Far more often than seems reasonable, there’s not even a full cup of coffee in the coffee pot! Once again, someone has left a nearly empty pot with no regard to the next person (me, whine)!

This seems to happen so often I began to wonder what kind of boors I’ve been working with all these years. They seem like nice people and all, but…?

And then I realized: there’s a perfectly logical reason, a mathfactor puzzle, if you will, that explains this phenomenon perfectly, no boors required, no special tricks, just sensible activity by all. My faith in my colleagues has been restored.

Why is it that on average I see an emptier rather than fuller coffee pot?

PS let us know what works for when we return…

Benford’s Law is really quite amazing, at least at first glance: for a wide variety of kinds of data, about 30% of the numbers will begin with a 1, 17% with a 2, on down to just 5% beginning with a 9. Can you spot the fake list of populations of European countries?

 Looking at these lists we have a clue as to when and how Benford’s Law works. [spoiler]

In one of the lists, the populations are distributed more or less evenly in a linear scale; that is, there are about as many populations from 1 million to 2 million, as there are from 2 million to 3 million, 3 million to 4 million etc. (Well, actually the distribution isn’t quite linear,  because the fake data was made to look similar to the real data, and so has a few of its characteristics.)

The real list, like many other kinds of data, is distributed in a more exponential manner; that is, the populations grow exponentially (very slowly though) with about as many populations from 100,000 to 1,000,000; then 1,000,000 to 10,000,000; and 10,000,000 to 100,000,000. This is all pretty approximate, so you can’t take this precisely at face value, but you’ll see in the list of real data that, very roughly speaking, in any order of magnitude there are about as many populations as in any other– at least for a while. 

Data like this has a kind of “scale invariance”, especially if this kind of pattern holds over many orders of magnitude. What this means is that if we scale the data up or down, throwing out the outliers, it will look about the same as before. 

The key to Benford’s Law is this scale invariance. Data that has this property will automatically satisfy his rule. Why is this? If we plot such data on a linear scale it won’t be distributed uniformly but will be all stretched out, becoming sparser and sparser. But if we plot it on a logarithmic scale, (which you can think of as approximated by the number of digits in the data), then such data is smoothed out and evenly distributed. 

But presto! Look at how the leading digits are distributed on such a logarithmic scale!

That’s mostly 1’s, a bit fewer 2’s, etc. on down to a much smaller proportion of 9’s.

[/spoiler]

In the puzzle you have five different tasks.  On each day one of these tasks is given at random.  How long do you expect it to take to get all five tasks?

First consider a simple case.  Suppose some event has a probability, p, of happening on any one day.  Let’s say that E(p) is the expected number of days we have to wait for the event to happen.  For example if p=1 then the event is guaranteed to happen every day and so E(p)=1.

How can we calculate E(p)? 

Andy does an experiment.  He will wait for the event to happen and record how many days it took.  He will do this several times, for long enough to ensure that he gets an answer that is as accurate as he needs.  He will keep going for N days in total.  Afterwards he will take the average of all of his wait times to get an estimate for E(p).

The average he calculates is the total of all the wait times divided by the number of occurrences of the event.  But we can estimate both of these values and therefore estimate his value for E(p).  The total of all of the wait times is going to be about N.  Since the event has a probability of p of occurring on any particular day the number of occurrences will be about p times N.

So Andy’s average will be about N/(pN) which will be about 1/p.

We can make N as large as we like to make this result as accurate as we like.  So we can confidently say that E(p) = 1/p.

Let’s get back to our puzzle about tasks.  We need to wait for the first task, then the second, then the third and so on.  When there are t tasks left then the chance of getting a new one is t/5.  

So the total waiting time is

    E(5/5) + E(4/5) + E(3/5) + E(2/5) + E(1/5) = 5/5 + 5/4 + 5/3 + 5/2 + 5/1

       = 5(1/5 + 1/4 + 1/3 + 1/2 + 1) = 11 ⁵/₁₂ = 11.4166666… days

You may recognise the harmonic series!

So that’s the proof, now where’s the paradox?

Bob visits Andy when he is able and waits with him until the event occurs.  Of course Andy may well have been waiting for some time.

If Bob turns up just after the event has happened then he will wait for the same time as Andy.  If he turns up just before the event he will wait for one day.  On average he will wait for about half the time that Andy does.

When the event occurs Bob disappears and comes back when he is next able to.

At the end of the experiment Bob averages all of his wait times to get an estimate for E(p).  He gets an answer which is half of Andy’s!

That is the paradox!

Now Carol thinks she understands what is going on here.  The problem is that Andy is distorting his results by always starting the clock straight after an event has occurred.  That guarantees him to get the longest possible wait times.

She thinks the only way to get an accurate answer is to look at the wait time starting from each of the N days and then average these N wait times.

To calculate this she needs to make an assumption.  She knows that the event occurs every 1/p days on average.  She assumes they happen regularly every 1/p days.

 Let’s say they happen every n days where n = 1/p.  Remember Andy’s value for E(p) is 1/p = n.  Bob’s value is half that, so about n/2. 

The wait time will vary between 1 and n.  The average wait time will be n(n+1)/2n = (n+1)/2.

So Carols estimate for E(p) is about n/2 whereas Andy’s was 1/p = n.  So Carol is agreeing with Bob.

I can tell you that Andy had the right value and that the logic I used for Bob and Carol was flawed. 

Can you see where?

I was wondering what is the theoretical ‘area’ of contact between two spheres in contact with each other. I was unfortunately not able to locate much (if any) information on this. After some thought into this I’ve realised that the spheres would meet at a single ‘point’ however what would the area of this ‘point’ be? The only source related to this claimed the area of contact, the point, has no area. How can a point have no area? If the spheres touch, musn’t there be an area shared between them? Even if only one atom?

Hi, the issue here is that there is a vast difference between physical, real things and the mathematical ideas that model them.

Real, mathematical spheres don’t exist, plain and simple! Never could, even as a region of space— space itself has a granularity (apparently) at a scale of about 10^-33 meters. There simply cannot exist a perfectly spherical region in physical space, much less a perfectly spherical body.

But as an abstraction, the idea of a sphere is very useful: lots of things, quite evidently, are spherical for all practical purposes.

For that matter, “points” don’t exist either, and are also a mathematical abstraction. (So, too, is “area”. Real things are rough, bumpy and not at all like continuous surfaces, on a fine enough scale) But again, these _ideas_ are very good at getting at something important about lots and lots of physical things, and so have proved useful.

Tangent spheres do indeed meet in a single point, which has no area.

Spherical things meet in some other, messier way.

Hope this helps!

1 + 1/2 + 1/3 + 1/4 + . . .
diverges to infinity, growing as large as you please!

If you try adding terms up on a calculator, this scarcely seems possible! By the time you have added the hundredth term, you will have a reached only a whopping 5.187… (and each new term will be less than .01).

After adding up a million terms, you will have made it only to about 14.39272672… — and each new term will be less than .000001. Does the series really diverge?

The eighteenth century mathematician Jacob Bernoulli gave a very elegant proof that it does:

1/2 is at least 1/2

1/3 + 1/4 is at least 1/4 + 1/4 = 1/2

1/5 + 1/6 + 1/7 + 1/8 is at least 1/8 + 1/8 + 1/8 + 1/8 = 1/2

1/9 + . . . + 1/16 is at least 8 x 1/16 = 1/2

etc. So the result of adding up the first 2ⁿ terms 1/2 + 1/3 + . . . + 1/2ⁿ is at least n/2, and in particular, can be as large as we please.

But this does take a long time to get anywhere. To add up to, say, 100, Bernoulli’s proof shows us that 2¹⁹⁸ (about 4×10⁵⁹) terms will suffice. But maybe this is more than we actually need.

A basic fact from calculus is that the area under the curve y = 1/x, from x = 1 to x = N is exactly ln N.

Now the area of a box 1 unit wide and 1/n units tall is 1/n, and boxes of width 1 and heights 1, 1/2, 1/3, . . . altogether have area 1 + 1/2 + 1/3 . . .

Here we see that these boxes can be arranged to show that

1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 > ln 8

Shifting the boxes over, we see that

1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 < ln 8

This gives us a much better bound on the harmonic series. Generally, we have that

1 + 1/2 + . . . + 1/n is between ln (n+1) and (ln n) + 1.

So to be sure that the series sums to at least 100, we can be sure that e^100-1 (about 2.7×10⁴³) terms will suffice!

So the sum of the first million terms is about (ln 1,000,000) + γ, and if we want to sum to 100, we need to have n such that ln n + γ is greater than 100; in other words, e ^{(100 – γ)} (about 1.5×10⁴³) terms will do.

The series Σ 1/(n ln n) diverges even more slowly still, taking about e^e^n terms to sum to n (!!) The series Σ 1/(n (ln n) (ln ln n)) takes e^e^e^n terms to sum to n. Etc!!

Dana Richards, editor of The Colossal Book of Short Puzzles and Problems discusses the amazing Martin Gardner and his legacy!

Faster than an exponential! More powerful than double factorials!! The Busy Beaver Function tops anything that could ever be computed– and we mean ever