At least, that is what the media would have us believe.  For the past several weeks, Paul the Octopus has captured the hearts, minds, and stomachs of people around the world.  He’s a charming octopus, to be sure, but it isn’t his good looks that have gotten him this far.  Instead, it’s his seeming ability to correctly predict the outcome of soccer matches.  As time has gone on and Paul’s predictions have continued to prove themselves accurate, the amount of press he has received has only increased.  Articles about him are everywhere on the internet: here‘s one discussing public outrage after he correctly predicted Spain to defeat Germany in the semifinals, and here‘s an article discussing his preference for Spain over the Netherlands in the finals.  Search for “Paul the Octopus” in Google News and you will find thousands of results.

This is one popular mollusk.

Of course, I suppose it’s possible that Paul really can see into the future, at least as far as soccer is concerned.  After all, he did correctly predict the winner of every game asked of him; an impressive feat, seeing as how his advice was requested a total of 8 times.  However, Occam’s Razor suggests that we should look for a simpler explanation.

A natural first choice is to guess that Paul was simply guessing randomly, and is very lucky.  The odds of this happening are small – assuming he has a 50% chance of picking correctly, the odds of him being right each one of the 8 times he predicted a winner in this World cup would be 1/2⁸, which is only approximately .39%. Very low odds indeed.

This analysis ignores biases that may be present – in particular, Paul’s octopus vision may bias him towards certain flag designs (which, given the fact that he frequently chooses Germany, seems plausible).  The Wikipedia article I linked to discusses other sorts of possible bias.  However, these biases would only influence which box he selected – they wouldn’t necessarily affect the odds that his selection would correspond to the winning team (although it is possible that he is being persuaded to throw his chips in for the favored team, which would increase the likelihood of his success).  Either way, questions concerning how Paul makes his selection are more interesting to me, so let me focus for the moment on that.

First of all, one could easily argue that unlike flipping a coin, the trials here (i.e. Paul’s selections) are NOT independent. Indeed, what’s going on here may be very similar to an article in the New York Times a couple of years ago that discussed the lurking presence of the Monty Hall Problem in a classic experiment from psychology (which I discussed here).

The idea is quite simple.  Paul chooses between two opponents in the World Cup by selecting a piece of food from one of two boxes.  Each box is labeled with a country’s flag, and this is the most obvious distinction between the boxes.  Suppose, for the sake of argument, that Paul has in his mind a ranking of his preferences for the flags, starting with the one he likes the most, and ending with the one he likes the least.  Assuming between any pair of flags Paul prefers one to another, one could theoretically determine his preferences by giving him sufficiently many pairings of different flags.  Moreover, assuming he does have preferences, the game selections are no longer independent, because each game gives us some information about his preferences.

Let’s dig deeper and look at his selections throughout the World Cup.  Paul gave predictions for 8 games, and those 8 games involved 9 separate teams (only one game did not involve Germany).  In the first game, Germany versus Australia, Paul selected Germany.  Let us note that as: Now, already this gives us information about Paul’s preferences.  There are 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 9! = 362,880 possible ways to order these 9 teams, since you can choose any one of the 9 teams to be your favorite, any one of the 8 remaining to be your second favorite, and so on.  But given the above information, we already know that any choice with Australia ranked higher than Germany can not match Paul’s preferences.  This eliminates a surprising number of possible outcomes – half, in fact, since for every list in which Germany is ranked higher than Australia, we can flip these two countries to obtain a ranking in which Australia is higher than Germany.

This then affects the probability of all subsequent pairings!  Let’s take a look at the next game.  In the second game, Paul correctly predicted Serbia to defeat Germany.  We can write this as: This gives us even more information!  Now, not only do we know that Paul prefers Germany to Australia, we also know he prefers Serbia to Germany.

Moreover, suppose (as seems reasonable) we assume the probability that he prefers Germany to Australia is 50%.  Now, after the first match, we know he prefers Germany to Australia – the right followup question to ask is, what is the probability he prefers Serbia to Germany GIVEN that he prefers Germany to Australia?  In particular, we have a conditional probability question here – these two facts are not independent!  For example, the fact that Paul prefers Germany to Australia means that Germany cannot be last in his rankings – this should decrease the probability that he prefers Serbia to Germany.  And indeed it does: supposing that each of the 362,880 rankings is equally likely, the probability that someone will prefer Serbia to Germany given that they prefer Germany to Australia is not 50%, but is only 33%!

To see why, consider an arbitrary ranking of the 9 teams.  If we only consider the relative placement of Serbia (S), Germany (Ge), and Australia (A), there are 6 possible rankings among these three:

1. S > Ge > A
2. S > A > Ge
3. Ge > S > A
4. Ge > A > S
5. A > Ge > S
6. A > S > Ge.

However, if we also know that Ge > A, then rankings 2, 5, and 6 are eliminated, leaving us only with

1. S > Ge > A
2. Ge > S > A
3. Ge > A > S.

In particular, of these three remaining, Serbia is ranked higher than Germany only once.  Therefore, the probability that S > G GIVEN that G > A is only 1/3, or 33%.  (If you prefer, you can use a counting argument to show this as well.)

Of course, just as the first match gave us information, so did the second.  Therefore, when it comes to the third match, we have even more information at our disposal.  The third match was between Germany and Ghana, and Paul correctly identified Germany.  In other words: Now the appropriate question to ask, of course, is: what is the probability that Paul prefers Germany to Ghana, given that he prefers Serbia to Germany and Germany to Australia?  Well, we know that Germany can’t be his first or his last choice, because it must be preceded by Serbia and followed by Australia.  Therefore, among these four countries, Germany must rank second or third.

If Germany ranks second, Serbia must be first, but we are free to make Australia third or fourth.  Similarly, if Germany is third, then Australia must be last, and we are free to make Serbia first or second.  In other words, we have four outcomes:

1. S > Ge > Gh > A
2. S > Ge > A > Gh
3. S > Gh > Ge > A
4. Gh > S > Ge > A.

Among these four choices, only 2 have Germany preferred over Ghana.  Thus the conditional probability that one would prefer Germany to Ghana is again 50%.

The interested reader can easily continue on in this fashion.  If you’re impatient, however, you can calculate the probability that Paul would have made the selections he did more directly.  All we need to know is who Paul selected in each match.  I’ll tell you that in the subsequent matches, Paul picked Germany over England, Germany over Argentina, Spain over Germany, Germany over Uruguay, and Spain over the Netherlands.

Given this information, suppose you want to know the probability that Paul selects Germany over Australia, Argentina, Uruguay, Ghana, and England, while he selected Spain over Germany and the Netherlands and Serbia over Germany.  As stated before, there are 9! possible lists of preferences.  In this case, it’s not hard to determine how many would lead to the behavior seen in this year’s World Cup.  Since Paul picked Germany over 5 teams, but behind 2 teams, we know that Germany can only be ranked 3rd or 4th (any higher and there wouldn’t be room for Spain and Serbia above, any lower and there wouldn’t be room for the 5 teams below).

If Germany is ranked 3rd, then we can choose to put either Spain or Serbia in 1st place.  Whichever one we don’t put in first place will then need to go in 2nd place, so that both countries are ranked higher than Germany.  After that, we are free to order the remaining countries however we like.  In other words, we see that there are 2 x 6! = 1,440 possible lists of preferences if Germany is ranked 3rd:

1. 2 choices
2. 1 choice
3. Germany
4. 6 choices
5. 5 choices
6. 4 choices
7. 3 choices
8. 2 choices
9. 1 choice.

Meanwhile, if Germany is in 4th place, we need to figure out how many ways there are to choose the three teams above it.  Notice that since Australia, Argentina, Uruguay, Ghana and England must all be ranked lower than Germany, the three countries ranked above it must be Spain, Serbia, and the Netherlands.  However, since we also need the Netherlands to be ranked below Spain, this only gives us three possibilities for the ranking of the first three teams: Spain, Serbia, Netherlands; Spain, Netherlands, Serbia; and Serbia, Spain, Netherlands.  Once we have made that selection, however, we are free to choose the 5 teams below Germany however we please.  In other words, if Germany is 4th, there are 3 x 5! = 360 possible lists of preferences.

Combining these, we see that there are 1,800 possible lists of preferences that would lead to the behavior shown by Paul.  Since the total number of outcomes is 9!, this gives a probability of only 1,800/9!, or roughly .49%.  This is the same value you will get if you calculate the remaining conditional probabilities and multiply them together.

Of course, if one wants a more impressive number, one can always try to correct for bias in Paul’s selection.  For example, suppose we assume that Paul will never choose England or Australia if given the option of a different flag – this seems reasonable, given experiments on how his species sees (the other flags have more contrast and are more focused on horizontal shapes, which apparently his species is drawn to).  If we make this assumption, the number of potential preference lists drops to 7! x 2 = 10,080, in which case the probability that Paul would choose as he did jumps up to 17.9%!

There are valid concerns about this model, though.  For instance, given the choice between two flags, why should we assume that Paul will always choose the same one over the other?  Equivalently, why should we believe that Paul’s decisions follow a prescribed preference list?  Indeed, when Paul was used to make predictions in 2008, he selected Germany over Spain, unlike his selection of Spain over Germany in 2010.  In 2008, it was Germany who was the victor, and so Paul guessed incorrectly – perhaps he learns from his mistakes, after all.

Whatever the case, I doubt this octopus has any special ability.  And even if he does, I don’t know that that would necessarily be a good thing.  For we all know that when 8 limbs are combined with super powers, nothing good can come of it.

Is this the future that Paul's powers portend? I believe it is.

Forget about the Tuesday fact for a moment – if you have a friend with two children, one of whom is a boy, what is the probability that the other child is a boy?  You might expect that the answer should be 50%, since the sex of one child shouldn’t affect the sex of the other.  But this is not quite right, because you’re not told whether the boy is the older or younger child.

There are only four possibilities when one has two children, so the situation is easy to analyze.  With two kids, the four possibilities are boy boy, boy girl, girl boy, and girl girl.  If you know that one of the kids is a boy, this eliminates girl girl from the list of potential combinations, leaving us with the three outcomes boy boy, boy girl, and girl boy.  Of these three outcomes, we see that only the first has two boys, and so we conclude that the probability of the second child being a boy is 1/3, NOT 1/2!

There’s another way to answer this question, one that generalizes nicely to the more complicated question asked by Foshee.  There are two cases to consider: either the boy you know about is the younger child, or the older child.  If the boy you know about is the younger child, there are two possibilities for the older child (girl or boy).  Similarly, if the boy you know about is the older child, there are two possibilities for the younger child (girl or boy).  This gives four outcomes, but you have counted the boy boy outcome twice.  In other words, we see there are only three distinct outcomes, and only one of them has two boys, so again we see that the probability is 1/3.

Now let’s return to the original question.  Again, it seems like the fact that the boy was born on a Tuesday shouldn’t matter, but if we do the same analysis as above, we’ll see that this is not the case.  The information about the day does make the number of outcomes larger, however, so it’s easier to get mixed up.

As before, let’s split into two cases, depending on whether the Tuesday boy is younger or older.  If the younger child is the Tuesday boy, then there are 14 possible outcomes for the older child, since there are 2 choices for the sex of the child and 7 choices for the day of the week on which the child was born.  Similarly, if the older child is the Tuesday boy, there are once again 14 possible outcomes for the younger child.  However, notice that we have counted the outcome of two boys both born on Tuesday twice, just as we counted the outcome of two boys twice in the simpler problem.  Correcting for this double counting, we see that there are 14 + 14 – 1 = 27 possible outcomes.  Of these outcomes, 13 of them correspond to having a two boys – if the younger child is the Tuesday boy, there are 7 possible outcomes that will give us two boys, and if the older child is the Tuesday boy, there are again 7 possible outcomes.  As before, though, we’ve counted both children being Tuesday boys twice, so we subtract 1 to correct for this double-counting, which leaves us with a total of 7 + 7 – 1 = 13 desired outcomes.  This means that the probability of the second child being a boy is 13/27.  While still not equal to 1/2, this is much closer to 1/2 than 1/3.

Much like the Monty Hall problem, however, one can understand the mathematical reasoning behind this problem and still have trouble with the intuition.  After all, why should the day on which a child was born have any bearing on the sex of the other child?  On the face of it, the solution to this problem doesn’t make any sense.  One way to try and marry this solution to our intuition is to assign some numbers and explore the data, seeing where our intuition diverges from this picture.

Suppose we take a survey of 19,600 families with two children (you will understand why I’ve chosen this seemingly random number in a moment).  Of those families, suppose they are evenly divided among boy boy, boy girl, girl boy, and girl girl households.  In particular, we see that there are 4,900 families with two boys, 4,900 families with two girls, and 9,800 families with a boy and a girl (in half of these families the boy is older, and in half the boy is younger).

Now suppose we further subdivide the data according to the day of the week in which the child was born.  Suppose that a child is equally likely to have been born on any day of the week.  This means that, for example, among the 4,900 families with two boys, 4,900 ÷ 49 = 100 will have boys who were both born on Monday, 100 will have boys for which the oldest was born on Monday and the youngest Tuesday, 100 will have boys for which the oldest was born on Monday and the youngest Wednesday, and so on.  In other words, for every possible combination of sexes and days of birth, there will be 100 families that match that combination.

Suppose, now, that you take a random family in which one of the children is a boy born on Tuesday.  There will be 1,400 families for which the younger child is a born boy on Tuesday, and 1,400 families for which the older child is a boy born on Tuesday, which means there are 2,700 families total families under consideration (note that we have counted the 100 families with two boys born on Tuesday twice).  Of those 2,700 families, 1,300 will have two boys, for the same reason as above (you can also look at the diagram above – imagine each square as representing 100 families in the survey).  So, if we pick a FAMILY at random with a boy born on Tuesday, the probability that the other child is a boy is 1,300 out of 2,700, or 13/27.

Now, instead of picking a family at random, suppose we pick a BOY at random who was born on Tuesday, and ask for the probability that the boy’s sibling is also a boy.  In this case, note that there are 2,800 boys in our sample who were born on Tuesday – 1,400 who are the younger sibling and 1,400 who are the older sibling.  In particular, note that we don’t subtract out the families with two boys both born on Tuesday for double counting in this case; this is because we are choosing the boy, not the family, and choosing the younger boy born on Tuesday is different from choosing the older boy born on Tuesday.  Because there’s nothing to subtract out, we see that of these 2,800 boys, 1,400 have sisters and 1,400 have brothers.  Therefore the probability that the BOY has a brother is 1/2, the answer our intuition gave us from the beginning!

In other words, you can try to understand this seeming paradox as a difference in perspective.  If we look from the perspective of the family unit, the probability that a two-child FAMILY with one son born on Tuesday will have two sons is 13/27, slightly less than one half.  However, from the perspective of the children, the probability that a BOY born on Tuesday in a two-child family will have a brother as opposed to a sister is 1/2.  In this latter case, the day of the week really is irrelevant.

For some, however, this explanation still may not seem like enough.  I’d encourage anyone with a different approach to sound off below!

How can it be that despite the best intentions of many couples, such a significant proportion will not endure?  As one always should, we can turn to mathematics for possible answers.  In fact, José-Manuel Rey of the Department of Economic Analysis at the Universidad Complutense in Madrid has done just that, by proposing a mathematical model to explain the dynamics of long term relationships.  His model is based on the following assumptions: first, that the individuals in the couple have similar traits (this is spelled out more precisely in Rey’s paper, but basically this is so that he can consider one utility function for the couple rather than separate utility functions for each individual), and second, he assumes a so-called law of thermodynamics for sentimental relationships.  This law can be stated as follows: “There is tendency for the initial feeling for one another to fade away. This kind of inertia must be counteracted by conscious practices.”  In other words, the natural deterioration in a relationship must be counteracted by continual effort, such as a romantic evening, a heartfelt conversation, or listening to some slow jams by R. Kelly.

Given these assumptions, Rey constructs a model to try and explain the paradox (which he terms the failure paradox) in the fact that people who believe they will be together forever will, with a high degree of probability, end up separating.  And sure enough, his model offers us an explanation.  His work is perhaps best summarized by the following picture:

The graph above is of feeling (x axis) versus effort (y axis) – in other words, how satisfied is the couple in their relationship compared to how much effort they are willing to put into it?  Rey’s model asserts that there is a minimum happiness (denoted x_min above) below which a relationship cannot survive.  In other words, if your happiness crosses the red vertical line, your relationship is doomed.  The couple’s initial happiness is given by the value x₀ – note that this is greater than the happiness of the equilibrium point, reflecting the common observation that happiness within a couple frequently decreases from its initial state (you can think of this as the honeymoon period, if you like).  This isn’t to say that couples in long term relationships are unhappy, for indeed the happiness at the equilibrium point is still above x_min; all this is saying is that happiness is lower at the equilibrium stage than it was initially.

What about the effort involved?  Heuristically speaking, the amount of effort you put into a relationship should increase your own happiness, up to a point, but then should begin to negatively affect your happiness.  Driving your partner to the airport every once in a while will probably make you feel good about yourself, but driving your partner to the airport every weekend will probably start to wear on you.  On the graph above, c* represents the turning point from when the amount of effort you put in positively affects your happiness to when it negatively affects your happiness.

Here is where an important feature of the model comes into play: the amount of effort at the equilibrium point is greater than c*!  In other words, in order to maintain equilibrium in your relationship, you need to be putting in more effort than you would optimally choose to.  It is this s0-called “effort gap” that Rey identifies as being the cause of so many problems.  Note that if the effort gap is large enough, then maintaining the appropriate level of effort may drive happiness below the minimum required value, and even if this is not the case, maintaining the appropriate level of effort will still require some loss of the couple’s happiness, since the effort required will always be greater than c*.  To put it more bluntly, relationships require sacrifice.

I’d encourage you to keep this in mind as you navigate the unpredictable seas of love.  When it comes to the effort involved in keeping your relationship afloat, mathematics warns you against complacency.  When in doubt, it’s always best to go the extra mile.  And if you need help wooing your special someone, you can always call on R. Kelly to help set the mood.

Aziz Ansari does a mighty fine impression of Kells.

I'm fairly confident that this is the first time a mathematician has been branded with the Perez Hilton logo.

The Poincaré conjecture, first posed by Poincaré over 100 years ago, is a question about conditions under which an object is essentially a hypersphere, that is, a sphere sitting inside 4 dimensional space.  More specifically, it asks whether or not every simply connected, closed 3-manifold is homeomorphic to the 3-sphere (the answer is affirmative).  Believe it or not, there is a fairly accessible article on Wikipedia which describes these terms and briefly discusses Perelman’s methods.  The $1 million prize stems from the fact that the Clay Institute dubbed this one of the 7 Millennium problems in mathematics.  Each of the 6 remaining problems also has a $1 million prize associated with it (and who said mathematics didn’t pay?).

It should be mentioned that although Perelman is credited with the proof of the Poincaré conjecture, his papers do omit some of the details.  However, his methods were all correct, and Kleiner and Lott were able to use his detailed blueprints to give a full proof using his methods.  I’m sure they would be happy to split the cash, should Perelman not recant on his wishes.

It’s also worth noting that this isn’t Perelman’s first exposure in the realms of pop culture.  In fact, when Perelman won the Fields Medal 4 years ago for solving the Poincare conjecture, Stephen Colbert took him to task, although I’d argue Colbert’s jealous outburst stems only from the fact that he didn’t win the prize himself.

It’s quite clear, however, that Dr. Perelman does not welcome all this attention.  In fact, recent media request have been met with the following response: “You are disturbing me.  I am picking mushrooms.”  Could this be an indication of where he gets his mathematical ideas?  Perhaps.

It is clear that Perelman has no desire to be a mathematical ambassador to the general population.  But like it or not, the general population seems to be fascinated by him.  Here’s a tip for someone trying to be inconspicuous: the best way to minimize publicity when someone offers you a million dollars probably doesn’t involve rejecting the offer.  Certainly if the Institute is looking for places to distribute that cash, I would welcome some summer funding.

Last year, Professor Steven Strogatz of Cornell University wrote a series of op-eds for the New York Times that discussed the presence of mathematics in unlikely places. I discussed one of these columns here.  Now, either those articles were well-received, or Professor Strogatz is well-connected, because this year he’s back in the Times with a much more ambitious series of articles.  This time around, Strogatz is attempting to “[write] about the elements of mathematics, from preschool to grad school, for anyone out there who’d like to have a second chance at the subject.”

Preschool to grad school is a significant amount of ground to cover, but thus far Strogatz has used his articles to assault this goal with gusto.  To date, he has tackled counting, patterns in addition, negative numbers, division, and basic high school algebra.  This doesn’t really do justice to his content, though.  Along the way he gives the reader some Sesame Street, and discusses a number of tangential topics, including the inability of Verizon employees to do math, the half-your-age-plus-seven rule, and pre-WWI European history.  The latter comes about in a discussion of that old adage which is familiar to anyone who saw the first Alien vs. Predator movie: the enemy of my enemy is my friend.

Predators must be awesome at math.

While some of Professor Strogatz’s explanations are a bit hand wavy (in particular, his explanation of why (-1) x (-1) = 1 is a lacking), on the whole they are quite good.  In particular, he offers a nice explanation of what it is for a mathematical argument to be “elegant.”  But even more impressive than his writing is its location – to have a discussion of mathematics with as wide an audience as the New York Times readership is commendable.  Even if people are not inspired to learn more mathematics after reading these pieces, hopefully they will have at least learned something.  As with exercise, a little mathematics is better than no mathematics at all.

Moreover, these articles highlight aspects of math not usually seen in popular discourse.  Much like Paul Lockhart’s A Mathematician’s Lament (which Strogatz references), these snack-size essays are focused on simple mathematical ideas, and the beautiful (and sometimes unexpected) results that follow.  Nowhere here does Professor Strogatz multiply two really big numbers together; in fact, he’s quite sympathetic to the fact that for many people, there is nothing more tedious than calculation.  By leading the conversation in this way, he’s hopefully able to give a taste of what makes math beautiful to an audience for whom such a statement might otherwise be labeled heresy.

I don’t know where this series of articles is headed, but I look forward to finding out, and hope you do to.  Professor Strogatz’s articles are grouped together here.

(Hat tip to dad for sending me a few of these articles.)

Some time ago, I wrote an article on the optimal way to select a mate, assuming you know how many eligible partners exist, and that once you’ve dated someone, you can’t go back and date them again (sorry, Drew Barrymore and that dude from the Apple commercials).  This is less romantically known as the secretary problem.  Let me briefly recall the problem and its solution: suppose you have n candidates, from which you want to pick the best one.  This applies to a variety of situations, from hiring a secretary to finding a girlfriend to apartment hunting.  In either case, the outcome is the same: you should look at roughly the first n/e of them (yes, that e), and then select the first one after those n/e which is better than all that you have seen so far.  While this strategy won’t guarantee you get the best choice, it will give you the best choice around 37% of the time.

The major problem with this model is that in many situations, the value of n is unknown.  There are ways to circumvent this problem, which I will not discuss here.  Instead, in the context of finding a mate, I offer the following method to calculate the number of partners you could reasonably expect to find in your area.  This method recently gained some attention when Peter Backus, a Ph.D. candidate from the University of Warwick wrote a paper titled “Why I Don’t Have a Girlfriend.”

The basic technique involves modifying the Drake Equation, an equation used to estimate the number of potential extraterrestrial civilizations in our galaxy.  For those who have never been introduced to this equation, it asserts the following:

N = R · f_p · n_e · f_l · f_i · f_c · L.

According to Wikipedia, these variables represent the following quantities:

R = the average rate of star formation per year in our galaxy,
f_p = the fraction of those stars that have planets,
n_e = the average number of planets that can potentially support life per star that has planets,
f_ℓ = the fraction of the above that actually go on to develop life at some point,
f_i = the fraction of the above that actually go on to develop intelligent life,
f_c = the fraction of civilizations that develop a technology that releases detectable signs of their existence into space,
L = the length of time such civilizations release detectable signals into space.

Given estimates for all of these parameters, one could then estimate the number of civilizations in our galaxy.  Since we don’t know the values of any of these parameters, however, this is more of a thought experiment than anything else.

Nevertheless, the idea can be easily modified to try and find the number of eligible mates in a given area.  Peter Backus’ approach is fairly specific to him, but he links to a more general approach discussed here, in which the following equation is presented:

n = P · f_t · f_o · f_c · A · R

In this case, the parameters are given by

P = the population.  This could be the population of your university, your city, or your country, depending on how ambitious you are.
f_t = the fraction of that population which you would want to mate with, in broad terms.  If you’re a straight male, this would be the fraction of females.  If you’re a gay male, it would be the fraction of males, and so on.
f_o = the fraction of the population you’d want to mate with which wants to mate with you.  For example, if you’re a straight male who wants to mate with females, this will compensate for the fact that some females will be lesbian and therefore unwilling to mate with you.
f_c = author Raymond Francis labels this the out fraction, and describes it as the answer to the question “Of the people in your target gender and orientation, how many of them are open enough about their sexuality to engage in a relationship of the sort you’re hoping for?”  If you are straight, this value is likely 1, or very close to it.  If not, things can be a little bit fuzzier.
A = the fraction of those remaining who fall within your desired age range.  This is, of course, personal to you – if you’d like a socially acceptable age range, you could follow the “half your age plus seven” rule.
R = factors for any remaining filters you wish.  Do you want your partner to have a certain level of education, or a certain income?  Do you need a non-smoker, or demand a euphonium player?  Here’s where you can fold that into the mix.

Yes, Wikipedia has a graph illustrating the half your age plus seven rule. Amazing.

With all these parameters accounted for, N will give you the number of potential mates in your area.

Let’s take this equation for a spin, shall we?  Suppose you are a straight male living in Los Angeles, and looking for a girl to date in Los Angeles.  According to Wikipedia, the estimated population of LA as of 2008 was 3,833,995.  Of those, let’s say that 51% are female, and of the females, let’s posit that 90% are straight or bisexual.  f_c should be high in this case – to be conservative, let’s put it at 95%.

To estimate the age filter, one can obtain some data from this site.  Suppose you are a 25 year old man – then, absent any personal preference, the socially acceptable age range of women for you to date is between 19.5 and 36.  According to census data, in 2000 there were 3,694,820 people in Los Angeles, and of them, 974,004 were between the ages of 20 and 34.  Additionally, there were 251,632 people between the ages of 15 and 19, and 584,036 people between the ages of 35 and 44.  If we make the assumption that ages are roughly uniformly distributed within these brackets, this gives us an additionlal 141,970 people either between the ages of 19.5 and 20, or between the ages of 35 and 36.  Combining this gives a total of 1,115,974 people between the ages of 19.5 and 36 in Los Angeles in 2000, or roughly 30% of the population.  Let’s use this for our value A.

Assuming you have no other restrictions (i.e. taking R = 1), this gives us n = 3,833,995 · .51 · .9 · .95 · .3 = 501,544.  That’s a lot of ladies out there for the taking.  Of course, taking R = 1 is probably unrealistic.  It’s unlikely you want to date women who are married, for example, and everyone has their own personal taste that will decrease the pool even further.   Once you’ve calculated your personal value for R, however, you then know how many eligible mates will be in your area.  Given that, you’ll know how large n/e is, and then you’ll know how many people you should date before you think about settling down.

Although Peter Backus has received a fair amount of buzz for the short paper he has written on this idea, he readily admits that he is not the first to think of applying Drake’s equation in this situation.  I’ve discussed mostly Raymond Francis’ approach here, but Backus has links to many other people that have discussed the idea on his website.  In particular, here’s an exchange from CBS’s The Big Bang Theory (don’t worry, there’s a laugh track so you know when things are supposed to be funny).

In summary, not only can the Drake Equation be used to consider the existence of extraterrestrial life, it can also be used to consider potential mates right here on Earth.  The next step, of course, is obvious: we must combine these two equations to calculate the number of potential extraterrestrial mates.  Undoubtedly the number will be small, but one should never underestimate the power of love.

Interstellar love is a wonder to behold.

Late last year, a study was published in Proceedings of the National Academy of Sciences which tried to pin down origins for the gender gap in mathematics education.  As I’ve discussed before, the gender gap in math education is shrinking, and has been shown to be less about biology and more about culture – in cultures where gender equality is weaker, the gender gap is stronger.  Nevertheless, even in American culture, the gender gap still persists, and this study by Sian Beilock and others has tried to figure out how, if the gender gap is culturally based, it comes about in young students.  The original study can be found here, while a discussion of the study that was featured in the news can be found here.

Professor Beilock and her colleagues tried to correlate young students’ math anxiety with the math anxiety of their teachers.  In particular, they looked at 1st and 2nd grade students, of whom a vast majority (over 90%) have teachers who are female.  The study assessed the math anxiety of the teachers and measured the math achievement of the students at the beginning and end of the year.  Here are the results, taken from the introduction to the paper:

There was no relation between a teacher’s math anxiety and her students’ math achievement at the beginning of the school year. By the school year’s end, however, the more anxious teachers were about math, the more likely girls (but not boys) were to endorse the commonly held stereotype that “boys are good at math, and girls are good at reading” and the lower these girls’ math achievement. Indeed, by the end of the school year, girls who endorsed this stereotype had significantly worse math achievement than girls who did not and than boys overall.

These findings make intuitive sense, and lend further support for the need to better our mathematics education at all levels, or at the very least require primary educators to study mathematics more seriously.  Teaching mathematics with confidence is not something that comes automatically, even for those who may have been good at math in their early years.

It’s interesting that boys weren’t more likely to endorse the view that boys are good at math and girls are good at reading if their teacher had math anxiety.  I’d be curious to see what the case is in a classroom led by a male teacher, both with and without math anxiety.  Given the dearth of male primary educators, however, this type of data may be harder to acquire.  In any event, the lesson here is clear: if you want your daughter to not fear math, it wouldn’t hurt to demand that her teachers not fear it either.  Or at the very least, demand that any math fear be exhibited only by male teachers.  That may be a cheaper solution.

I’d also be interested in knowing whether this trend can be reversed by a suitably competent teacher.  If a group of 2nd grade girls is taught math by a woman who is unqualified, but in 6th grade is taught by a woman who is exceptional, can this help undo the damage that the 2nd grade teacher has done?  I would hope so.

It's too bad Prezbo doesn't have lady parts.

An unlucky coin? Unlikely.

Um…what?  Who cares?  While 20/43 is slightly less than the expected 50%, this difference is not even close to being statistically significant.  Actually, the fact that this ratio is only 1 1/2 games shy of the mean is pretty good.  Matt Springer has posted an article that discusses why we shouldn’t really care about this difference.