Navigation: Jump to content areas:


Pro Quality. Fan Perspective.
Login-facebook
Around SBN: The Worst Team Ever Projected?

In Support of Random Play Calling

Chris at Smart Football has discussed the merits of calling plays randomly on a couple occasions, including earlier this month where he found some support for it from Bill Walsh. Essentially, the argument in favor of it breaks down like this: in a game where everyone is trying to out-think one another, the only way to eliminate the advantage of superior thinkers is to do things randomly.

I've been vaguely in favor of this, but I saw something this morning on Slashdot that really helped the principle hit home. It's the Tuesday Birthday Problem, and it has generated a lot of heated discussion among mathematicians. Simply put, it is this:

I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?

The classic answer to this problem is 1/3. There are four equally likely possibilities for the birth order of any two children: boy-boy, boy-girl, girl-boy, and girl-girl. The last one is impossible because one child is a boy, so three possibilities remain. Among those only one has a boy-boy outcome, therefore the probability is 1/3.

A Stanford mathematician went all-out and took the day of the week into account. He used the assumption that by specifying the day of the week the one child was born on, the other must not have been born on a Tuesday. In that scenario, the probability of two boys is 13/27, which is almost 1/2. Looking at it that way, it suggests that the classic answer to the puzzle is wrong.

There's a good reason for that: it is wrong*. Assuming that the there's an equally likely chance of a boy or girl birth (and that the two kids aren't twins), the probability of the second child being a boy is 1/2. The probability breakdown of the three options listed in the classic answer really isn't equal; it's 50% for boy-boy and 50% for not boy-boy, with a 25% chance each for the boy-girl and girl-boy orders. The gender of the unknown child is completely independent of your knowledge of the other child's gender, and it conforms to the 50-50 chances of all children.

So what does this have to do with play calling? I think this problem, along with Car Talk's Monty Hall problem, shows that our minds are not wired for dealing with true randomness. In fact, sometimes an answer that feels elegant (like the classic solution above) can be flat-out wrong debatable. It takes study and discipline to deal with it properly, and in the heat and high emotional level of a game, it's likely that opposing coaches and especially players will not have that discipline at all times.

Star-divide

They'll likely even try to read into things and find patterns that aren't there, like a person fully expecting to see tails after flipping a coin and getting heads ten times in a row. The coin isn't due for tails; the probability for tails on every flip is still 50%. People are hard wired for finding patterns even when they're not there, and it's difficult to overcome that.

As Chris mentions, doing things truly randomly is probably not a good idea. He points out that you'd want to pick randomly from plays that complement and counter each other. I'd also add that true randomness might get you calling a Hail Mary play on first-and-goal from the two yard line. You might catch the defense off guard with that, but your likelihood of scoring would be much less than if you called the utterly predictable halfback dive.

Ultimately I doubt we'll ever see anyone call plays with true randomness, as you'd need a Mike Leach-sized playbook or an even smaller one to make it manageable. Perhaps just flipping a coin for running and passing could work, but as someone who's never coached, I can't tell you how well that would work out.

If it ever does happen, it probably would occur in the one of the vast laboratories known as high school football and small college football. Many of the innovations in offense of the past couple decades have come from those locales. I think it's worth a shot; you literally can't know for sure what will happen.

*Note: This is a riddle with a lot of different right answers. The one I posted is the one I think is most correct (because I view the children's births not as a set of probabilistic events but as two independent events), but it is by no means the only answer. I should not have made it sound like it was the only one, because it's not.

Comment 28 comments  |  0 recs  | 

Do you like this story?

More from Team Speed Kills

Offense or Defense?

Aug 2009 by Year2 - 3 comments

Comments

Display:

OK, I read thru this and got that Monty Hall was born 1/3 on a Tuesday and is 1/2 a girl, and believes half back dive plays are better than hail mary’s. God can the season come soon!!!!

by oldtimegator on Jun 29, 2010 11:26 AM EDT reply actions   1 recs

LSU's playcalling last year

Seemed like an exercise in randomness at times, so I can vouch for that not working…

by Billy Gomila on Jun 29, 2010 11:50 AM EDT reply actions  

Hard as this may be to believe

That was an exercise in over thinking at it’s best!

by blackertai on Jun 29, 2010 2:22 PM EDT up reply actions  

The Gary Crowton principle of awfulness.

It strikes again.

"Everyone counted us out. I don't know why they keep doing that." -- Kyle Wilson

by Loque on Jun 30, 2010 2:15 AM EDT up reply actions  

Football is the most over-analyzed sport in the world

and let’s say you can somehow randomize your playbook. What good will it do you? You get some advantage of suprise, but as you pointed out you might get a lot of inappropriate calls.

Despite the innovations of the spread, the West Coast Offense, the veer, the wishbone, etc, etc, etc, the game of football is basically a brutal one of blocking, tackling, and running. Since college football only has a few hours to drill the team, the team with the best fundamentals usually wins. If you’re spending all your time crafting elaborate playbooks, you are spending less time teaching the basics.

Which would you rather have, a playbook with 100 plays, only 10 of which you run well, or a playbook with 10 plays, all of which you run well? In college, I don’t believe you can have both.

"It's not the size of the dog in the fight, it's the size of the fight in the dog." - Bear Bryant

by NJBammer on Jun 29, 2010 11:51 AM EDT reply actions  

Like I said...

You can’t just take a regular playbook and go random from there. You’d have to craft things with the knowledge you’d be going random, meaning a smaller number of plays that explicitly counter each other. You’d also have to have different subsets of plays appropriate for the different situations you face. Just because the calls would come at random doesn’t mean you don’t apply some thinking in your preparation.

The point is that, given the incredible amount of film analysis that goes on, it’s possible to find teams’ tells. The anecdote from Bill Walsh tells of how he was able to predict when teams were going to blitz with high accuracy. One of the sites Chris linked to shows that teams are more likely to run on second-and-10 than on second-and-nine or -11, because second-and-10 usually comes after an incomplete pass and coaches tend to call runs after first down incompletions. You can find all kinds of things like that and game plan for them. Randomizing calls takes that away because it becomes impossible to know what’s coming next.

The opposite end of the spectrum is what I think you’re trying to describe and is what Alabama has traditionally done: we don’t care if you know what’s coming because we’re going to out-block, out-muscle, and out-execute you. That works well if you have a talent advantage, but few teams have the personnel to make it win games for them. Most teams have to catch the defense off guard in some way beyond just play action, and randomized play calling would be one way to accomplish that.

Team Speed Kills
SBNation's SEC Blog

by Year2 on Jun 29, 2010 12:17 PM EDT up reply actions  

This is the Paul Johnson offense

Totally random choice from him. There are what, 3 plays in his playbook?

by blackertai on Jun 29, 2010 2:23 PM EDT up reply actions  

I see your point

I do. However in the college game there is so little time to prepare your players, IMHO it is far more important to work on execution than scheme or playcalling. If you have a talent deficit, you would tend to go with higher risk/higher reward schemes to maximize your odds of winning on any given sunday, if you have a talent surplus you run safe and conservative to try to grind your opponent down.

But in either case, my real point is a team should figure out who they are and then work on being that to the best of their ability, rather than waste time worrying about randomizing their playcalling.

I grant that this doesn’t apply as well in the pros, where the teams spend thousands of hours a year teaching execution, and have the luxury of gunning for any advantage they can in the area of playcalling, schemantic advantage, whatever. But in college most teams really could just tell the other teams what they are about to run and I doubt it would make a heck of a lot of difference.

"It's not the size of the dog in the fight, it's the size of the fight in the dog." - Bear Bryant

by NJBammer on Jun 29, 2010 2:55 PM EDT up reply actions  

If playcalling is randomized, it isn't really a waste of time.

And if you’re not wasting your time worrying about play calling, since it is going to be generated by a Keno computer or American Legion Bingo ball, that would leave you with plenty of time to work on execution and fundamentals, right?

by marktheshark on Jun 29, 2010 3:11 PM EDT up reply actions  

Exactly

You don’t run through play calling during practice, and you can’t work on execution outside the NCAA-mandated 20 hours of practice a week during certain times of the year. Putting together sets of plays for certain situations is something coaches already do.

The difference is between calling them based on a strategy or a “feel for the game” and calling them based on randomness. There very well could be something to be said for going the random route because it’s impossible to accurately predict what a random team will do next.

Team Speed Kills
SBNation's SEC Blog

by Year2 on Jun 29, 2010 4:06 PM EDT up reply actions  

a few points

1) the benefit at the college level of randomizing your playcalling is almost zero in terms of lift you get for the unexpected.
2) Coaches need to call plays based on their own understanding otherwise they will learn less about coaching. i.e. if you play poker letting your computer call the bets randomly you may win a few hands but you won’t learn about poker.
3) Related to #2, sometimes you want your opponent to think he can predict your call, so your suprise is a greater one. Sort of how a great poker player will be caught bluffing intentionally to set up his opponent for later.

But then again, I could be wrong. I just don’t think at the end of the day randomizing is worth it at the college level.

"It's not the size of the dog in the fight, it's the size of the fight in the dog." - Bear Bryant

by NJBammer on Jun 29, 2010 4:36 PM EDT up reply actions  

Um, the classical answer is the classical answer because it's right.

The point of the problem is that most people say “1/2” when the true answer is “1/3” because people, for fairly valid reasons, don’t think about it right.
The question as properly stated, is the same as “Of the two-child families with at least one boy, how many have two boys?” and the answer is “1/3 of them.”
The question, as people tend to interpret it, is “You find out the gender of one random child in a two-person family, and that child is male. What are the chances that both children are male?” The answer to this is “1/2.”

The way it’s supposed to work, people ask something which is essentially the first question and the human mind naturally views it as the second (because the second version is the way things usually work in real life, and people are notoriously bad at adjusting their thinking and will think about the 10%-of-the-time first situation in a way that’s appropriate for the 90%-of-the-time second question situation). But often, the person asking the question will get confused, ask a question which is essentially the second version, and then tell the listener that the answer is “1/3.” So, sometimes “as people interpret it” is due to faulty human intuition, sometimes it’s due to them correctly interpreting the incorrect question.

"This World Cup has turned out like World War 2! The French & Ialians surrendered early, the USA arrive at the last minute and the English are left to fight the Germans!"

by AllSaintsDay on Jun 29, 2010 5:53 PM EDT reply actions  

It all depends on interpretation

If the children are twins, then the probabilities change (see the link). If the family has more than two children (because it doesn’t say the person has two and only two children), the probability changes. If you interpret it as the Stanford guy did, you don’t get 1/2 or 1/3.

That’s what makes it such a great riddle. You can read into it whatever you want. For my money, I still say the births are independent events and the outcome of one doesn’t affect the other. The flip of that coin is still a 50-50 affair no matter what happened on the other one. But I’ve always been one to compartmentalize.

Team Speed Kills
SBNation's SEC Blog

by Year2 on Jun 29, 2010 11:40 PM EDT via mobile up reply actions  

The Stanford guy didn't interpret anything differently.

He would tell you that the probability in the all-you-know-is-gender case is 1/3. In fact, he did. His name is Keith Devlin. The article you mention is here, and in talking about the original puzzle he says

Another probability question that causes many people difficulty is the children’s gender puzzle: I tell you I have two children and that (at least) one of them is a boy, and ask you what you think is the probability that I have two boys. Many people, when they hear this puzzle for the first time, give the answer 1/2, reasoning that there is an equal likelihood that my other child is a boy or a girl. But this is not correct. Based on what you know, you should conclude that I am actually twice as likely to have a boy and a girl as I am to have two boys. So your right answer to my question is not 1/2 but 1/3.

He then goes on to say, about the “boy born on a Tuesday” variation, that

The correct answer to the new puzzle is 13/27, just slightly less than 1/2, and not at all close to 1/3.
If you say that the day-of-the-week thing is a new analysis of the old situation, you’re arguing against the Stanford mathematician, because he said it was a new puzzle.

by AllSaintsDay on Jun 30, 2010 1:02 PM EDT up reply actions  

And I’m sorry if that seems a bit snippy. I’m just very certain that there are two different puzzles, whose answers are indisputably 1/3 and 13/27, and I want to make the point that I am very certain. (And I am certain of this because I’ve taken two undergraduate and one graduate course on probability, and I’m already well aware of the correct analysis of each problem because I’ve read that column before.)

by AllSaintsDay on Jun 30, 2010 1:10 PM EDT up reply actions  

The help me understand

I’m not looking at this as a set of two children. I’m looking at the births of the children as two separate events. Let’s even call them child A and child B.

If the gender of both children is unknown, then what’s the probability that child B is a boy? It’s 1/2, given the generally held assumptions about the problem. I think we can both agree on that.

Now, I tell you that child A is a boy. What is the probability that child B is a boy? If it’s something other than 1/2, that means making an observation had changed the probability of child B’s gender. If the answer then becomes a 1/3 probability that child B is a boy and 2/3 that it’s a girl, it implies that a family having a boy as their first child is twice as likely to have a girl than a boy as their second child. Given that the gender of the second child is completely independent from the gender of the first child’s, how can that be?

What am I doing wrong here? I’d like to know because I have not taken as many classes on probability as you have.

Team Speed Kills
SBNation's SEC Blog

by Year2 on Jun 30, 2010 3:18 PM EDT up reply actions  

The nuances of probability

One problem is your view of them as sequential events. The question isn’t “My first child is a boy, what is the probability that my second child will be a boy;” it’s “One of my 2 children is a boy; what is the probability that they are both boys.”

In your terminology, consider that child A and child B are indistinguishable. In other words, child A isn’t necessarily the older one, but we do know child A is a boy. Important: Child B is already born. So it’s “Knowing child A is a boy, what is the probability that child B is also a boy.” Since both children have already been born, there are indeed four possibilities: B-B, B-G, G-B, G-G, of which the last is ruled out.

by duker on Jun 30, 2010 7:24 PM EDT up reply actions  

The part about “it implies that a family having a boy as their first child is twice as likely to have a girl than a boy as their second child” involves sequence; the rest of my reply does not.

Put it this way:

What is the probability that a particular child is male? It’s 1/2. Now I will give you a piece of information unrelated to that child. What is the probability that the child is male? Still 1/2.

In this case, the piece of unrelated information is the gender of the child’s sibling. It could be an older sibling; it could be a younger sibling. Either way, the gender of that sibling has zero effect on the gender of the child in question. It’s still 1/2.

Team Speed Kills
SBNation's SEC Blog

by Year2 on Jun 30, 2010 9:26 PM EDT up reply actions  

You know there are 2 kids. Possible combinations: B-B, B-G, G-B, G-G.

P(both boys) = 1/4. P(not both boys) = 3/4.

By Bayes’ Theorem:

P(both boys | at least 1 boy)
= P(at least 1 boy | both boys) * P(both boys) / P(at least 1 boy)
= 1 * (1/4) / (3/4) = 1/3

In minimal probability terminology:

With no info, the probability that in a pair of children both are boys is 1/4. The probability that not both are boys is 3/4. If I reveal that at least 1 is a boy, then that rules out G-G, so the probability that not both are boys is now 2/3. So the probability that both are boys is 1/3.

Not sure if it’s affecting the thinking here, but the Monty Hall problem is different than this one.

by duker on Jun 30, 2010 10:56 PM EDT up reply actions  

I've come up with the best explanation of my answer yet

This is what I’ve been understanding intuitively but haven’t been able to put into words until now. Let’s call the son who was born on Tuesday “Charles” for simplicity. There are four possibilities for the births of the two children:

Charles – Boy
Charles – Girl
Boy – Charles
Girl – Charles

The probability of having two boys is 1/2. There isn’t just one single boy-boy possibility; there are two because Charles could have come either first or second.

Team Speed Kills
SBNation's SEC Blog

by Year2 on Jun 30, 2010 11:29 PM EDT up reply actions  

I did some reading

A hotly debated topic indeed. It really depends on how the question is asked and sometimes even how it’s interpreted. For example, the way you pose it in your post, I think “of all 2 children families, if I choose one at random, the prob of getting one with 2 boys is 1/4.” Similarly, of all families with 2 boys and in which at least 1 is a boy, the prob of getting one with 2 boys is 1/3. However, if you view it as independent events, the prob is 1/2, as you say. Apparently, both are accepted and backed by statisticians unless the question is posed in a way that inhibits one interpretation from being reasonable.

by duker on Jun 30, 2010 11:22 PM EDT up reply actions  

Well, for one you don’t “view it as independent events.” The events either are or aren’t independent. In this case, they are independent if we start off talking about all two-child families.

The situation is the same as “I’ve flipped two coins. Given that a coin landed heads, what’s the probability they both landed heads?” And you can actually do this for yourself and see that if you flip a pair of coins a bunch of times, about a third of the “at least one heads” flips will have two heads, even though the two coin flips are, in one sense, independent.

The reason is that you’re designating a first child and a second child, even though the problem doesn’t do that. It doesn’t matter if you designate them “Child A and Child B” or “oldest and youngest” or “Charles and not-Charles.” Designating one child as “the boy he told me about” or “the Tuesday boy” when the problem just says “at least one boy” rather than “this particular child is a boy” changes the problem, and the fact that people insist on designating them is why it has a counter-intuitive answer.

Maybe I can explain it better if you can tell me this: If I flip two coins, get two heads, and tell you “Yeah, I got at least one heads,” which coin was I telling you about? You aren’t allowed to say something like “Either one, it doesn’t matter.” You have to point to one particular coin and say “That one.”

Or, to make it topical, if I tell you that UCLA didn’t sweep USC (i.e., the Cocks got at least one win), which particular South Carolina win was I telling you about?

Prone to asking "Who Dat Say Dey Gonna Beat Dem Saints!?", waving my arm in a tomahawk fashion and doing the War Chant, yelling "Tiger Bait" at passersby, and throwing up the O.

by AllSaintsDay on Jul 1, 2010 2:55 AM EDT up reply actions  

From the article
Gardner himself tripped up on his simpler Two Children Problem. Initially, he gave the answer as 1/3, but he later realized that the problem is ambiguous in the same way that Peres argues that the Tuesday Birthday Problem is. Suppose that you already knew that Mr. Smith had two children, and then you meet him on the street with a boy he introduces as his son. In that case, the probability the other child is a son would be 1/2, just as intuition suggests. On the other hand, suppose that you are looking for a male beagle puppy. You want a puppy that has been raised with a sibling for good socialization but you are afraid it will be hard to select just a single puppy from a large litter. So you find a breeder who has exactly two pups and call to confirm that at least one is male. Then the probability that the other is male is 1/3.

In the scenario of Mr. Smith, you’re randomly selecting a child from his two children and then noticing his sex. In the puppy scenario, you’re randomly selecting a two-puppy family with at least one male.

Since the Mr. Smith scenario is the most likely situation for running into this in the real world, that’s what I go with. But it could be either one depending on your interpretation. Both 1/2 and 1/3 are acceptable answers depending on your interpretation of the problem.

Team Speed Kills
SBNation's SEC Blog

by Year2 on Jul 1, 2010 7:47 AM EDT up reply actions  

While I agree that 1/2 is a legitimate answer in some cases, I don’t agree with the Mr. Smith case. Since you know he has 2 children, his family is one of many 2-children families. When you meet a son, you know that his family is one of many 2-children families with 1 son. Of those, 1/3 have a 2nd son while 2/3 have 1 child of each gender.

However, if Mr. Smith introduced his son as his “younger” or “older” son, then I agree that the probability would be 1/2.

As with many probability problems, I find frequency analysis to help a lot. Consider (from another article…please forgive the article’s misuse of the superlative):

Begin with a group of 100 families, each with two children, distributed as follows:

25 families with oldest child a boy and youngest child a boy.
25 families with oldest child a boy and youngest child a girl.
25 families with oldest child a girl and youngest child a boy.
25 families with oldest child a girl and youngest child a girl.

Of this group, there are 50 families in which the oldest child is a boy. Of those 50 families, there are 25 families in which the youngest child is also a boy. In other words, out of the group of families in which the oldest child is a boy, 50% have two boys.

From the same group, there are 75 families in which at least one child is a boy. Of those 75 families, there are 25 families in which the other child is also a boy. In other words, out of the group of families in which at least one child is a boy, only 33% have two boys.

In my opinion, the Mr. Smith example you reference is clearly the latter.

by duker on Jul 1, 2010 10:21 AM EDT up reply actions  

But the Two-Child Problem isn’t about the Mr. Smith scenario. Asking someone a question to which they’ll probably answer 1/2, and then saying “Aha! The real answer is…1/2!” is dumb.

But there is a question out there, which will crop up occasionally (but less than half the time), where the answer is 1/3. And the point of the Two-Child Problem (if it’s correctly stated) is that if you insist on saying “I’m going to take the interpretation which is correct, say, 90% of the time, and act like it’s correct 100% of the time and the other interpretation simply doesn’t exist.” then you should not claim that you understand probability/randomness/&c.

The correctly asked Two-Child Problem seems like the answer should be 1/2, but it’s actually 1/3. If the correct answer is 1/2, it’s because someone asked it wrong, because the entire point of the Two-Child Problem is to make it that 10% situation.

Prone to asking "Who Dat Say Dey Gonna Beat Dem Saints!?", waving my arm in a tomahawk fashion and doing the War Chant, yelling "Tiger Bait" at passersby, and throwing up the O.

by AllSaintsDay on Jul 1, 2010 1:50 PM EDT up reply actions  

Nerd Alert!

As I was reading this article, I became fixated on the “riddle”, and very little else. The answer is “1/3” only if the children are indistinguishable, in a quantum mechanical sense. In fact, it was a problem very similar to this one (except involving photons instead of children) that led to the development Bose-Einstein statistics (a central concept in QM).
There, that’ll teach you to post riddles.

by dxf04 on Jun 29, 2010 9:52 PM EDT reply actions   1 recs

in a game where everyone is trying to out-think one another, the only way to eliminate the advantage of superior thinkers is to do things randomly.

If this is true, shouldn’t only about half of the teams benefit from random play calling since the other half would have better than average play anticipation abilities?

"Infield hits are sexy, because they require technique."

-Ichiro

by VivaLosBravos on Jun 29, 2010 11:34 PM EDT reply actions  

It depends

You could have a lot of guys that are at roughly the same ability level. Coaches have good days and bad days, and it’s nearly impossible to predict when each of those happen. Scheduling isn’t balanced, so if a team with, say, the 60th best thinker plays a large number of teams with coaches ranked below him (like if it was the best team in the Sun Belt), then going random might not be the best for it.

It also should be noted that game planning would be incredibly important to any random play calling scheme, and the ability of coaches to do proper game planning is decidedly nonrandom.

Team Speed Kills
SBNation's SEC Blog

by Year2 on Jun 30, 2010 7:51 AM EDT up reply actions  

Comments For This Post Are Closed


User Tools

Welcome to the SB Nation blog about the SEC

FanPosts

Community blog posts and discussion.

Recent FanPosts

Hatbeard__2__small
More Rumors on LSU Locker Room
X2_6e41244_small
Finally some proof that the SEC is the best conference
Small
College Revenues 2011
4238784107_small
Richt the Rule Breaker?
Small
A Fair Way to Determine the National Champion
Dool-aid_small
With Due Respect to Rick Reilly
Small
Playoff Idea: The World Cup of College Football
Coffee_small
Why is the BigXII (-4) getting so much love by the BcS computers?

+ New FanPost All FanPosts >


Managers

Gabalogo2_small cocknfire

Gator-f__custom__small Year2

Authors

Kleph_logo_copy_small kleph