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
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?
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
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,
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
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.
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
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
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
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