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.

My love asked me about a claim about this made on a Facebook discussion. The writer asserted that six heads was exactly as likely as *any* other outcome of six coin tosses. My love wondered: is that true?

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.

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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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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