Two point or not two point

This NFL season, Bill Barnwell is preaching quant analysis to the masses over at Grantland. My friend Chris linked me to his article yesterday, where he tackles a couple of topics: icing the kicker with a time out and going for a two point conversion. Coaches face these decisions a lot, so I appreciate the effort to tackle them rigorously. However, Chris and I found some flaws in Barnwell’s take.

Warmup: Icing the Kicker 

Barnwell shows some probabilities from Tobias Moskowitz’s book, Scorecasting, on iced vs. non-iced kickers. We don’t see the statistical significance of the differences (so it’s hard to know whether small differences are due to icing or due to chance), but it appears icing doesn’t matter. Icing slightly improves field goal percentage with under 15 seconds remaining and decreases field goal percentage with 15 seconds to 2 minutes remaining, but never by more than 2 or 3 percentage points either direction.

So if icing doesn’t matter, why do we care if coaches do it? Barnwell says they shouldn’t bother; I say, let the poor guy call a timeout if he wants. It’s all he’s got, and there’s no evidence that it hurts. Let’s talk about more important decisions.

The Main Event: Two Pointers

Barnwell tackles the specific situation faced by the Vikings (vs. Broncos) and Packers (vs. Giants) last Sunday. The Wisconsesotans both scored fourth quarter TDs, putting them up 7 before the conversion. They both lined up for extra points, giving their opponents a chance to score a TD and two point conversion for the tie. A two point conversion would have made the lead 9, putting the game out of reach for the Packers (the Vikings still had 9 minutes of clock to kill). A two point miss would have still assured a seven point lead.

Should they have gone for two? Barnwell says yes for the Packers at least, but these guys have fancy charts that tell them what to do, so I’m going to guess no. The probability analysis here is deeper than a simple comparison, and there are lots of different outcomes. When in doubt, write it all out.

Let’s stick with the Packers’ late game scenario, which has fewer possible outcomes. Assume the Packers and Giants always convert one pointers and convert two pointers with probability p. Since you either succeed or fail and probabilities must add up to one, the probability of failure on a two pointer is 1 – p. If the Packers go up by 9, it’s safe to assume that they always win. Give the Giants probability q of scoring a TD (probability 1 – q that they fail). Let’s say it’s 50/50 who wins if the Giants tie, because of the importance of the coin flip.

If the Packers go for one (up by 8), they can still win in three ways: failed TD drive, failed two point, or overtime/last second win (the Packers pulled off a crazy last second win, but we’ll lump that in with OT for simplicity). Here are the probabilities of those outcomes:

P(failed Giants TD drive) = 1 – q
P(Giants TD drive but failed two point) =  q(1 – p)
P(Giants TD drive and two point but Packers OT win) = 0.5qp

The total probability of a Packers win — if they go for one — is the sum:

P(Packers win | go for one) = 1 – q + q(1 – p) + 0.5qp
                                             = 1 – qp + 0.5qp
                                             = 1 – 0.5qp

If they go for two, things depend on whether they make it. Get the two, and the Packers win for sure. Miss, and they can still win if the Giants fail to score a TD, or score but lose in OT. New probabilities:

P(Packers make two point) = p
P(Packers miss two point and failed Giants TD drive) = (1 – p)(1 – q)
P(Packers miss two point and Giants TD drive but Packers OT win) = 0.5(1 – p)q

The sum is the probability of a Packers win if they go for two:

P(Packers win | go for two) = p + (1 – p)(1 – q) + 0.5(1 – p)q
                                             = 1 – q + pq + 0.5q – 0.5pq
                                             = 1 – 0.5q + 0.5pq
                                             = 1 – 0.5q(1 – p)
Which one of these is bigger? It all boils down to p versus 1 – p. Or, in English, it only matters how good the Packers (and Giants, since we assumed the same probability) are at two point conversions. This result is intuitive. In the outcomes above, it’s clear that only one team is going to go for two. Whichever team goes for two faces relatively good odds to win if they succeed, but relatively poor odds to win if they fail. The Packers get to decide first, so if p is high, they should be happy to take on the two point risk — they would be more likely to land in the good odds. If not, they should skip the two pointer and force that risk on the Giants. Specifically, if p < 0.5, then p < 1 – p, and the Packers should go for one.
Barnwell does a good job discussing the odds of converting two pointers. However, the general conclusion is that teams convert less than 50 percent (perhaps they would do better if they ran more, and Tebows and Newtons help too). I bet that Mike McCarthy looked down at his chart, and it said, “go for one,” so he did.
It makes you wonder how they calculate these charts! I’ve looked into it a little bit, and they are based on long probability branches, but chart-makers have to calibrate all those probabilities. If I were a coach, I’d probably follow my gut outside of 5 or 6 minutes to go, when there are even more uncertain outcomes.
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One response to “Two point or not two point

  1. Pingback: Sunday night madness and just a little Tebow | Causal Sports Fan

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