Recently I gave the playoff records of home teams in the various rounds of the playoffs. Here is the table from that earlier blog post:
Post merger (since 1970) records 
home win percentage 
regular season 
.576 
playoffs overall 
.679 
wildcard round 
.630 
division round 
.713 
conference round 
.681 
I also gave some probabilities based on those records of the various seeds getting to the Superbowl, but those probabilities were based on some further assumptions:
“The table below shows the various probabilities of making it to the Superbowl based on the above records and based on the assumption the 5 and 6 seeds travel in every round, the 3 and 4 seeds host in the first round and travel the rest, the 2 seed hosts in all but the conference round, and the 1 seed hosts every round. 1 and 2 also don’t have to play in the wildcard round.”
That was intellectually lazy, bad as I hate to do it to myself. I decided to break it down into getting to the conference championship game (ccg) as both host and visitor, and then calculate it from there, which is fairly straightforward.
First, let’s setup a few variables. Let divHome be the home records of division home games in the playoffs (.713). Let divAway be 1 – divHome. Similarly, wcHome = .630, wcAway = 1 – .630, confHome = .681, and confAway = 1 – .681.
Seed1’s probability of hosting the ccg is straightforward. It just needs to win its home division game.
seed1Host = divHome
Seed2 needs to win its division home game, but it also needs for seed1 to lose its division home game (or else seed1 would be host).
seed2Host = divHome * divAway
For seed3 to be host of the ccg it needs to win a wildcard home game + a division away game + it needs the other division away team to win (to knock off either seed1 or seed2 in the other division game).
seed3Host = wcHome * divAway * divAway
For seed4 to be host of the ccg it needs to win its wildcard home game against seed5 + it needs seed6 to beat seed3 in its wildcard away game + it needs to beat seed1 in its division away game + it needs seed6 to beat seed2 in its division away game.
seed4Host = wcHome * wcAway * divAway * divAway
For seed5 to host the ccg it needs pretty much the same thing to happen for it that seed4 needed to happen for it except the only difference is seed5 is playing a wildcard away team that it needs to win whereas seed4 is playing a wildcard home game that it needs to win.
seed5Host = wcAway * wcAway * divAway * divAway
Seed6 cannot host the ccg under any scenario.
seed6Host = 0
Now, we work on how the various seeds can be the ccg away team.
Seed1 cannot under any scenario be the ccg away team because if it gets there it will host over any other opponent.
seed1Away = 0
Seed2 can be the away team at the ccg game by winning its division home game + having seed1 win its division home game.
seed2Away = divHome * divHome
For seed3 to be the away team at the ccg it needs to win its wildcard home game + it needs to win its division away game + it needs the other division home team (seed1 or seed2) to win.
seed3Away = wcHome * divAway * divHome
For seed4 to be the away team at the ccg is where it starts getting complicated. Seed4 needs to (obviously) get to the ccg, but the ccg needs to be against a higher seed (not against seed5 or seed6). Seed5 is easily taken care of because that’s seed4’s wildcard round opponent, but seed6 could get there, in which case seed4 would be the host. Seed4 needs either seed6 to lose to seed3 in the wildcard round *or* seed6 to lose to seed1 in the division round. The probability of seed6 losing one or the other is the same as the probability of it *not* winning both (1(wcAway*divAway)).
seed4Away = wcHome * divAway * (1(wcAway *divAway))
Seed5’s road to getting to the ccg as visitor is almost the same for seed4 except its wildcard game is an away game instead of a home game.
seed5Away = wcAway * divAway * (1(wcAway*divAway))
Seed6 cannot ever host the ccg because no matter what happens in the other games it will always be the lower seed in the ccg if it gets that far, which merely involves winning 2 away games (wcAway and divAway)
seed6Away = wcAway * divAway
So, now we know the probabilities for each of the 6 seeds of both hosting the ccg and being visitor in the ccg. We also know the historical home records in the ccg, so it’s really just a matter of adding them up.
seed1Win = seed1Host * confHome + seed1Away*confAway
seed2Win = seed2Host * confHome + seed2Away*confAway
seed3Win = seed3Host * confHome + seed3Away * confAway
seed4Win = seed4Host * confHome + seed4Away*confAway
seed5Win = seed5Host * confHome + seed5Away * confAway
seed6Win = seed6Host * confHome + seed6Away * confAway
wcHome 
0.63 
divHome 
0.713 
confHome 
0.681 
wcAway 
0.37 
divAway 
0.287 
confAway 
0.319 


seed1Host 
0.713 
seed2Host 
0.204631 
seed3Host 
0.05189247 
seed4Host 
0.0192002139 
seed5Host 
0.0112763161 
seed6Host 
0 


seed1Away 
0 
seed2Away 
0.508369 
seed3Away 
0.12891753 
seed4Away 
0.1616097861 
seed5Away 
0.0949136839 
seed6Away 
0.10619 


seed1Win 
0.485553 
seed2Win 
0.301523422 
seed3Win 
0.0764634641 
seed4Win 
0.0646288674 
seed5Win 
0.0379566364 
seed6Win 
0.03387461 
link to csv file (rightclick, save as, rename from png to csv and import into spreadsheet)
Here are the formulas for Seeds 1 through 6 getting to the Superbowl in case anybody wants to write a program to do it using different values for wcHome, divHome, and confHome:
S1[] := confHome*divHome;
S2[] := divHome*(confHome + (1 – confHome)*divHome – confHome*divHome);
S3[] := (1 + divHome)*(confHome*(1 + divHome) – (1 – confHome)*divHome)*wcHome;
S4[] := ((1 + divHome)*wcHome*(confHome*(1 + divHome)*(1 + wcHome) + (1 – confHome) * (divHome + wcHome – divHome*wcHome)));
S5[] := (1 + divHome)*(1 + wcHome)*(confHome*(1 + divHome)*(1 + wcHome) + (1 – confHome)*(divHome + wcHome – divHome*wcHome))
S6[] := (1 – confHome)*(1 + divHome)*(1 + wcHome)
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