OT: Interesting article on clock management failures

[QuantumMechanic]

A fascinating article from a Buffalo writer (!) on clock management issues. Worth the read!

For NFL teams, it's a game of beat the clock - BN Blitz

 

[Tony2046]

That was worth the read. Ernie Adams reputation continues to grow. Now if we could get an article that explained Pink Stripes. haha

The Columbia Ph.D., who has exchanged game theory notes with Patriots analytics guru Ernie Adams but doesn’t hear from anybody else aside from the occasional reporter, ran the calculations and was amazed.

Adams is a material component to the Patriots’ reputation as the best clock managers. He is said to have a photographic memory and is in Bill Belichick’s ear at every critical juncture of a game.

 

[Fencer]

Pretty good article. Hints at a kind of "state machine" model of football games, which is the right way to think about the issues.

I've had some of the same thoughts over the years ... but then, my PhD thesis was on zero-sum stochastic games.

 

[DarrylS]

When I watch other teams am pained about how poor their clock management is.. they have little concept of how to use the clock to their advantage, many do not even know how to use the 2 minute warning..

When I watch the Patriots am amazed at how well they use the clock.. BB/Brady are geniuses in this area.

Last Sunday night thought they seems a little sloppy with their clock management.. the five run plays they ran in the last series before the half before the incomplete pass to LaFell seemed uncharacteristic for this team.. thought they left possible points on the board there, from my couch looked like they were running the clock down. A field goal would have put them up 17-7...

 

[PP2]

Great find, QM. I enjoyed reading that article.

 

[RobertWeathers]

Great read!

 

[PatsWickedPissah]

A fascinating article from a Buffalo writer (!) on clock management issues. Worth the read!

For NFL teams, it's a game of beat the clock - BN Blitz

Great info! Why would someone Dislike this information? Maybe a mis-hit on a cellphone. It's happened to me.

 

[Ice_Ice_Brady]

A fascinating article from a Buffalo writer (!) on clock management issues. Worth the read!

For NFL teams, it's a game of beat the clock - BN Blitz

Thanks for posting this..great read! I found this part absolutely fascinating, and it confirms some more abstract, skeptical thoughts I've had about the ball-control offense:

While operating a ball-control offense keeps the other quarterback off the field, it reduces your offense’s possessions just the same. Meanwhile, your offense has taken itself out of its regular system, thereby playing at less-than-optimal efficiency.


“If you weaken your best offense just for the sake of holding the ball longer,” Sackrowitz said, “your probability of scoring goes down a bit. If they maintain their regular scoring rate on fewer possessions, but you’ve lowered your scoring rate, the math just doesn’t work out.


“The number of possessions you cut to shorten the game becomes too much to accomplish. You may end up with closer games, but you will lose more often.”


Sackrowitz’s research showed teams using a “time-consuming” offense decreased the chances of winning by a remarkable 12 percent against an opponent that remains in its regular offense.

 

[slam]

Thanks for posting this..great read! I found this part absolutely fascinating, and it confirms some more abstract, skeptical thoughts I've had about the ball-control offense:

While operating a ball-control offense keeps the other quarterback off the field, it reduces your offense’s possessions just the same. Meanwhile, your offense has taken itself out of its regular system, thereby playing at less-than-optimal efficiency.


“If you weaken your best offense just for the sake of holding the ball longer,” Sackrowitz said, “your probability of scoring goes down a bit. If they maintain their regular scoring rate on fewer possessions, but you’ve lowered your scoring rate, the math just doesn’t work out.


“The number of possessions you cut to shorten the game becomes too much to accomplish. You may end up with closer games, but you will lose more often.”


Sackrowitz’s research showed teams using a “time-consuming” offense decreased the chances of winning by a remarkable 12 percent against an opponent that remains in its regular offense.

Could this be correlation but not causation? A team trying to "shorten" a game is probably the objectively worse team that's the underdog.

 

[TB_Helmet]

Thanks for posting this..great read! I found this part absolutely fascinating, and it confirms some more abstract, skeptical thoughts I've had about the ball-control offense:

While operating a ball-control offense keeps the other quarterback off the field, it reduces your offense’s possessions just the same. Meanwhile, your offense has taken itself out of its regular system, thereby playing at less-than-optimal efficiency.


“If you weaken your best offense just for the sake of holding the ball longer,” Sackrowitz said, “your probability of scoring goes down a bit. If they maintain their regular scoring rate on fewer possessions, but you’ve lowered your scoring rate, the math just doesn’t work out.


“The number of possessions you cut to shorten the game becomes too much to accomplish. You may end up with closer games, but you will lose more often.”


Sackrowitz’s research showed teams using a “time-consuming” offense decreased the chances of winning by a remarkable 12 percent against an opponent that remains in its regular offense.

Yeah I always had trouble understanding the 'keep the other QB off the field' theory that people will often bring up when playing against Brady or Pink Head teams. If a ceratin QB normally scores 70% of the time and you score 50% of the time, then you're still going to lose no matter whether you limit that QB (and yourself) to 10 drives a game vs 6.

I guess fewer drives introduces even more variation to that 70%, but that variation could go either way (maybe the other guy scores on 5 of the 6 drives and you score on only 2 -- you still lose).

The only advantage to 'keeping the other QB off the field' that I can see is that it tires out the other team's defense.

 

[QuantumMechanic]

Yeah I always had trouble understanding the 'keep the other QB off the field' theory that people will often bring up when playing against Brady or Pink Head teams. If a ceratin QB normally scores 70% of the time and you score 50% of the time, then you're still going to lose no matter whether you limit that QB (and yourself) to 10 drives a game vs 6.

I disagree. If one team is inferior than I think it makes sense to use a ball-control offense even if it hurts your own offense, so long as it doesn't hurt it too much.

Let's assume a simple game where you can only score TDs. One Team P has a p chance of scoring a TD on a given drive and the other Team Q has a q (where p > q) chance of scoring a TD on a given drive. Under that assumption, the number of TDs a team scores in a game where they have N drives is given by the binomial distribution -- the chance of P scoring k TDs on N drives is:
(N choose k) p^k (1 - p)^(N - k)

Then you can do a brute-force sum to figure out what the chance is that in those N drives Team P will score more TDs than Team Q. I wrote a quick & dirty program to do that and here's those probabilities where p=0.50 and q=0.35:
p scores TD on 0.5 of drives, q scores TD on 0.35 of drives
2 drives, p wins: 0.43062500000000004
3 drives, p wins: 0.49196874999999995
4 drives, p wins: 0.535686328125
5 drives, p wins: 0.56991365234375
6 drives, p wins: 0.5981763696289063
7 drives, p wins: 0.6223186502685546
8 drives, p wins: 0.6434266423286438
9 drives, p wins: 0.662195075346031
10 drives, p wins: 0.6790962421555876

And here's the results where p=0.50 and q=0.30:
p scores TD on 0.5 of drives, q scores TD on 0.3 of drives
2 drives, p wins: 0.4724999999999999
3 drives, p wins: 0.5442499999999999
4 drives, p wins: 0.5954812499999999
5 drives, p wins: 0.6354818749999999
6 drives, p wins: 0.6683020312499997
7 drives, p wins: 0.6960855781249999
8 drives, p wins: 0.7201125298828128
9 drives, p wins: 0.7412131175585933
10 drives, p wins: 0.7599595403490231

Now, pretend that if Team Q uses its regular q=0.35 offense each team gets 8 drives but if Team Q uses its inferior q=0.30 ball control offense each team only gets 4 drives. In that case, Team P will win 64.3% of the time if Q plays "normal" but will only win 59.5% of the time if Q plays "ball control".

So even though Q is less efficient in "ball control" mode it will actually win more often than if it used its regular offense.

On the other hand, if Q's "ball control" mode drops its efficiency too much, then Q's chance of winning will be worse than if they had just left things alone.

Note - remember that there can be ties, which are not being counted in P's win percentage above -- those are the percentages for outright wins. So that's why Team P's outright win percentage is lower for small number of drives than you might think. For example, in the first 2-drive case Team P only wins approx 43% of the time. That does not mean Team Q won 57% of the time. In actuality, Q won even fewer times than P did, with the remainder being ties.

 

[zydecochris]

I disagree. If one team is inferior than I think it makes sense to use a ball-control offense even if it hurts your own offense, so long as it doesn't hurt it too much.

Let's assume a simple game where you can only score TDs. One Team P has a p chance of scoring a TD on a given drive and the other Team Q has a q (where p > q) chance of scoring a TD on a given drive. Under that assumption, the number of TDs a team scores in a game where they have N drives is given by the binomial distribution -- the chance of P scoring k TDs on N drives is:
(N choose k) p^k (1 - p)^(N - k)

Then you can do a brute-force sum to figure out what the chance is that in those N drives Team P will score more TDs than Team Q. I wrote a quick & dirty program to do that and here's those probabilities where p=0.50 and q=0.35:
p scores TD on 0.5 of drives, q scores TD on 0.35 of drives
2 drives, p wins: 0.43062500000000004
3 drives, p wins: 0.49196874999999995
4 drives, p wins: 0.535686328125
5 drives, p wins: 0.56991365234375
6 drives, p wins: 0.5981763696289063
7 drives, p wins: 0.6223186502685546
8 drives, p wins: 0.6434266423286438
9 drives, p wins: 0.662195075346031
10 drives, p wins: 0.6790962421555876

And here's the results where p=0.50 and q=0.30:
p scores TD on 0.5 of drives, q scores TD on 0.3 of drives
2 drives, p wins: 0.4724999999999999
3 drives, p wins: 0.5442499999999999
4 drives, p wins: 0.5954812499999999
5 drives, p wins: 0.6354818749999999
6 drives, p wins: 0.6683020312499997
7 drives, p wins: 0.6960855781249999
8 drives, p wins: 0.7201125298828128
9 drives, p wins: 0.7412131175585933
10 drives, p wins: 0.7599595403490231

Now, pretend that if Team Q uses its regular q=0.35 offense each team gets 8 drives but if Team Q uses its inferior q=0.30 ball control offense each team only gets 4 drives. In that case, Team P will win 64.3% of the time if Q plays "normal" but will only win 59.5% of the time if Q plays "ball control".

So even though Q is less efficient in "ball control" mode it will actually win more often than if it used its regular offense.

On the other hand, if Q's "ball control" mode drops its efficiency too much, then Q's chance of winning will be worse than if they had just left things alone.

Note - remember that there can be ties, which are not being counted in P's win percentage above -- those are the percentages for outright wins. So that's why Team P's outright win percentage is lower for small number of drives than you might think. For example, in the first 2-drive case Team P only wins approx 43% of the time. That does not mean Team Q won 57% of the time. In actuality, Q won even fewer times than P did, with the remainder being ties.

QM, during all of the time that I've been on this board, this may be the most interesting and well thought out post I have ever seen, kudos to you for this extremely thought provoking analysis. I agree with your conclusion, if one team P is more likely to score touchdowns than the other team Q, then the best strategy for Q to have the best chance for winning is to "keep the sample size small", and have as few possessions for both teams as possible. It is somewhat similar to the strategy that some well-coached college basketball teams used years ago before the adoption of the shot clock: when playing more talented teams, keep the number of possessions to a minimum to keep the "sample size" of shots down and hope for a few lucky bounces of the ball.

The other take-away I came away with is that on your top data set the P team scored 50% of the time while the Q team scored 35% of the time, which would seem to indicate that Team P is a much, much stronger team than Team Q. Nevertheless, even with 10 possessions for both teams (N=10), the Team P probability to win is only 68% (about 2 out of 3). It is something to think about next time we play a team that we are supposed to blow out, there is always a decent chance the ball could bounce the wrong way and we could lose. What does your model say for 14-15 possessions (is that a typical number for a real game?).

As N goes up the win percentage for Team P goes up, and it seems to me that trend should continue. Suppose we build a robot team P that scores on 50% of their possessions and a robot team Q that scores on 35% of their possessions. If these robot teams play continuously for a year or two and have, say, N=100,000 possessions, does your computer model show that the win percentage for P becomes almost 100%? It seems to me that should be the case, no?

 

[QuantumMechanic]

For the same 0.5/0.35 numbers:
10 drives, p wins: 0.6790962421555876
11 drives, p wins: 0.6944666378066617
12 drives, p wins: 0.7085550971928166
13 drives, p wins: 0.7215512815129138
14 drives, p wins: 0.7336034158735196
15 drives, p wins: 0.7448298075999882
16 drives, p wins: 0.7553265943976104
17 drives, p wins: 0.7651731153994633
18 drives, p wins: 0.77443573215711
19 drives, p wins: 0.7831706090174656
20 drives, p wins: 0.7914257768824896
(skipping)
100 drives, p wins: 0.9816787528443454

It's like being at a casino -- the more wagers you place, the more certain the casino will make money off you.

 

[BradyFTW!]

That was worth the read. Ernie Adams reputation continues to grow. Now if we could get an article that explained Pink Stripes. haha

Very interesting. I also remember that 7 or 8 years ago, reading some article about an MIT guy who published a paper about the expected points of going for it on 4th down vs. kicking/punting at various distances and field positions. He said that the Patriots (Adams) were the only team that contacted him, and that it led to a very long, in-depth conversation on exactly how the Patriots could optimize their strategy there too.

 

[BradyFTW!]

I disagree. If one team is inferior than I think it makes sense to use a ball-control offense even if it hurts your own offense, so long as it doesn't hurt it too much.

Absolutely true. It's the law of large numbers in action, essentially. Same reason why upsets happen a lot more frequently in the NFL playoffs (single elimination) than the NBA playoffs (best of seven). The more data points you have, the more likely it is that you'll get the overall result that you 'should' get.

 

[zydecochris]

It's like being at a casino -- the more wagers you place, the more certain the casino will make money off you.

Exactly. If the odds are against you, it is best to minimize the sample size, whether it is wagers at a casino or football or basketball possessions against a superior team.

For casinos, there are evidently a very few card countering experts who can swing the odds in their favor in Blackjack (I think they are banned at the casinos) and perhaps a few Sharps know so much about sports betting that they can swing the odds in their favor. However, in all other cases the odds are in favor of the casino as opposed to the casino visitor. Thus, as you say, the more wagers you make (i.e., the greater the sample size) the more certain you will lose money.

Thus, if the goal is only to maximize your chances to win money, it is best to minimize the sample size to one throw or roll of the dice. For example, if you have some stake you can afford to lose (say its one million dollars), the most likely way to double it would be to put it all on one throw. For the case of roulette, you could put all million dollars on red, and then you would have 47.3% chance of doubling your money (to two million dollars), and either way you walk away subsequently.

In contrast, as you point out, the way you would almost certainly lose money is to have a gigantic sample size and make a gigantic number of wagers. For the same roulette example with a million dollar stake, if you put one dollar on red a million times, with such a large sample size, not only would there be virtually zero chance of doubling your money, there would also be virtually zero chance that you would not have lost part of your original stake (probably about 5%).

Of course, the strategy of putting all of your stake on one throw or roll probably wouldn't be the most entertaining way to spend a Las Vegas weekend, but it would maximize the probability of winning.

....I guess that is why I only go to Las Vegas for the shows.

 

[TB_Helmet]

I disagree. If one team is inferior than I think it makes sense to use a ball-control offense even if it hurts your own offense, so long as it doesn't hurt it too much.

Let's assume a simple game where you can only score TDs. One Team P has a p chance of scoring a TD on a given drive and the other Team Q has a q (where p > q) chance of scoring a TD on a given drive. Under that assumption, the number of TDs a team scores in a game where they have N drives is given by the binomial distribution -- the chance of P scoring k TDs on N drives is:
(N choose k) p^k (1 - p)^(N - k)

Then you can do a brute-force sum to figure out what the chance is that in those N drives Team P will score more TDs than Team Q. I wrote a quick & dirty program to do that and here's those probabilities where p=0.50 and q=0.35:
p scores TD on 0.5 of drives, q scores TD on 0.35 of drives
2 drives, p wins: 0.43062500000000004
3 drives, p wins: 0.49196874999999995
4 drives, p wins: 0.535686328125
5 drives, p wins: 0.56991365234375
6 drives, p wins: 0.5981763696289063
7 drives, p wins: 0.6223186502685546
8 drives, p wins: 0.6434266423286438
9 drives, p wins: 0.662195075346031
10 drives, p wins: 0.6790962421555876

And here's the results where p=0.50 and q=0.30:
p scores TD on 0.5 of drives, q scores TD on 0.3 of drives
2 drives, p wins: 0.4724999999999999
3 drives, p wins: 0.5442499999999999
4 drives, p wins: 0.5954812499999999
5 drives, p wins: 0.6354818749999999
6 drives, p wins: 0.6683020312499997
7 drives, p wins: 0.6960855781249999
8 drives, p wins: 0.7201125298828128
9 drives, p wins: 0.7412131175585933
10 drives, p wins: 0.7599595403490231

Now, pretend that if Team Q uses its regular q=0.35 offense each team gets 8 drives but if Team Q uses its inferior q=0.30 ball control offense each team only gets 4 drives. In that case, Team P will win 64.3% of the time if Q plays "normal" but will only win 59.5% of the time if Q plays "ball control".

So even though Q is less efficient in "ball control" mode it will actually win more often than if it used its regular offense.

On the other hand, if Q's "ball control" mode drops its efficiency too much, then Q's chance of winning will be worse than if they had just left things alone.

Note - remember that there can be ties, which are not being counted in P's win percentage above -- those are the percentages for outright wins. So that's why Team P's outright win percentage is lower for small number of drives than you might think. For example, in the first 2-drive case Team P only wins approx 43% of the time. That does not mean Team Q won 57% of the time. In actuality, Q won even fewer times than P did, with the remainder being ties.

Hey thanks for doing that -- definitely answered my original question! I'm a stats guy too so no arguments here especially after you broke it down with a binomial distribution and I assume some python coding.

A couple notes:

-- I'm sure that when NFL analysts use the same tired cliche of 'you have to keep Brady off the field in order to win', they have no idea why, they're just repeating what they've been told. They certainly aren't thinking about weighted coin flips.

-- The original article says that the ball control strategy is not a good idea, but the argument it makes is that the team is running an offense it's not used to -- so if q is 0.35 in that team's normal offense, it could be a lot less in their modified offense. At some point q drops low enough that the overall chances of winning decrease relative to the inferior team's normal offense.

-- None of this accounts for giving the other team a short field, special teams TDs, or 99 yard interception returns

 

[PatsWickedPissah]

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