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!