The playoffs are a funny bit of business, where people tend to assume the #1 seed has a really good chance of making it to the Super Bowl. That is, unfortunately, not even close to the truth. If you ignore home field advantage, then it becomes easy to see that in these circumstances, the #1 and #2 seeds have 1 chance in 8 of winning (0.125), whereas seeds 3-6 have a 1 in 16 chance of winning (0.0625). But since in the playoffs, there is a home field advantage (at least until you reach the Super Bowl), the actual odds from Seeds 1 to 6 vary quite dramatically.

For now, we’re going to assume a home field advantage of 0.60. From 2001 to 2010, 100 non-Super Bowl playoff games were played, and the home team won 60 of them. This year, the home team won every time, unless the visitor was named the New York Giants, leading to a record of 8-2. So, I guess, the running total now, from 2001 to 2011,  has to be 68/110, or 61.8% or so.

That said, I’m still going to use 60% in my calculations below.

For the sake of making it easier to turn any calculations into code, we’ll assign the home field advantage to the variable U (for “upper”), and to 1 – U, we will assign the variable L (for “lower”). Given these assignments, we now have:

Temporary variables:

LL = L*L
T23 = U*L + L*U
T45 = LL*U + (1. – LL)*L

Calculations of playoff odds

Seed 1 = U*U*0.50
Seed 2 = U*T23*0.50
Seed 3 = U*L*T23*0.50
Seed 4 = U*L*T45*0.50
Seed 5 = L*L*T45*0.50
Seed 6 = L*L*L*0.50

T23 is necessary to calculate the second game of Seed 2 or the third game of Seed 3. In this game, these two teams could face Seed 1, Seed 4, Seed 5, or Seed 6. Critically, they will either face Seed 1, for which they would be the visiting team, or all others, for which they would be the home team. The odds therefore become (odds of Seed 1 winning)(vistor’s odds) + (1 – odds of Seed 1 winning)(home team odds).

T45 is necessary to calculate the third game of Seed 4 or 5. In this game, these two teams could face Seed 1, Seed 2, Seed 3, or Seed 6. As Seed 6 is the only team for which Seeds 4 and 5 would be the home team, it is easiest to calculate the odds of Seed 6 making it to the third game, and then subtract those odds for the probability of playing as the visitors. Since the odds of Seed 6 arriving at game 3 are L*L, you end up with the formula given above.

Choosing a value of 0.60 for the home field advantage, we end up with:

Seed 1 : 0.18
Seed 2 : 0.144
Seed 3 : 0.0576
Seed 4 : 0.05184
Seed 5 : 0.03456
Seed 6 : 0.032

The range, from 18% to about 3%, is considerably more broad than the naive 1/8 to 1/16 values. Home field has a marked effect on the ability of teams to reach and win the Super Bowl. But the sheer number of teams involved, 12, and the arrangement of the playoffs, means that a #1 seed has, with a HFA of 60%, about a 36% change of making it to the Bowl, and a 18% chance of winning.

Note: this link has a coded version of the calculations above.

After the Giants victory over the Packers, I finally got up the nerve to say what my system has been saying from the start, that my predictive system markedly favors the Giants throughout the entire playoffs.

Going all the way?

The deal, of course, is a heavily favored team can lose. A team seeded 1 or 2 and favored by 70% in every game only has a 34% chance of making it through 3 games. The nature of the playoffs make it difficult for any team, even a really good team, to win it all.

That said, the Giants are favored by 75% over the San Francisco 49ers. The only advantage the 49ers hold is home field advantage. The Giants have to be considered a playoff experienced team, and they have a massive strength of schedule advantage, the same advantage that will give them precendence over either New England or Baltimore. If you choose to treat the Giants as having no playoff experience, that lowers their odds to win to a mere 58%.

Favored in the Conference Championship Round:

Giants over 49ers: 75%
NE over Ravens: 59%

Favored in the Super Bowl:

Giants over NE: 66%
Giants over Ravens: 64%
NE over 49ers: 64%
Ravens over 49ers: 65%

Odds of winning the Super Bowl:

Giants: 49%
NE: 24%
Ravens: 18%
49ers: 9%

For contrast, we’ll calculate the Pythagorean odds for these teams as well, ignoring the effects of strength of schedule, and playoff experience.

49ers over Giants: 86%

NE over Ravens: 61%

49ers over NE: 61%

And the 49ers are favored to win the Super Bowl, via Pythagoreans, by 52%.

Of course, if you’re taking these kinds of offensive metrics seriously, please note the odds of the Giants having made it this far was only 7.4% (Originally calculated as 5.4%). Consider those odds, please, before writing my little predictive system off.

Both the Giants and Denver have won today, eliminating all wild cards and leading to two #4 seeds playing at the #1 seeds. In the case of the Giants, using my formula, we have the question of whether they truly have playoff experience. If they do not, then Green Bay is favored, on average, by 56%, though the relative error of strength of schedule results allow for Green Bay being favored by as much as 73% to the Giants being favored by 63%. If the Giants are treated as if they have playoff experience, then there is a wide range of results, from Green Bay being favored by 55% to the Giants being favored by 78%, with the average result being the Giants favored by 63%. Note that home field plus Pythagoreans would favor Green Bay by 83%.

In the Case of Denver versus New England, New England has playoff experience and home field in their favor, and Denver played a tougher schedule. New England is favored by my scheme by 69%. Home field plus Pythagoreans would favor New England by 88%.

The wins by Houston and New Orleans ensure that the #3 NFC and AFC seeds will be playing the #2 seeds, and that the #1 seeds will be playing the winner of the #4-#5 game. For now we’ll simply ask: if a team has playoff experience, but a rookie quarterback, does the rookie negate that experience advantage? Houston certainly looked good in their game.

Odds:

In San Francisco-New Orleans, the Saints have the advantage of playoff experience, but San Francisco has home field and a tough schedule. My code suggests the odds in this game are 50-50. In Baltimore-Houston, Baltimore has all three advantages, and is favored to the tune of a 81% chance to win.

Playoff experience is a potent effect, enough to overcome Denver’s advantages in home field and tougher schedule.

Steelers: Super Bowl last year, Away, SOS = -0.84, Pythagorean = 71.8%

Broncos: Last in playoffs 2005, Home, SOS = -0.23, Pythagorean = 35.3%

Typically in playoff games, you don’t see huge differences in offensive stats, because the teams that make it in the modern NFL tend to be good offensive teams.But Denver is nearly as bad this year as Seattle was last year (Seattle actually was worse, with a Pythagorean of 32.7%).  Treating this as a regular season game, instead of a playoff game would give PIT a 76% edge. Instead, using the playoff formula, PIT would be favored by 54%.

This playoff game is one of the more unique battles, as Houston is lacking playoff experience, as well as Matt Shaub. But with this post we’re going to introduce mods to the code we presented in our previous article, to allow us to set relative bounds on strength of schedule. The SOS effect has a large relative error, about 80%, so what happens to the odds in this matchup when we do exactly that?

Bengals: Playoff Exp 2 yrs ago, Away, SOS = -0.85, Pythagorean = 54.1%

Texans: No Playoffs ever, Home, SOS = -1.90, Pythagorean = 69.5%

Plugging these numbers into my formula, you get CIN favored by 66%. On the low end of the SOS relative, you get CIN favored by 60% and at the high end, CIN favored by 71%.  Given that 68% is what you get from playoff experience proper, the effect of Houston’s better record (and thus HFA) is roughly cancelled out by CIN’s better SOS.

So why isn’t Houston  favored  more, given their powerful offense? As stated previously, offensive metrics aren’t predictive to p = 0.05, more like p = 0.15 or so. Further, Houston had the easiest schedule in all of football. Cinncinnati also had an easy one, but not the easiest one.

Much as in the previous series, we’re going to analyze the playoff prospects of New Orleans and Detroit. We’re also going to post the code (very hacky) that I’ve been using to study playoff teams. The code (2 pics required) is as follows:

Now one thing about this code, because it’s using Getopts::Long, numbers have to be positive or else this code will think that the number is an option. The simple fix is to find  the value of the most negative SOS and add a positive number equal in magnitude to both SOSs. As the only important  value is the difference, this is a valid form of data entry.

Ok, the significant factors, plus Pythagoreans:

Detroit: No playoff exp, Away, SOS = 0.63, Pythagorean 62.9%

New Orleans: Won Super Bowl 2 years ago, Home, SOS = -1.60, Pythagorean 77.7%

Because NO’s SOS is negative, just let it equal zero and add 1.60 to the SOS of Detroit, yielding 2.23. That’s the info you would pump into the calculator above. And it gives you the  following results:

New Orlean’s advantage due to playoff experience alone give NO a 68% chance of winning.

Adding in home field advantage give New Orleans a 76% chance of winning.

Adding in strength of schedule reduces New Orleans chances to 69%. New Orleans is heavily  favored.

By comparison, after all is said and done, had Atlanta been slotted into this game, the playoff calculator gives Atlanta a 51% chance of winning. Atlanta has a slightly better SOS than Detroit, and it also has recent playoff experience.

Given how powerful the New Orleans offense is, should Atlanta have sought out a team with a weaker offense, such as New York? That’s one of the counterintuitive points of my previous playoff analysis. Offensive metrics tend to yield a p of 0.15, not 0.05. They’re suggestive, not etched in stone advantages. New Orleans’ powerful offense may come into  play, but then again, it may not.

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