I plan to continue my discussion of Boij–Söderberg theory from last time, but I’d like to make a quick detour first. This came up today in the office when I was talking with yanzhang.

Question: If I randomly pick numbers (with a uniform distribution) from the unit interval [0,1], what is the expected number of numbers that I need to pick before their sum is at least 1?

The answer for some strange reason is the number e, and I’ll illustrate this with two approaches. Both give different expressions of e, so this problem will show an equivalence between two definitions of e.

The first approach is to use calculus. For , let f(r) be the expected number of randomly selected numbers in [0,1] so that their sum is at least r, and define f(r) = 0 if r is negative. Then we have the following integral equation

.

To see this, pick a number t randomly in [0,1], and then add f(r-t). Since we need to account for all such t, we just integrate over [0,r] (I could integrate over [0,1], but remember that f(r-t) = 0 if t>r), and divide this integral by the length of [0,1], which is 1. The second inequality follows from a change of variables s=r-t. Taking derivatives of both sides, we get by the fundamental theorem of calculus, so for some constant c. Of course, f(0) = 1, so c=1. Our original question was for r=1, in which case f(1) = e, as we promised.

The second approach is the same, except we discretize everything. Let’s ask an analogous question. First fix k. If I randomly pick integers uniformly from [0,k], what is the expected number that I need before their sum is at least n, where ? Let g(n) be this expected value. Then again, we have

by the same reasoning as before. Let be the first difference operator, i.e., . Now apply to the above equation to get

.

Of course, g(-1) = 0, and rewriting this last equation gives

assuming that n>0. Since g(0) = 1, this means that . Now set k=n: If we rephrase this question, we also see that g(n) is the expected number of random selections of rational numbers with denominator n from [0,1] such that their sum is at least 1. Letting n go to infinity, we should get the answer from the previous method since the rationals form a dense subset of the reals.

So these two approaches show the equivalence of two definitions of e: one as the unique number such that , and the other as the limit .

Do you know of any other solutions to the original question?

-Steven

Problem: we have n people sitting around in a circle. I tap one of them, and then go on a drunken/random walk going in either direction with equal probablility, tapping people as I go on my merrily drunken way. For each person besides the first, what is the probability he/she will be the last one I tap?

Consider the probability of not hitting the guy to A’s right. Clearly it is equal to the probability that he is the last one hit. Now, for any dude in the middle X, the probability that he is the last one to be hit is equal to the sum of: “we go and hit the guy to X’s left and then never hit X” and “we go and hit the guy to X’s right and then never hit X.” Notice that because you have to hit one of them first at some point, the sum of these two is just: “we go and tap a person adjacent to X and then never hit X.” But note this is exactly equal to the probability of not hitting the guy to A’s right. Thus, the probability of being tapped last is equal for everyone besides A, and is 1/(n-1).

I thank Boris Alexeev for giving me this problem.

-YZ

Today we have a M-2… but it feels more like a M-3 or M-4. I’ll explain after the two problems.

The first problem is fairly classical. It appears in many texts, and was first brought to my attention by Chris Evans:

N people stand in line, all facing right, where each person can only see the people in front. They each get a hat (red or blue), and in order (from left to right), they are asked to guess the color of their own hat. Find a strategy to maximize the number of people who guess their hat colors correctly. Note that the strategy can depend on what people further right hear from the guesses of people further left.

Solution:

The funny thing about this problem is that you can ask it as a stronger statement (“find a strategy so that N-1 people guess their hat color correctly”), which makes the problem a lot easier =) This is why I ask the problem in the above form.

Now that I’ve said that, you may want to take a moment to try to do it again.

Give up? Do this: Think of a red hat as a 0 and a blue hat as a 1. The first person guesses the color corresponding to the sum of everyone else (mod 2). Note that this gives each person after him enough information; the second person can just subtract off what he sees from what the first person says. The third person, after hearing the second person’s guess, knows the sum of all hats after and including the third, so he can just subtract off what he sees to get his own hat. This gets at least N-1 right, and half of the times gets all N right. Note we can’t possibly do anything better since the first person can only ever get his own hat right with probability 1/2.

I heard the second problem recently from Nick Rosenbloom and Dustin Clausen:

Countably infinite people stand in line, all facing right… the rest of the setup is the same. Find a strategy to maximize the cardinality of people who guess their hat colors correctly.

Sketch:

The awesome thing about this problem is there is also a way to ask this problem that hints at the answer. What is slightly different about this formulation of asking the problem is that it sounds like a strictly harder problem has been asked: one can restrict the problem so that the strategy cannot rely on hearing what people left of each person answers.

This looks kind of impossible, which is perfect for things like the Axiom of choice (or equivalently, Zorn’s Lemma). Consider all countably infinite binary strings and divide them up into equivalence classes, where two strings A and B are equivalent if they share a truncation (i.e. the bits after the k_th bit of A are equal to the bits after the l_th bit of B). Note it takes a bit of work to show that this is an equivalence class, which I’ll leave as an exercise. You might want to do the case of the string being periodic after a certain point differently from the case where it is not.

Now, just pick a representative from each equivalence class. I leave it as another exercise to see that you are done =) It basically involves each person to try to fit himself into a string which has the same equivalence class at what he sees. The earliest people may get this wrong, but the later people (once you get to the common truncation) will get them all right. So you get all but finitely many guesses correct.

I think the coolest thing about this set of problems is the way you can rephrase each one and still ask the problem at the same strength. Note this becomes a M-3 if there was a solution to the second problem that breaks the restriction given at the beginning of the proof. So if any of you have ideas, let me know!

-Y

Source: Chris Evans, Nick Rosenbloom, Dustin Clausen

Here is another Matryoshka problem – well not really, since I can’t prove that the second problem is “strictly harder” than the first in any meaningful way, but it does have the property that the most intuitive trick needed to solve the first one fails on the second one:

Timothy Gowers posted a fairly classical “paradox” in probability involving two boxes:

original post

My analysis:

Note that the guy who gives you the box must pick (a, 2a) with equal probability for every a – i.e., he needs to be able to pick a real number uniformly. Unfortunately, this is impossible.

This is actually a common sleight of hand that many probability “paradoxes” use at some level. You can increase your “defense against probability paradoxes” by adding this to the list of things to check against every time you find a strange probability situation, in the same way that recognizing Russell’s Paradox counters many logic “paradoxes.”

—

Now the inner doll. Prof. Gowers attributes this one to Noga Alon:

original post

These two problems are similar in spirit. However, the trick that works for the first one no longer works here: we have a very well-defined distribution! So we have to dig a little deeper. Here is my take:

In the paradox, we considered the random variable of the values of the two envelopes (call them and , and proceeded to divide the probability space into a series of sums x_i (where ), seeing that for all i. Then we did an implicit sum of all of these equations to conclude that .

The problem is in the sum – we cannot do this summation when the expectation of X and Y are infinite (we can easily check, by the way, that and are both infinite), so we cannot conclude that one is larger than the other. Here is a clearer example: consider the list of inequalities , , … each one of these statements is true, but it makes no sense to then argue that , because the sums diverge and cannot be defined. If you track the list of inequalities in the problem, you will notice the exact analogous chicanery going on, just with different numbers.

The cool lesson form this M-2 pair is that I am reminded again to appreciate why math books always mention “boundary cases” like infinity and deal with them very carefully. Usually we see them as just stupid limitations for corner cases, but this is a case where not thinking about the clauses on these border cases can mislead (to be honest, I never remember these boundary cases, like the algebra theorems that hold for every ring except the zero ring).

By the way, I bet there is some way to turn each one of these into a “paradox” in their respective fields. Probability is just riper for these conundrums since the problems offer more intuitive statements.

-Y

Source: Gowers’ Weblog