = p > 1/2$ (even though they are dependent), and yet (2) a longer series would give no advantage (nor disadvantage) to team A, because $P(A_{series}) = p$ regardless of $k$. However, I want a scenario where a longer series actually decreases $P(A_{series})$.
My 2nd attempt is something convoluted: the first 7 games ($k=4$) are played "normally" (i.i.d.) but then games 8 and 9 are always won by the loser of the best-of-first-7 series. I have not worked this out fully, but while this may give an example where $P(A_{series} | k = 4) > P(A_{series} | k = 5)$, I think this also implies the marginal probabilities $P(A_8), P(A_9) < 1/2$. So this isn't satisfying as a counterexample, since not only $P(A_i)$ are non-constant, team A actually becomes the worse team in a sense.
My 3rd attempt is a "surgical tweak" to the 2nd attempt: (1) if the first 7 games include exactly 4 wins by A, then games 8 & 9 are won by team B (therefore making B the overall winner), but (2) if the first 7 games have any other result, then games 8 & 9 are played i.i.d. but with an enhanced $P(A_8)=P(A_9) = p' > p$. By restricting the special dependence to an event of small enough probability, and balancing it out with $p' > p$ in the case (2), I think I can manage $P(A_8) = P(A_9) =p $. I have not worked out whether $P(A_{series} | k = 4) > P(A_{series} | k = 5)$ for some choice of $p, p'$. Also, this counterexample is too "artificial" for my taste.
What I seek is a probability model where:
For every game $i$, the marginal probability $P(A_i)$ is the same, i.e. $\forall i: P(A_i) = p > 1/2$.
There exists $k_0$ s.t. $P(A_{series} | k = k_0) > P(A_{series} | k = k_0 + 1)$.
- Bonus if this is true for all $k_0$, or all sufficiently large $k_0$.
Aesthetic requirement :) - Any dependence is as "simple" as possible, i.e. I prefer not to have something like my 3rd attempt (or even more convoluted). I know this isn't a math requirement, and people can have different tastes... comments on this point are welcome.
