Wednesday, April 24, 2013

The Wolfpack and the Lone Wolves

So a friend of mine posted a link to this story, and because it involves game theory (even though it wasn't actually a game theory course) and I do, in fact, work like that, I immediately started thinking about a way to analyze it.  Actually, I think thirty students is way too many to get some of the more interesting interactions going; the wolfpack is almost certainly the way to go, especially if you're not one of the brighter bulbs.  Thirty is probably too many to deal with analytically anyway.  So let's start with three.

Suppose the three students A, B, and C have (possibly) different aptitudes, represented by a, b, and c, respectively.  These three numbers represent the probability with which each of the students answers questions correctly.  (We'll assume that questions have two answers, one right and one wrong.)  Without loss of generality, let's say that a b c.  Under which conditions will two or more of these students collude?  Without explicitly prescribing a curve, let us say that the aim of any of the students is to improve their own grade; specifically, there is no benefit to philanthropy.

We can fairly quickly conclude that two students will not collude.  Consider A and B, and suppose first that a > b, and that A and B know that (that A is a better answerer than B).  A and B compare answers.  If they coincide, then of course they both answer that way, but if they differ, they'll choose A's answer (since it's more likely to be correct than B's).  But if that's the case, then they'll both answer correctly only if A already had the right answer.  That is to say, both A and B will answer correctly with probability a.  Well, there's no reason for A to collude with B, since it helps B without helping A.

The situation is not helped even if a = b, since the only difference is that some other means must be used for breaking the tie.  No matter how the tie is broken, the answer that is chosen cannot have a greater probability of being correct than a = b, so there is no benefit to collusion for either A or B.

A similar line of reasoning applies to any other pair of students.  Well, then how about all three students colluding?  That will only happen if all three students are benefited, and A, with the highest aptitude, is the standard here.  Let's consider how A's answer would be affected by the collusion.  The first way is that A's initially correct answer would be made incorrect by collusion.  That happens if A would have answered correctly, but B and C would not.  That happens with probability a (1 - b) (1 - c).

The second way to affect A's answer is to change an initially incorrect answer into a correct one.  That happens with probability (1 - a) b c.  So, on balance, A has an incentive to collude (and therefore all students do) if

(1 - a) b c > a (1 - b) (1 - c)

For instance, if the three students respectively have 90, 80, and 70 percent probabilities of answering questions correctly, then we have

(0.1) (0.8) (0.7) = 0.056 > 0.054 = (0.9) (0.2) (0.3)

and it makes sense for all three to collude, by this metric.

Why by this metric?  What other metric could there be?  Suppose we now introduce an explicit curve: The students receive, as their final grade, not their actual raw score, but a ranking-scaled score.  The top raw score earns three points, the second best raw score earns two points, and the lowest raw score earns one point.  Two students tying at the top both earn 2.5 points, while two students tying at the bottom earn 1.5 points, and finally if all three students tie, they all earn two points.

Under these conditions, the three students will not all collude.  A, as the best student, is the most likely of the three to earn three points, and the more questions there are, the more certain that is.  If A, B, and C all collude, they will all three earn two points (since their answers will be identical).  So chuck out three-way collusion.

But two-way collusion is now even less likely than before.  As we observed, it only improves the accuracy of the inferior student.  Before, that at least did not hurt the superior student, but now it improves the inferior student's scaled score at the expense of the superior student's scaled score.  So two-way collusion is out, too.

Shall we move on to four students?  I'll save that for a later post.

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