## Reading the Comics, April 15, 2016: Remarkably, No Income Tax Comics Edition

I’m as startled as you are. While a couple comic strips mentioned United States Income Tax Day, they didn’t do so in a way that seemed on-point enough for this Reading The Comics post. Of course, United States Income Tax Day happens to be the 18th this year. I haven’t seen Sunday’s comics yet.

David L Hoyt and Jeff Knurek’s Jumble for the 11th of April one again uses arithmetic puns for its business. Also, if some science fiction writer doesn’t take hold of “Gribth” as a name for something they’re missing a fine syllable. “Tahew” is no slouch in the made-up word leagues either.

David L Hoyt and Jeff Knurek’s Jumble for the 11th of April, 2016. The link will probably expire sometime before the year 2112.

Ryan North’s Dinosaur Comics for the 12th of April obviously originally ran sometime in mid-March. I have similarly ambiguous feelings about the value of Pi Day. I suppose it’s nice for people to think of “fun” and “mathematics” close together. Utahraptor’s distinction between “Pi Day” of March 14 and “Approximate Pi Day” of the 22nd of July s a curious one, though. It’s not as though 3.14 is any more exactly π than 22/7 is. I suppose you can argue that at some moment on 3/14 between 1:59:26 and 1:59:27 there’s some moment, 1:59:26.5358979 et cetera going on forever. But that assumes that time is a continuous thing, and it’s not like you’ll ever know what that moment is. By the time you might recognize it, it’s passed. They are all Approximate Pi Days; we just have to decide what the approximation is.

Bill Schorr’s The Grizzwells for the 12th is a silly-homework problem question. I know the point is to joke about how Fauna misunderstands a word. But if we pretend the assignment is for real, what might its point be? To show that students know the parts of a right triangle? I guess that’s all right, but it doesn’t seem like much of an assignment. I don’t blame her for getting snarky in the face of that.

Rick Kirkman and Jerry Scott’s Baby Blues for the 13th is a gag about picking random numbers for arithmetic homework. The approach is doomed, surely, although it’s probably not completely doomed. I’m not sure Hammie’s age, but if his homework is about adding and subtracting numbers he probably mostly gets problems that give results between zero and twenty, and almost always less than a hundred. He might hit some by luck.

Rick Kirkman and Jerry Scott’s Baby Blues for the 13th of April, 2016. It’s only after Hammy walks away that Zoe wonders why he needs five random numbers?

I’ve mentioned some how people are awful at picking “random” numbers in their heads. Zoe shows off one of the ways people are bad at it. People asked to name numbers “randomly” pick odd numbers more than even numbers. Somehow they just feel random. I doubt Kirkman and Scott were thinking of that; among other things, five numbers is a very small sample. Four odds out of five isn’t peculiar, not yet. They were probably just trying to pick numbers that sounded funny while fitting the space available. I’m a bit surprised 37 didn’t make the list.

Mark Anderson’s Andertoons for the 13th is Mark Anderson’s Andertoons entry for this essay. I like the teacher’s answer, though.

Patrick Roberts’s Todd the Dinosaur for the 14th just uses arithmetic as the most economic way to fit several problems on-screen at once. They’ve got a compactness that sentence-diagramming just can’t match.

Patrick Roberts’s Todd the Dinosaur for the 14th of April, 2016. No fair wondering why his more distant eye is always the larger one.

Greg Cravens’s The Buckets for the 15th amuses me with its use of coin-tossing as a way of making choices. I’m also amused the coin might be wrong only about half the time.

John Deering’s Strange Brew for the 15th is a visual puzzle. It’s intending to make use of a board full of mathematical symbols to represent deep thought. But the symbols aren’t quite mathematics. They look much more like LaTeX, a typesetting code used to express mathematics in print. Some of the symbols are obscured, so I can’t say exactly what’s meant. But it should be something like this:

$F = \{F_{x} \in F_{c}: (is ... (1) ) \cap (minPixels < \|s\| < maxPixels ) \\ \partial{P} \\ (is_{connected}| > |s| - \epsilon) \}$

At the risk of disappointing, this appears to me gibberish. The appearance of words like ‘minPixels’ and ‘maxPixels’ suggest a bit of computer code. So does having a subscript that’s the full word “connected”. I wonder where Deering drew this example from.

• #### Jacob Kanev 9:46 pm on Tuesday, 19 April, 2016 Permalink | Reply

Nice. On a totally unrelated note, my favourite comic about random numbers is from Dilbert, when he visits the accounting department: http://dilbert.com/strip/2001-10-25

Regards from Jacob.

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• #### Joseph Nebus 2:08 am on Friday, 22 April, 2016 Permalink | Reply

Ha ha! Thank you. That’s a strip I had forgotten. It’s true, though; it’s so very hard to say what randomness is, or really pin down whether we’ve ever seen it.

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• #### elkement (Elke Stangl) 12:50 pm on Wednesday, 27 April, 2016 Permalink | Reply

I wonder why dinosaurs are so popular as characters? ;-)

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• #### Joseph Nebus 6:47 pm on Friday, 29 April, 2016 Permalink | Reply

Oh, dinosaurs have a lot going for them. They’ve got a great visual style and there’s at least one to fit any mood you might have. I’m a little surprised there are so few comic strips that have them. But modern comic strips have a strange aversion to funny-looking characters.

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## Ensembled

A couple weeks back voting in the Democratic party’s Iowa caucus had several districts tied between Clinton and Sanders supporters. The ties were broken by coin tosses. That fact produced a bunch of jokes at Iowa’s expense. I can’t join in this joking. If the votes don’t support one candidate over another, but someone must win, what’s left but an impartial tie-breaking scheme?

After Clinton won six of the coin tosses people joked about the “impartial” idea breaking down. Well, we around here know that there are no unfair coins. And while it’s possible to have an unfair coin toss, I’m not aware of any reason to think any of the tosses were. It’s lucky to win six coin tosses. If the tosses are fair, the chance of getting any one right is one-half. Suppose the tosses are “independent”. That is, the outcome of one doesn’t change the chances of any other. Then the chance of getting six right in a row is the chance of getting one right, times itself, six times over. That is, the chance is one-half raised to the sixth power. That’s a small number, about 1.5 percent. But it’s not so riotously small as to deserve rioting.

Yes and no. It depends on what you mean by “any other outcome”. Grant that heads and tails are equally likely to come up. Grant also that coin tosses are independent. Then six heads, H H H H H H, are just as likely to come up as six tails, T T T T T T. I don’t think anyone will argue with me that far.

But are both of these exactly as likely as the first toss coming up heads and all the others tails? As likely as H T T T T T? Yes, I would say they are. But I understand if you feel skeptical, and if you want convincing. The chance of getting heads once in a fair coin toss is one-half. We started with that. What’s the chance of getting five tails in a row? That must be one-half raised to the fifth power. The first coin toss and the last five don’t depend on one another. This means the chance of that first heads followed by those five tails is one-half times one-half to the fifth power. And that’s one-half to the sixth power.

What about the first two tosses coming up heads and the next four tails? H H T T T T? We can run through the argument again. The chance of two coin tosses coming up heads would be one-half to the second power. The chance of four coin tosses coming up tails would be one-half to the fourth power. The chance of the first streak being followed by the second is the product of the two chances. One-half to the second power times one-half to the fourth power is one-half to the sixth power.

We could go on like this and try out all the possible outcomes. There’s only 64 of them. That’s going to be boring. We could prove any particular string of outcomes is just as likely as any other. We need to make an argument that’s a little more clever, but also a little more abstract.

Don’t think just now of a particular sequence of coin toss outcomes. Consider this instead: what is the chance you will call a coin toss right? You might call heads, you might call tails. The coin might come up heads, the coin might come up tails. The chance you call it right, though — well, won’t that be one-half? Stay at this point until you’re sure it is.

So write out a sequence of possible outcomes. Don’t tell me what it is. It can be any set of H and T, as you like, as long as it’s six outcomes long.

What is the chance you wrote down six correct tosses in a row? That’ll be the chance of calling one outcome right, one-half, times itself six times over. One-half to the sixth power. So I know the probability that your prediction was correct. Which of the 64 possible outcomes did you write down? I don’t know. I suspect you didn’t even write one down. I would’ve just pretended I had one in mind until the essay required me to do something too. But the exact same argument applies no matter which sequence you pretended to write down. (Look at it. I didn’t use any information about what sequence you would have picked. So how could the sequence affect the outcome?) Therefore each of the 64 possible outcomes has the same chance of coming up.

So in this context, yes, six heads in a row is exactly as likely as any other sequence of six coin tosses.

I will guess that you aren’t perfectly happy with this argument. It probably feels like something is unaccounted-for. What’s unaccounted-for is that nobody cares about the difference between the sequence H H T H H H and the sequence H H H T H H. Would you even notice the difference if I hadn’t framed the paragraph to make the difference stand out? In either case, the sequence is “one tail, five heads”. What’s the chance of getting “one tail, five heads”?

Well, the chance of getting one of several mutually exclusive outcomes is the sum of the chance of each individual outcome. And these are mutually exclusive outcomes: you can’t get both H H T H H H and H H H T H H as the result of the same set of coin tosses.

(There can be not-mutually-exclusive outcomes. Consider, for example, the chance of getting “at least three tails” and the chance of the third coin toss being heads. Calculating the chance of either of those outcomes happening demands more thinking. But we don’t have to deal with that here, so we won’t.)

There are six distinct ways to get one tails and five heads. The tails can be the first toss’s result. Or the tails can be the second toss’s result. Or the tails can be the third toss’s result. And so on. Each of these possible outcomes has the same probability, one-half to the sixth power. So the chance of getting “one tails, five heads” is one-half to the sixth power, added to itself, six times over. That is, it’s six times one-half to the sixth power. That will come up about one time in eleven that you do a sequence of six coin tosses.

There are fifteen ways to get two tails and four heads. So the chance of the outcome being “two tails, four heads” is fifteen times one-half to the sixth power. That will come up a bit less than one in four times.

There are twenty, count ’em, ways to get three tails and three heads. So the chance of that is twenty times one-half to the sixth power. That’s a little more than three times in ten. There are fifteen ways to get four tails and two heads, so the chance of that drops again. There’s six ways to get five tails and one heads. And there’s just one way to get six tails and no heads on six coin tosses.

So if you think of the outcome as “this many tails and that many heads”, then, no, not all outcomes are equally likely. “Three tails and three heads” is a lot more likely than “no tails and six heads”. “Two tails and four heads” is more likely than “one tails and five heads”.

Whether it’s right to say “every outcome is just as likely” depends on what you think “an outcome” is. If it’s a particular sequence of heads and tails, then yes, it is. If it’s the aggregate statistic of how many heads and tails, then no, it’s not.

We see this kind of distinction all over the place. Every hand of cards, for example, might be as likely to turn up as every other hand of cards. But consider five-card poker hands. There are very few hands that have the interesting pattern of being a straight flush, five sequential cards of the same face. There are more hands that have the interesting pattern of four-of-a-kind. There are a lot of hands that have the mildly interesting pattern of two-of-a-kind and nothing else going on. There’s a huge mass of cards that don’t have any pattern we’ve seen fit to notice. So a straight flush is regarded as a very unlikely hand to have, and four-of-a-kind more likely but still rare. Two-of-a-kind is none too rare. Nothing at all is most likely, at least in a five-card hand. (When you get seven cards, a hand with nothing at all becomes less likely. You have so many chances that you just have to hit something.)

The distinction carries over into statistical mechanics. The field studies the state of things. Is a mass of material solid or liquid or gas? Is a solid magnetized or not, or is it trying to be? Are molecules in a high- or a low-energy state?

Mathematicians use the name “ensemble” to describe a state of whatever it is we’re studying. But we have the same problem of saying what kind of description we mean. Suppose we are studying the magnetism of a solid object. We do this by imagining the object as a bunch of smaller regions, each with a tiny bit of magnetism. That bit might have the north pole pointing up, or the south pole pointing up. We might say the ensemble is that there are ten percent more north-pole-up regions than there are south-pole-up regions.

But by that, do we mean we’re interested in “ten percent more north-pole-up than south-pole-up regions”? Or do we mean “these particular regions are north-pole-up, and these are south-pole-up”? We distinguish this by putting in some new words.

The “canonical ensemble” is, generally, the kind of aggregate-statistical-average description of things. So, “ten percent more north-pole-up than south-pole-up regions” would be such a canonical ensemble. Or “one tails, five heads” would be a canonical ensemble. If we want to look at the fine details we speak of the “microcanonical ensemble”. That would be “these particular regions are north-pole-up, and these are south-pole-up”. Or that would be “the coin tosses came up H H H T H H”.

Just what is a canonical and what is a microcanonical ensemble depends on context. Of course it would. Consider the standpoint of the city manager, hoping to estimate the power and water needs of neighborhoods and bringing the language of statistical mechanics to the city-planning world. There, it is enough detail to know how many houses on a particular street are occupied and how many residents there are. She could fairly consider that a microcanonical ensemble. From the standpoint of the letter carriers for the post office, though, that would be a canonical ensemble. It would give an idea how much time would be needed to deliver on that street. But would be just short of useful in getting letters to recipients. The letter carrier would want to know which people are in which house before rating that a microcanonical ensemble.

Much of statistical mechanics is studying ensembles, and which ensembles are more or less likely than others. And how that likelihood changes as conditions change.

So let me answer the original question. In this coin-toss problem, yes, every microcanonical ensemble is just as likely as every other microcanonical ensemble. The sequence ‘H H H H H H’ is just as likely as ‘H T H H H T’ or ‘T T H T H H’ are. But not every canonical ensemble is as likely as every other one. Six heads in six tosses are less likely than two heads and four tails, or three heads and three tails, are. The answer depends on what you mean by the question.

• #### Ken Dowell 3:52 am on Tuesday, 23 February, 2016 Permalink | Reply

So it occurs to me that problem is not the coin toss but rather the system that demands a winner even though there really isn’t one.

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• #### Joseph Nebus 4:12 am on Tuesday, 23 February, 2016 Permalink | Reply

Well, it is. But using any kind of voting scheme to make decisions leaves you vulnerable to problems. What to do in case of a tie is an obvious one. Once there’s three possible choices in play (and at least three voters) it becomes impossible to guarantee a perfectly fair outcome.

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