Weighted Fair Division

I'm sure this is an old puzzle somewhere in the world, but it came upon us a few years ago here at work in connection with driving to lunch.

Where I work, the company provides a cafeteria where one may purchase lunch.  Unfortunately, the lunch is either too expensive or not good enough, depending on your point of view, so we generally eat out.  We're lucky that we can do that.  Anyway, in general, we try to take turns driving so that we're all likely to drive about equally often.  It doesn't always work out that way, but that's the aim.

If we all ate out every meal, it'd be simple; we'd all drive with equal probability.  But what happens if, as has been the case occasionally throughout the years I've worked here, one of us can only eat out once per week?  How often should that person drive, when they do eat out?

To make things simpler, let's assume that there are two of us daily diner (five times per week), and one single-day diner.  Four days out of the week, there are only two diners.  If each one drives one-half of the time, then both of them end up driving two days out of the four.

On the last remaining day of the week, when there are three diners, should each drive one-third of the time?  Well, if we do things that way, then each of the two diners drives 2-1/3 days, on average, whereas the single-day diner drives just 1/3 day per week, on average.  That's not fair, because the two daily diners drive seven times as much as the single-day diner, even though they only eat five times as often.  The single-day diner should have to shoulder more of the driving burden on that one day.

Let's denote by p the probability that the single-day diner drives on that day.  Then the two daily diners drive on that day with probability (1-p)/2, and over the course of the week, they drive (5-p)/2 days, on average.  According to our fairness metric, we must find p such that (5-p)/2 = 5p, which yields

5-p = 10p

11p = 5

p = 5/11

So the single-day diner should drive nearly half of the time, on those days when he or she joins the two daily diners.  By a similar line of reasoning, if there are three daily diners, that probability drops to 5/16, and in general, with n daily diners, the probability is 5/(1+5n), with each of the daily diners driving five times more often, or 25/(1+5n).

What happens if there are m one-day diners (each of them eating on the same day)?  Then the probability p that any of the one-day diners should drive on that one day drops even further, to 5/(m+5n).

One might well consider (providing one is still reading) extending these to k-day diners, and whether the results depend on the k-day diners eating on the same k days, or if the results are insensitive to the distribution of those k days.

Roll Over, You Pats!

This past weekend's Super Bowl XLVI (that's forty-six) provided yet another confluence of probability, tactics, and sports.  That's never a bad thing.

I'm speaking, of course, of the decision on the part of Patriots coach Bill Belichick to permit the Giants to score on second down and goal from the Patriots' six-yard line, with about a minute left in the game.  The Patriots put up only token defense, so that when Ahmad Bradshaw took the handoff from Eli Manning, he was able to waltz into the end zone.  Almost literally: Bradshaw had a moment of indecisiveness as he reached the one-yard line, but soon backed into the end zone for the touchdown.

Even before that play began, color commentator Cris Collinsworth had already suggested that the Patriots might permit the Giants to score easily, because the Patriots only had one timeout remaining.  They would therefore be able to stop the clock after second down, but not after third down.  Since the play clock starts at forty seconds once the ball is set, the Giants would attempt a field goal on fourth down with only about ten to fifteen seconds remaining on the game clock.  Collinsworth reasonably contended that the Patriots should prefer trying to score a touchdown with a minute left (plus their one timeout) over trying to score a field goal with ten to fifteen seconds left (without any timeouts).

(It's worth pointing out that then-Packers coach Mike Holmgren had been roundly criticized for making a similar tactical decision fourteen years earlier, in Super Bowl XXXII against the Broncos.  Times change.)

And now, once Bradshaw had scored, Collinsworth decried Bradshaw's touchdown as a tactical error.  Well, setting aside the tendency of sports broadcasters to exaggerate practically anything, was it a tactical error?  Which outcome is better for each team?

Well, first of all, there's the intuitive argument that if one team wants you to do something, then your best strategy ought to be to resist that.  So if the Patriots are parting the Red Sea, maybe your best bet is to lie down.  And indeed, the Giants had considered that.  Manning later reported that he was telling Bradshaw to go down in the field of play.  The Patriots, for their part, said that it was immaterial, that they would have shoved Bradshaw into the end zone, but that tactic would not have worked if Bradshaw had taken a knee: Any subsequent bump by a defender, even the lightest touch, would have made Bradshaw down by contact at the one-yard line.

But let's not let psychological ploys decide the question.  Which tactical choice is the right one here?

The Patriots have two choices—allow the touchdown, or play straightforward defense—but there are more than two possible outcomes.  If the Patriots play defense, there are still multiple possibilities:

• The Giants might score on second down anyway.
• Or they might score on third down.
• Or they might score a field goal on fourth down.  (We'll assume they wouldn't try to score a touchdown.)
• Or they might fail to score at all, either because of a turnover or a missed field goal.

If the Patriots allow the touchdown, and we assume for the time being that the Giants don't refuse that touchdown, then the Patriots would have to score a touchdown in about a minute, with one timeout remaining.  Let's say they're able to do that with some probability qTD.

On the other hand, if the Patriots play defense, then there are those four possibilities:

• If the Giants score on second down, the Patriots still have to score a touchdown with about a minute remaining, and one timeout.
• If the Giants score on third down, the Patriots have to score a touchdown with about a minute remaining, but no timeouts.
  
• If the Giants score a field goal, the Patriots have to score a field goal with ten to fifteen seconds left, and no timeouts.
• If the Giants fail to score at all, the Patriots can simply run out the clock.

If the Giants score on second or third down against straightforward defense, the Patriots are left in pretty much the same situation as if they just let them score on second down, modulo that timeout.  So as it stands, they're just a bit worse off if they play defense.

Now let's take a look at those last two cases.  If they don't score on second or third down, the remaining possibilities are a turnover, a missed field goal, or a made field goal.  Out of those, I'd guess the made field goal happens nineteen times out of twenty.  In the remaining cases, the Patriots just have to sit on the ball, which I'd also guess would happen nineteen times out of twenty (remember, they might have to avoid the safety).  So the question roughly boils down to, which is more likely: Scoring a touchdown in a minute, or one of the following happening—scoring a field goal in ten to fifteen seconds, securing a turnover, or the Giants missing a field goal?  If it's the touchdown, the Patriots should let the Giants score.  If it's any of the remaining three choices, they should play straightforward defense.

Given that Lawrence Tynes hadn't missed a field goal of thirty yards or less in forever, the Giants were going to play possession football, and the Patriots would have no timeouts left for a field goal attempt, I'd go with letting them score, just as Belichick did.  But there's no way this is a foregone conclusion.  Sometimes, it's just a close call.

The Philosophy of Probability

I've previously written about the importance of context in statistical analysis, though only as a prelude to saying something about game theory, and how it can influence one-dimensional performance measures in obscure ways.  Well, hold onto your hats, because now I'm going to talk directly about context.

As a simple, frivolous example of context, consider the claim (possibly apocryphal—it's such a good story) of an ESP researcher who apparently did a comprehensive study with a large number of volunteers.  A thousand or so, in fact.  And he said that although most of the volunteers lacked any notable ESP talent, one in particular seemed to have it in spades.  In fact, he said, he had done a statistical analysis on the results, and this volunteer had scored "at the three-sigma level."

What does that mean?  It means that the results of his volunteers followed a bell-curve sort of distribution, which has a standard deviation.  The standard deviation is a measure of the spread or width of the bell curve, and is denoted by the Greek letter sigma (σ).  So a result that is 3σ above the average is very unusual indeed.

How unusual?  Well, a 1σ result is high enough that we'd expect only one volunteer in six to score that high, just out of random chance.  A 2σ result is high enough that we'd expect only one volunteer in forty to score that high.  And a 3σ result is high enough that we'd expect only one volunteer in a thousand to score that high.  So that must be significant, right?  I mean, only one in a thousand volunteers could be expected to score that high by chance.  Oh, except that there were a thousand volunteers...so...uhh, maybe it's not so significant after all.

OK, maybe that story is apocryphal; I couldn't find it in a Google link.  (Except now that I've posted this story, I'll be able to find it—on my blog.)  But there is this excellent xkcd comic, which makes exactly the same point. 

The first time I was directly confronted with this was many years ago, when I was tutoring a family friend in probability and statistics.  Her teacher had assigned her a worksheet, and one of the problems concerned airplane accidents:

A survey of U.S. airplane accidents included seven major accidents involving fatalities.  Although the survey covered five airlines, four of the accidents involved a single airline: US Airways.  Is this statistically significant at the 5% level?

(There may really have been such a survey: There was a period in the early-to-mid 1990s in which US Airways did in fact have a slew of major accidents.)  Now, I should say something about that last sentence, because it sounds like we're asking what the probability is that the observation is just a result of random chance, and whether that probability is less than 5 percent.  But that's not actually quite right.

When we say that something is statistically significant at a given probability level, that refers to something called the null hypothesis, a central notion in statistics, and the inspiration for the name of this blog.  The exact interpretation of the null hypothesis depends on the kind of problem you're examining, but roughly speaking, it asserts that there is no correlation, that there is no effect to be measured, that everything observed is the result of random chance variation.  One doesn't—in fact, can't—prove the null hypothesis; in a sense, it is not even really assumed.  We just compare other hypotheses to it.

So, in this case, the null hypothesis is that US Airways is not in fact more likely to have accidents than any other airline, that each accident is equally likely to involve any of the airlines.  What we do then is to compute the probability that the pattern observed—four out of seven accidents involving US Airways—if the null hypothesis is presumed for the sake of argument to hold.  If the resulting probability is less than 5% (1 in 20), then the observation is statistically significant at the 5% level.  If the resulting probability is less than 1%, then the observation is statistically significant at the 1% level.  And so on.

Well, let's go through the exercise.  If there are five airlines, and seven accidents, and each accident is equally likely to involve any one of the five airlines (we'll assume for the time being that no accident involves more than one of the airlines), the probability we want is given by the binomial theorem:

P[US Airways is involved in four of seven accidents] = C(7,4) (1/5)^4 (4/5)^3 = 0.0287-

Since the probability is 2.87% < 5%, the observation is significant at the 5% level.  Even if you add in the probability that they're involved in five or more accidents out of the seven, that probability only swells to 3.33%, so it's still significant at the 5% level.

Or so the teacher claimed.  I wasn't so sure.  I would say it depends a lot on what you're trying to determine, and here we get into an area where statistics is as much philosophy as it is mathematics and science.

The question is, why are you asking this question about US Airways?  Is it because you have some other, material reason for doubting the safety of their flights?  Or is it just the fact that four out of seven accidents involved them?  This may seem like arguing about the number of angels that can dance on the head of a pin, but in truth, your approach to the question depends vitally on which it is.  If it's because you have some other reason for suspecting US Airways—say, that they have shoddy maintenance records—then your line of reasoning is perfectly valid.

But if it's just the latter—if it's just a matter of noticing a cluster of US Airways accidents—then any airline at all might be the target of such a statistical analysis.  We should then be asking what the probability is that any of the five airlines was involved in four (or more) of seven accidents.  Since only one airline can be involved in four or more of the accidents, we can determine that probability very simply, by multiplying the single-airline probability by five, in which case we get 16.66%, which is decidedly not statistically significant.  There's actually a one-in-six chance that some airline would be involved in four or more of seven accidents.

Why, that's only 1σ!  Big deal!

Shot Selection and the Secretary Problem

One of my favorite problems in all of recreational mathematics is the so-called secretary problem.  In this problem, you are interviewing a hundred candidates for a secretarial position.  For the purposes of discussion, we'll assume that the various candidates have a precise suitability rating, and of course, you want to maximize this rating for your hire.  Ideally, then, you'd interview all hundred candidates first, get their ratings, and then hire the best one.

Unfortunately, that's not the way things work in this problem.  You only get the candidates one at a time, and you have to decide then and there whether or not to hire them or not.  Once you've rejected a candidate, they're lost to you forever.  One could, theoretically, lose the best candidate on the very first interview.

The question then is, what is your best strategy, and what is your probability of making the best possible hire using that strategy?

(By the way, if you think this problem is formulated in a politically incorrect way, I first encountered it in a form called the sultan's dowry, in which a suitor for the sultan's daughters had to select the one with the largest dowry.  If he picked the right one, he got to marry her, but if he didn't—well, let's just say an unsuccessful suitor and his head are soon parted.  But it wasn't entirely unproductive; I eventually formulated a variation called the iterated sultan's dowry, in which a second suitor, seeing the first suitor's unsuccessful head roll down the hill, gets to use the information in choosing a prospective mate, and then the third, the fourth, etc.  This variation has an interesting solution which is unfortunately too large to fit into this parenthetical comment.)

It can be shown, fairly easily, that the best strategy must be of the form "Skip the first n candidates, recording their suitability ratings.  Then choose the next candidate whose rating exceeds theirs."  The reason is that as you plow through the candidates, the probability that the best one is yet to come never increases, whereas the probability that you've already encountered the best one never decreases.  So the question reduces to figuring out what the right choice for n is.

Ultimately, following a strategy like this, you could end up choosing no candidate at all if the best candidate is already in the first n, since you've already skipped all of those.  But if you do pick a candidate, it will be number k > n.

For that one to be the best overall, the best must be in the last 100-n.  Furthermore, the second best of the first k (that is, the best before the ultimate choice) must belong in the first n.  Now, let's work out the probability that both of these happen.  In order to do this, we have to break down the possibilities into all the different cases.

The first case is that the best candidate is the very next one—candidate number n+1—which happens with probability 1/100.  You'll choose that one provided that there's no other candidate between the first n and number n+1 that is better than the first n.  Since there are no candidates in between, that probability is 1.  So the incremental probability for this case is 1/100 times 1, or just 1/100.

The second case is that the best candidate is the one after that—candidate number n+2—which again happens with probability 1/100.  You'll choose that one provided that there's no candidate between the first n and number n+2 that is better than the first one.  That will be true provided the best of the first n+1 happens within the first n, so the incremental probability for this case is 1/100 times n/(n+1).

Following the same line of reasoning, the third case—that the best candidate is number n+3—provides an incremental probability of 1/100 times n/(n+2), the fourth case provides an incremental probability of 1/100 times n/(n+3), etc., until the last case—that the best candidate is number 100—provides an incremental probability of 1/100 times n/99.

Putting all these cases together, this strategy "wins" with probability 

n/100 × [1/n + 1/(n+1) + 1/(n+2) + · · · + 1/99]

It can be shown, using relatively straightforward calculus, that this expression reaches a maximum when n = 37, and yields a probability of success of about 0.37.

That's not a coincidence, incidentally.  For large candidate pools (and a hundred candidates qualify as a large pool), of size N, the best strategy is to skip the first n = N/e, where e = 2.71828+ is the base of the natural logarithm, and to take the earliest best candidate thereafter.  The approximate probability of success (that is, choosing the very best candidate of them all) is very close to 1/e = 0.36787+.

For a lot of people (including myself), that's rather stunning.  It implies that even if you have a million candidates, you have a strategy that picks the very best one of them with better than a one-in-three chance.

The reason I'm putting basketball in the mix is that there's a fairly straightforward application to a vital aspect of scoring: shot selection.

Consider: A possession in basketball lasts for anywhere from 0 to 24 seconds (neglecting offensive rebounds).  You can't always guarantee that you'll make the shot; the next best thing (at least before the endgame) is to select the very best shot—that is, the shot that has the best probability of going in (neglecting fouls and three-point shots).

 

In other words, ahem, optimal shot selection.

But you don't always know when that best shot is going to come, especially when you're working out of a halfcourt set.  Is it the very first one?  Is it the next best one?  Maybe the best one will come at least twenty seconds into the possession.  You just don't know.  But maybe, now, you have a rule of thumb for selecting that best shot.  You skip the ones that come in the first 24/e = 9 seconds (approximately), and take the next best one that comes thereafter.

Obviously, this rule makes lots of assumptions, such as (a) the defense is equally tenacious across the entire possession, (b) the offense is equally productive of shot opportunities across the entire possession, (c) the best shot opportunity is equally likely to come at any time during the entire possession, etc.  But to the limited extent that these assumptions are approximately valid, it's not a bad rule of thumb.  It suggests that the Phoenix Suns of the early-to-mid-2000s were a bit hasty.

But not by much.  Just a second or two.

Constructions with Compass, Straightedge, and Flatiron

There may not be many of you out there who remember high-school geometry fondly, but those of you who do probably share my appreciation for compass-and-straightedge constructions.  I (re-)started thinking about this today, when someone mentioned, jocularly, preparing for travel at Warp Cube Root of Two, and it occurred to me that this was one of the three classical things that one cannot do with compass and straightedge.

Those strangely intermediate folks who are at once unfamiliar with compass-and-straightedge constructions and yet not intimidated by their spectre may find the Wikipedia page a reasonable starting place.  But what I'm proposing today is something different, which I'm a bit startled hasn't been discussed more prominently: three-dimensional constructions.  I've had this in my virtual back pocket for a while; this seems as good an opportunity as any to pull it out for a looksee, and figure out if anyone else has encountered anything like this.

The idea here is to extend to three dimensions what ordinary compass-and-straightedge constructions do in two dimensions.  The first thing is to define the tools and rules for their use.  For instance, in two dimensions, the tools are a compass and straightedge (like a ruler, but with only one edge and no markings), and with them, one may:

1. Draw a line between any two distinct points.
2. Draw a circle with one point as the center, and any other point on its circumference.
3. Draw an arbitrary point on a line or a circle, or off it.
4. Draw the point at the intersection of two lines (if they intersect).
5. Draw the point (or two) at the intersection of two circles (if they intersect).
6. Draw the point (or two) at the intersection of a line and a circle (if they intersect).

That's it; that's all you're allowed to do.  With these restrictions, the Greeks discovered that it is possible to construct an astonishing variety of geometrical objects, but they were unable to construct three famous examples.  They were unable to construct the square root of pi (also known as squaring the circle); they were unable to trisect arbitrary angles (that is, construct an angle with one-third the extent of a given angle); and they were unable to construct the cube root of two (also known as doubling the cube).

It turns out that these are impossible, and can be proved to be so, using some notions from field theory.  That has not, of course, stopped people from submitting reams upon reams of alleged constructions of one of these three objects, all of which (you may be assured) are somewhere bogus.

But enough of that for now.  In three dimensions, the canvas is not a flat plane, as it is in two dimensions, but all of space.  And we introduce a new tool, which I will call a flatiron, which permits you to draw planes.  The flatiron rules are as follows; in addition to the above, one may:

1. Draw the unique plane containing any three non-collinear points.
2. Draw a sphere with one point as the center, and any other point on its surface.
3. Draw an arbitrary point on a plane or a sphere, or off it.
4. Draw the line at the intersection of two planes (if they intersect).
5. Draw the circle (or point) at the intersection of two spheres (if they intersect).
6. Draw the circle (or point) at the intersection of a plane and a sphere (if they intersect).
7. Draw the point (or two) at the intersection of a line or circle with a plane or sphere (if they intersect).

As an example of what one might do in a three-dimensional construction, consider the following fairly simple task: Given points P and Q, construct a regular tetrahedron with PQ as edge.  We proceed as follows:

1. Draw spheres of radius PQ around both P and Q.
2. Draw the circle C at the intersection of spheres P and Q.
3. Draw R, an arbitrary point on circle C.
4. Draw a sphere of radius PR around R.
5. Draw S, one of the two points of intersection between circle C and sphere R.  PQRS is then a regular tetrahedron.

It should be straightforward to see that one may construct a regular cube as well.  Does this mean that one can construct the cube root of two in three dimensions?  I suspect not, although I cannot yet prove this.

So, a couple of questions, one easy, and one not so easy:

1. Suppose that indeed, the cube root of two is not constructible in three dimensions.  What about the fourth root of two?  In which dimensions might that be constructible?
2. Is it possible to construct all five of the regular polyhedra?  In addition to the tetrahedron and cube, these include the regular octahedron (eight faces), dodecahedron (twelve faces), and icosahedron (twenty faces).

 

It's True

This man has no gut.

Could You Be a Crackpot?

Of course not!  That's why I invented the following barometer for identifying crackpots, so you'll see how much you're not a crackpot.  This won't take long.  Just read the following essay, then answer the question afterward.

You have always, from an early age, been fascinated by the workings of the universe.  For longer than you can remember, the suspicion that something is amiss in the prevailing scientific theory has gnawed at you, but only recently, in your maturity, have you recognized how to rectify its flaws.

This is no patchwork repair—it is a fundamental revision in how to perceive the universe.  You have discovered this shift by virtue of your stark insight, uncluttered by years of outdated academic instruction.  You have been labelled a crackpot, but this has always been the mark of the true genius.  Witness Galileo: He had to withstand the prejudices and superstitions of his time, even amongst his colleagues, in his pursuit of the truth!  And you, like Galileo, are no crackpot.

Your breakthrough must be distributed personally, as part of a grass-roots campaign, in order to avoid the censorship that is part and parcel of the mainstream scientific press.  But you have no doubt that you will succeed; you are brave enough to face the naysayers who have shouted down other talented scientists who lacked your resolve.  Witness Einstein: His original mindset differed from the battered viewpoint that jealous criticism reduced him to!  You can quote effectively from his writings to support this.

The prevailing scientific theory requires—in fact, relies crucially on—a body of mathematics that was clearly invented ad hoc: for the purpose of showing exactly that for which it was invented.  This circular line of reasoning is a basic but crushing flaw.  Your alternative, for those who have the vision to properly appreciate it, is conversely a work of great elegance and beauty, which resonates with rather than challenges your acute intuition.

When at last it is properly recognized, your breakthrough will enable great boons for humanity.  You will be offered vast rewards for your work, but you will turn most of them down, asking only to continue your scientific investigations without the further distraction of interacting with the scientific community.  Those who doubted you will permit this reluctantly, out of respect for your talent, but all the while will admire your work from afar.

OK, now that you've read that (haven't you?!), here is your one question:

On a scale of 0 to 10—0 indicating complete disagreement and 10 indicating complete agreement—how well does the foregoing essay represent your viewpoint?

Scoring:

10: You are a crackpot.
0-9: You are not a crackpot.  True crackpots are in for a pound if they're in for a penny.