Epsilon Clue

There’s a stock market scam that goes something like this: make a list of 1024 people, and send them an “investment newsletter”. The copy sent to the first 512 people says that a particular stock will go up; the other 512 get a copy that says that that stock will go down. Let’s say it goes down. You throw away the list of people whom you told the stock would go up, divide the remaining 512 in two, and send them another “investment newsletter”. You tell the first 256 that some other stock will go up, and tell the other 256 that that stock will go down. Eliminate those to whom you gave a false prediction, divide the remaining ones in two, and send them another tip, as before. Do this ten times, and you’ll wind up with one person to whom, by sheer numbers, you’ve given ten good predictions in a row. You then tell that person that he’ll have to pay you to receive further stock tips.

One problem with this scam (from the scammer’s point of view) is that there’s a lot of waste: you have to start with over a thousand names and whittle them down to just one sucker. But what if you lowered your standards a bit? After all, if someone gets nine good predictions and one bad one, you can still say you have a 90% success rate, and that should help sell your nonexistent Wall Street wisdom. What about 80%? Or 70%? If you start with 1024 names, how many potential suckers will you have if you consider the ones to whom you sent seven or more correct predictions, and not just the one where you got all ten right?

Fig. 1. A grid of city streets

But first, let’s look at a different problem. Let’s say you’re standing at the corner of 1st Ave. and A St., and you want to get to 2nd and D. You’re on foot, so you want to take the shortest possible path, but there are several of these. How many? You could walk east, then turn north on D St.; or you could walk east for two blocks, north one block on C St., then east another block on 2nd Ave.; and so forth. As you can quickly check, there are four paths from 1st and A to 2nd and D.

But is there an easy way to calculate the number of paths to a given intersection, without having to enumerate them? As it turns out, there is:

If you’re going someplace on 1st Ave., there’s only one way to get there: walk east until you get where you’re going. And if you want to go someplace on A St., the only thing to do is to walk north until you get there.

If you’re going to, say, 4th and C, there are two main ways to arrive: either you can make your way (somehow) to 3d and C, then walk north; or you can make your way to 4th and B, and walk east. So the number of ways to get to 4th and C is the number of ways to get to 3d and C, plus the number of ways to get to 4th and B.

Fig. 2. The number of ways to get to each intersection from 1st and A.

So we can fill in the number of paths to get to each insersection: each number is the sum of the one south of it and the one west of it.

Fig. 3. The intersections you can get to by walking four blocks.

Another question we can ask is, which intersections can you get to by walking four blocks? You’re going to walk n blocks east and m blocks north, so you can get to any intersection where n+m = 4.

We can now return to our original question: you’re going to make ten stock market predictions for a bunch of people. Let’s say that making a correct prediction corresponds to walking one block east, and making an incorrect prediction corresponds to walking one block north. So we see that there’s only one way of getting ten correct predictions: get all ten of them right. There are also ten ways of getting nine correct predictions: get the first nine correct and miss the tenth; or get the first eight right, miss the ninth, and get the last one correct; and so forth. This is the same as asking how many ways there are of getting to Eye St. and 2nd Ave., which we now know how to calculate. And the number of ways of getting eight correct predictions is the number of ways of getting to H and 3d, and so forth.

                    1
                  1   1
                1   2   1
              1   3   3   1
            1   4   6   4   1
          1   5  10  10   5   1
        1   6  15  20  15   6   1
      1   7  21  35  35  21   7   1
    1   8  28  56  70  56  28   8   1
  1   9  36  84 126 126  84  36   9   1
1  10  45 120 210 252 210 120  45  10   1

If we turn our table of numbers so that 1st and A is at the top, and the line of places we can get to in 10 blocks is at the bottom, we get Pascal’s triangle. This is one of my favorite mathematical constructs, because it pops up everywhere: statistics, polynomial multiplication, and even discussions on how to defraud investors. (Let’s also note in passing that Pascal’s triangle is symmetrical, and that each row adds up to a power of 2.)

In row n, element m is (n+m)! / n! × m!. It’s easy to see why: let’s say you have a bag of cards that say things like “IBM will go up (incorrect)”, “Alcoa will go down (correct)” and so forth. Now let’s say that you’ve reached into the bag 10 times and pulled out 7 correct predictions (and 3 incorrect ones). There are 10! ways you could have pulled those particular ones out of the bag, so that gives us the (n+m)! in the numerator.

There are also 7! possible orders in which you could have pulled out the 7 correct predictions, but since we don’t care about the order, they’re all equivalent, so we divide by 7!. Likewise, the 3! orders in which you could have pulled out the 3 incorrect predictions are equivalent, so we divide by 3!. This gives us (7+3)! / 7! × 3! = 3628800 / (5040 × 6) = 120 ways to have 7 correct predictions out of 10.

So what does this all tell us? If you’re only going to try to sucker those to whom you’ve given a perfect 10 correct predictions in your scam, you’re going to have to start with 1024 names just to whittle them down to one lonely sucker. But if you want to scam those who got nine out of ten, you get an additional 10 suckers. Lowering your standards to 8 out of 10 gets you an additional 45 suckers (1+10+45 = 56), and reaching futher to those who got 7 out of 10 gives you 1+10+45+120 = 176 suckers out of your original 1024. A vast improvement over the original. Until, of course, the FCC sends you to FPMITA prison.

This entry was posted on Saturday, January 27th, 2007 at 2:04 pm and is filed under Math. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.