The answer would be “of course not”. I was playing against, mostly, the same people who were in the state finals. (A few who didn’t qualify for the finals joined the side tournament.) In that I had done well enough, winning seven games in all out of fifteen played. It’s implausible that I got significantly worse at pinball between the main and the side tournament. But can I make a logically sound argument about this?
In full, probably not. It’s too hard. The question is, did I win way too few games compared to what I should have expected? But what should I have expected? I haven’t got any information on how likely it should have been that I’d win any of the games, especially not when I faced something like a dozen different opponents. (I played several opponents twice.)
But we can make a model. Suppose that I had a fifty percent chance of winning each match. This is a lie in detail. The model contains lies; all models do. The lies might let us learn something interesting. Some people there I could only beat with a stroke of luck on my side. Some people there I could fairly often expect to beat. If we pretend I had the same chance against everyone, though, we get something that we can model. It might tell us something about what really happened.
If I play 16 matches, and have a 50 percent chance of winning each of them, then I should expect to win eight matches. But there’s no reason I might not win seven instead, or nine. Might win six, or ten, without that being too implausible. It’s even possible I might not win a single match, or that I might win all sixteen matches. How likely?
This calls for a creature from the field of probability that we call the binomial distribution. It’s “binomial” because it’s about stuff for which there are exactly two possible outcomes. This fits. Each match I can win or I can lose. (If we tie, or if the match is interrupted, we replay it, so there’s not another case.) It’s a “distribution” because we describe, for a set of some number of attempted matches, how the possible outcomes are distributed. The outcomes are: I win none of them. I win exactly one of them. I win exactly two of them. And so on, all the way up to “I win exactly all but one of them” and “I win all of them”.
To answer the question of whether it’s plausible I should have done so badly I need to know more than just how likely it is I would win only three games. I need to also know the chance I’d have done worse. If I had won only two games, or only one, or none at all. Why?
Here I admit: I’m not sure I can give a compelling reason, at least not in English. I’ve been reworking it all week without being happy at the results. Let me try pieces.
One part is that as I put the question — is it plausible that I could do so awfully? — isn’t answered just by checking how likely it is I would win only three games out of sixteen. If that’s awful, then doing even worse must also be awful. I can’t rule out even-worse results from awfulness without losing a sense of what the word “awful” means. Fair enough, to answer that question. But I made up the question. Why did I make up that one? Why not just “is it plausible I’d get only three out of sixteen games”?
Habit, largely. Experience shows me that the probability of any particular result turns out to be implausibly low. It isn’t quite that case here; there’s only seventeen possible noticeably different outcomes of playing sixteen games. But there can be so many possible outcomes that even the most likely one isn’t.
Take an extreme case. (Extreme cases are often good ways to build an intuitive understanding of things.) Imagine I played 16,000 games, with a 50-50 chance of winning each one of them. It is most likely that I would win 8,000 of the games. But the probability of winning exactly 8,000 games is small: only about 0.6 percent. What’s going on there is that there’s almost the same chance of winning exactly 8,001 or 8,002 games. As the number of games increases the number of possible different outcomes increases. If there are 16,000 games there are 16,001 possible outcomes. It’s less likely that any of them will stand out. What saves our ability to predict the results of things is that the number of plausible outcomes increases more slowly. It’s plausible someone would win exactly three games out of sixteen. It’s impossible that someone would win exactly three thousand games out of sixteen thousand, even though that’s the same ratio of won games.
Card games offer another way to get comfortable with this idea. A bridge hand, for example, is thirteen cards drawn out of fifty-two. But the chance that you were dealt the hand you just got? Impossibly low. Should we conclude from this all bridge hands are hoaxes? No, but ask my mother sometime about the bridge class she took that one cruise. “Three of sixteen” is too particular; “at best three of sixteen” is a class I can study.
Unconvinced? I don’t blame you. I’m not sure I would be convinced of that, but I might allow the argument to continue. I hope you will. So here are the specifics. These are the chance of each count of wins, and the chance of having exactly that many wins, for sixteen matches:
So the chance of doing as awfully as I had — winning zero or one or two or three games — is pretty dire. It’s a little above one percent.
Is that implausibly low? Is there so small a chance that I’d do so badly that we have to figure I didn’t have a 50-50 chance of winning each game?
I hate to think that. I didn’t think I was outclassed. But here’s a problem. We need some standard for what is “it’s implausibly unlikely that this happened by chance alone”. If there were only one chance in a trillion that someone with a 50-50 chance of winning any game would put in the performance I did, we could suppose that I didn’t actually have a 50-50 chance of winning any game. If there were only one chance in a million of that performance, we might also suppose I didn’t actually have a 50-50 chance of winning any game. But here there was only one chance in a hundred? Is that too unlikely?
It depends. We should have set a threshold for “too implausibly unlikely” before we started research. It’s bad form to decide afterward. There are some thresholds that are commonly taken. Five percent is often useful for stuff where it’s hard to do bigger experiments and the harm of guessing wrong (dismissing the idea I had a 50-50 chance of winning any given game, for example) isn’t so serious. One percent is another common threshold, again common in stuff like psychological studies where it’s hard to get more and more data. In a field like physics, where experiments are relatively cheap to keep running, you can gather enough data to insist on fractions of a percent as your threshold. Setting the threshold after is bad form.
In my defense, I thought (without doing the work) that I probably had something like a five percent chance of doing that badly by luck alone. It suggests that I did have a much worse than 50 percent chance of winning any given game.
Is that credible? Well, yeah; I may have been in the top sixteen players in the state. But a lot of those people are incredibly good. Maybe I had only one chance in three, or something like that. That would make the chance I did that poorly something like one in six, likely enough.
And it’s also plausible that games are not independent, that whether I win one game depends in some way on whether I won or lost the previous. But it does feel like it’s easier to win after a win, or after a close loss. And it feels harder to win a game after a string of losses. I don’t know that this can be proved, not on the meager evidence I have available. And you can almost always question the independence of a string of events like this. It’s the safe bet.
I have one last important thing to discuss before I finish my months spun off an offhand comment from The Price Is Right. There are a couple minor points I can also follow up on, but I don’t think they’re tied tightly enough to the show to deserve explicit mention or rate getting “tv” included as one of my keywords. Here’s my question: what’s the chance of winning an average pricing game, after one has got an Item Up For Bid won?
At first glance this is several dozen questions, since there are quite a few games, some winnable on pure skill — “Clock Game”, particularly, although contestants this season have been rotten at it, and “Hole In One … Or Two”, since a good miniature golfer could beat it — and some that are just never won — “Temptation” particularly — and some for which partial wins are possible — “Money Game” most obviously. For all, skill in pricing things help. For nearly all, there’s an element of luck.
I’m not going to attempt to estimate the chance of winning each of the dozens of pricing games. What I want is some kind of mean chance of winning, based on how contestants actually do. The tool I’ll use for this is the number of perfect episodes, episodes in which the contestant wins all six pricing games, and I’ll leave it to the definers of perfect such questions as what counts as a win for “Pay The Rent” (in which a prize of $100,000 is theoretically possible, but $10,000 is the most that has yet been paid out) or “Plinko” (theoretically paying up to $50,000, but which hasn’t done so in decades of playing).
One week, it seems, isn’t enough to tell the difference conclusively between the first bidder on Contestants Row having a 25 percent chance of winning — winning one out of four times — or a 17 percent chance of winning — winning one out of six times. But we’re not limited to watching just the one week of The Price Is Right, at least in principle. Some more episodes might help us, and we can test how many episodes are needed to be confident that we can tell the difference. I won’t be clever about this. I have a tool — Octave — which makes it very easy to figure out whether it’s plausible for something which happens 1/4 of the time to turn up only 1/6 of the time in a set number of attempts, and I’ll just keep trying larger numbers of attempts until I’m satisfied. Sometimes the easiest way to solve a problem is to keep trying numbers until something works.
In two weeks (or any ten episodes, really, as talked about above), with 60 items up for bids, a 25 percent chance of winning suggests the first bidder should win 15 times. A 17 percent chance of winning would be a touch over 10 wins. The chance of 10 or fewer successes out of 60 attempts, with a 25 percent chance of success each time, is about 8.6 percent, still none too compelling.
Here we might turn to despair: 6,000 episodes — about 35 years of production — weren’t enough to give perfectly unambiguous answers about whether there were fewer clean sweeps than we expected. There were too few at the 5 percent significance level, but not too few at the 1 percent significance level. Do we really expect to do better with only 60 shows?
The obvious thing to do is test: watch a couple episodes, and see whether it’s nearer to 1/6 or to 1/4 of the winning bids come from the first seat. It’s easy to tally the number of items up for bid and how often the first bidder wins. However, there are only six items up for bid each episode, and there are five episodes per week, for 30 trials in all. I talk about a week’s worth of episodes because it’s a convenient unit, easy to record on the Tivo or an equivalent device, easy to watch at The Price Is Right‘s online site, but it doesn’t have to be a single week. It could be any five episodes. But I’ll say a week just because it’s convenient to do so.
If the first seat has a chance of 25 percent of winning, we expect 30 times 1/4, or seven or eight, first-seat wins per week. If the first seat has a 17 percent chance of winning, we expect 30 times 1/6, or 5, first-seat wins per week. That’s not much difference. What’s the chance we see 5 first-seat wins if the first seat has a 25 percent chance of winning?
One assumption for a binomial distribution are that we have some trial, some event, which happens many times. Each episodes is the obvious trial here. The outcome we’re interested in seeing has some probability of happening on each trial; there is indeed some probability of a clean sweep each episode. The binomial distribution assumes that this probability is constant for every trial, that it doesn’t become more or less likely the tenth or hundredth or thousandth time around, and this seems likely to hold for The Price Is Right episodes. Granted there is some chance of a clean sweep in one episode; what could be done to increase or decrease the likelihood from episode to episode?
We have to decide that some outcomes have such a low probability of happening naturally that they represent something going on, and are not just the result of chance. How low that probability should be is our decision. There are some common dividing lines, but they’re common just because they represent numbers which human beings find to be nice round figures: five percent, one percent, half a percent, one-tenth of a percent. What significance level one picks depends on many factors, including what’s common in the field, how different outcomes are expected to be, even what one can afford. Physicists looking for evidence of new subatomic particles have an extremely high standard before declaring something is definitely a new particle, but, they can run particle detection experiments until they get such clear evidence.
To be fair, we ought to pick our significance level before we’ve worked out the probability of something happening, but this is the earliest I could discuss it with motivation for you to read about it. But if we take the five percent significance level, we see we know already that there’s a little more than a one and a half percent chance of there being as few clean sweeps as observed. The conclusion is obvious: all six winning contestants in an episode should have come from the same seat, over 6,000 episodes, more often than the one time Drew Carey claimed they had. We can start looking for explanations for why there should be this deficiency.
We became suspicious of the number of clean sweeps in The Price Is Right when there were not the expected six of them in 6,000 episodes. The chance there would be only one was about one and a half percent, not very high. But are there so few clean sweeps that we should be suspicious? That is, is the difference between the expected number of sweeps and the observed number so large as to be significant? Is it too big to just result from chance?
This is significance testing: is whatever quantity we mean to observe dramatically less than what is expected? Is it dramatically more? Is it at least different? Are these differences bigger than what could be expected by mere chance? For every statistician’s favorite example, a tossed fair coin will come up tails half the time; that means, of twenty flips, there are expected to be ten tails. But there being merely nine or as many as twelve is reasonable. Three or fifteen tails may be a little unlikely. Zero or twenty seem impossible. There’s a point where if our observations are so different from what we expect then we have to reject the idea that our observations and our expectations agree.
It’s not enough to say there’s a probability of only 1.5 percent that there should be exactly one clean sweep episode out of 6,000, though. It’s unlikely that should happen, but if we look at it, it’s unlikely there should be any outcome. Even the most likely result of 6,000 episodes, six clean sweeps, has only about one chance in six of happening. That’s near the chance that the next person you meet will have a birthday in either September or November. That isn’t absurdly unlikely, but, the person betting against it has the surer deal.
We worked out the likelihood that there would be only one clean sweep, with all six contestants getting on stage coming from the same seat in Contestants Row, out of six thousand episodes of The Price Is Right. That turned out to be not terribly likely: it had about a one and a half percent chance of being the case. For a sense of scale, that’s around the same probability that the moment you finish reading this sentence will be exactly 26 seconds past the minute. It’s pretty safe to bet that it wasn’t.
However, it isn’t particularly outlandish to suppose that it was. I’d certainly hope at least some reader found that it was. Events which aren’t particularly likely do happen, all the time. Consider the likelihood of this single-clean-sweep or the 26-seconds-past-the-minute thing happening to the likelihood of any given hand of poker: any specific hand is phenomenally less likely, but something has to happen once you start dealing. So do we have any grounds for saying the particular outcome of one clean sweep in 6,000 shows is improbable? Or for saying that it’s reasonable?
When last we talked about the “clean sweep” of winning contestants coming from the same of four seats in Contestants Row for all six Items Up For Bid on The Price Is Right, we had got established the pieces needed if we suppose this to be a binomial distribution problem. That is, we suppose that any given episode has a probability, p, of successfully having all six contestants from the same seat, and a probability 1 – p of failing to have all six contestants from the same seat. There are N episodes, and we are interested in the chance of x of them being clean sweeps. From the production schedule we know the number of episodes N is about 6,000. We supposed the probability of a clean sweep to be about p = 1/1000, on the assumption that the chance of winning isn’t any better or worse for any contestant. The probability of there not being a clean sweep is then 1 – p = 999/1000. And we expected x = 6 clean sweeps, while Drew Carey claimed there had been only 1.
The chance of finding x successes out of N attempts, according to the binomial distribution, is the probability of any combination of x successes and N – x successes — which is equal to (p)(x) * (1 – p)(N – x) — times the number of ways there are to select x items out of N candidates. Either of those is easy enough to calculate, up to the point where we try calculating it. Let’s start out by supposing x to be the expected 6, and later we’ll look at it being 1 or other numbers.
To find the probability of seeing some number, call it x since we don’t particularly care what it is, of successes out of some larger number, call it N because that’s a convenient number, of trials, we need to figure out how many ways there are to arrange x successes out of N trials. For small x and N values we can figure this out by hand, given time. For large numbers, we’d never finish if we tried by hand. But we can solve it, if we attack the problem methodically.
So, we calculated that on any given episode of The Price Is Right there’s around one chance of all six winners of the Item Up For Bid coming from the same seat. And we know there have been about six thousand episodes with six Items Up For Bid. So we expect there to have been about six clean sweep episodes; yet if Drew Carey is to be believed, there has been just the one. What’s wrong?
Possibly, nothing. Just because there is a certain probability of a thing happening does not mean it happens all that often. Consider an analogous situation: a baseball batter might hit safely one time out of every three at-bats; but there would be nothing particularly odd in the batter going hitless in four at-bats during a single game, however much we would expect him to get at least one. There wouldn’t be much very peculiar in his hitting all four times, either. Our expected value, the number of times something could happen times the probability of it happening each time, is not necessarily what we actually see. (We might get suspicious if we always saw the expected value turn up.)
Still, there must be some limits. We might accept a batter who hits one time out of every three getting no hits in four at-bats. If he got no runs in four hundred at-bats, we’d be inclined to say he’s not a decent hitter having some bad luck. More likely he’s failing to bring the bat with him to the plate. We need a tool to say whether some particular outcome is tolerably likely or so improbable that something must be up.
There are many hard things about teaching, although I appreciate that since I’m in mathematics I have advantages over many other fields. For example, students come in with the assumption that there are certainly right and certainly wrong answers to questions. I’m generally spared the problem of convincing students that I have authority to rule some answers in or out. There’s actually a lot of discretion and judgement and opinion involved, but most of that comes in when one is doing research. In an introductory course, there are some techniques that have gotten so well-established and useful we could fairly well pretend there isn’t any judgement left.
But one hard part is probably common to all fields: how closely to guide a student working out something. This case comes from office hours, as I tried getting a student to work out a problem in binomial distributions. Binomial distributions come up in studying the case where there are many attempts at something; and each attempt has a certain, fixed, chance of succeeding; and you want to know the chance of there being exactly some particular number of successes out of all those tries. For example, imagine rolling four dice, and being interested in getting exactly two 6’s on the four dice.
To work it out, you need the number of attempts, and the number of successes you’re interested in, and the chance of each attempt at something succeeding, and the chance of each attempt failing. For the four-dice problem, each attempt is the rolling of one die; there are four attempts at rolling die; we’re interested in finding two successful rolls of 6; the chance of successfully getting a 6 on any roll is 1/6; and the chance of failure on any one roll is —