## Reading the Comics, April 24, 2017: Reruns Edition

I went a little wild explaining the first of last week’s mathematically-themed comic strips. So let me split the week between the strips that I know to have been reruns and the ones I’m not so sure were.

Bill Amend’s FoxTrot for the 23rd — not a rerun; the strip is still new on Sundays — is a probability question. And a joke about story problems with relevance. Anyway, the question uses the binomial distribution. I know that because the question is about doing a bunch of things, homework questions, each of which can turn out one of two ways, right or wrong. It’s supposed to be equally likely to get the question right or wrong. It’s a little tedious but not hard to work out the chance of getting exactly six problems right, or exactly seven, or exactly eight, or so on. To work out the chance of getting six or more questions right — the problem given — there’s two ways to go about it.

One is the conceptually easy but tedious way. Work out the chance of getting exactly six questions right. Work out the chance of getting exactly seven questions right. Exactly eight questions. Exactly nine. All ten. Add these chances up. You’ll get to a number slightly below 0.377. That is, Mary Lou would have just under a 37.7 percent chance of passing. The answer’s right and it’s easy to understand how it’s right. The only drawback is it’s a lot of calculating to get there.

So here’s the conceptually harder but faster way. It works because the problem says Mary Lou is as likely to get a problem wrong as right. So she’s as likely to get exactly ten questions right as exactly ten wrong. And as likely to get at least nine questions right as at least nine wrong. To get at least eight questions right as at least eight wrong. You see where this is going: she’s as likely to get at least six right as to get at least six wrong.

There’s exactly three possibilities for a ten-question assignment like this. She can get four or fewer questions right (six or more wrong). She can get exactly five questions right. She can get six or more questions right. The chance of the first case and the chance of the last have to be the same.

So, take 1 — the chance that one of the three possibilities will happen — and subtract the chance she gets exactly five problems right, which is a touch over 24.6 percent. So there’s just under a 75.4 percent chance she does not get exactly five questions right. It’s equally likely to be four or fewer, or six or more. Just-under-75.4 divided by two is just under 37.7 percent, which is the chance she’ll pass as the problem’s given. It’s trickier to see why that’s right, but it’s a lot less calculating to do. That’s a common trade-off.

Ruben Bolling’s Super-Fun-Pax Comix rerun for the 23rd is an aptly titled installment of A Million Monkeys At A Million Typewriters. It reminds me that I don’t remember if I’d retired the monkeys-at-typewriters motif from Reading the Comics collections. If I haven’t I probably should, at least after making a proper essay explaining what the monkeys-at-typewriters thing is all about.

Ted Shearer’s Quincy from the 28th of February, 1978. So, that FoxTrot problem I did? The conceptually-easy-but-tedious way is not too hard to do if you have a calculator. It’s a buch of typing but nothing more. If you don’t have a calculator, though, the desire not to do a whole bunch of calculating could drive you to the conceptually-harder-but-less-work answer. Is that a good thing? I suppose; insight is a good thing to bring. But the less-work answer only works because of a quirk in the problem, that Mary Lou is supposed to have a 50 percent chance of getting a question right. The low-insight-but-tedious problem will aways work. Why skip on having something to do the tedious part?

Ted Shearer’s Quincy from the 28th of February, 1978 reveals to me that pocket calculators were a thing much earlier than I realized. Well, I was too young to be allowed near stuff like that in 1978. I don’t think my parents got their first credit-card-sized, solar-powered calculator that kind of worked for another couple years after that. Kids, ask about them. They looked like good ideas, but you could use them for maybe five minutes before the things came apart. Your cell phone is so much better.

Bil Watterson’s Calvin and Hobbes rerun for the 24th can be classed as a resisting-the-word-problem joke. It’s so not about that, but who am I to slow you down from reading a Calvin and Hobbes story?

Garry Trudeau’s Doonesbury rerun for the 24th started a story about high school kids and their bad geography skills. I rate it as qualifying for inclusion here because it’s a mathematics teacher deciding to include more geography in his course. I was amused by the week’s jokes anyway. There’s no hint given what mathematics Gil teaches, but given the links between geometry, navigation, and geography there is surely something that could be relevant. It might not help with geographic points like which states are in New England and where they are, though.

Zach Weinersmith’s Saturday Morning Breakfast Cereal for the 24th is built on a plot point from Carl Sagan’s science fiction novel Contact. In it, a particular “message” is found in the digits of π. (By “message” I mean a string of digits that are interesting to us. I’m not sure that you can properly call something a message if it hasn’t got any sender and if there’s not obviously some intended receiver.) In the book this is an astounding thing because the message can’t be; any reasonable explanation for how it should be there is impossible. But short “messages” are going to turn up in π also, as per the comic strips.

I assume the peer review would correct the cartoon mathematicians’ unfortunate spelling of understanding.

## How Much Might I Have Lost At Pinball?

After the state pinball championship last month there was a second, side tournament. It was a sort-of marathon event in which I played sixteen games in short order. I won three of them and lost thirteen, a disheartening record. The question I can draw from this: was I hopelessly outclassed in the side tournament? Is it plausible that I could do so awfully?

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:

Wins Percentage
0 0.002 %
1 0.024 %
2 0.183 %
3 0.854 %
4 2.777 %
5 6.665 %
6 12.219 %
7 17.456 %
8 19.638 %
9 17.456 %
10 12.219 %
11 6.665 %
12 2.777 %
13 0.854 %
14 0.183 %
15 0.024 %
16 0.002 %

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.

## So If You Can’t Win The Clock Game You Should Feel Bad

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).

## Proving Something With One Month’s Counting

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?

• #### Chiaroscuro 6:48 am on Monday, 20 February, 2012 Permalink | Reply

Impressive how looking at this smaller occurence (By bid rather than by show) yields such a more ready result than considering all the shows

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• #### Joseph Nebus 7:23 am on Monday, 20 February, 2012 Permalink | Reply

It’s a neat effect. It comes about from looking at something, the first bidder winning, that’s just so enormously likely to happen compared to the clean-sweep, though. One or two successes, more or less, doesn’t substantially change the fraction of wins for the first seat out of these 120 items up for bid. One or two successes, more or less, would be a big change in the number of clean-sweep episodes out of 6000.

So, roughly, it’s easier to tell the difference between something happening and just luck when it’s easy to get a lot of examples of the thing happening.

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## What Can One Week Prove?

We have some reason to think the chance of winning an Item Up For Bids, if you’re the first one of the four to place bids — let’s call this the first bidder or first seat so there’s a name for it — is lower than the 25 percent which we’d expect if every contestant in The Price Is Right‘s Contestants Row had an equal shot at it. Based on the assertion that only one time in about six thousand episodes had all six winning bids in one episode come from the same seat, we reasoned that the chance for the first bidder — the same seat as won the previous bid — could be around 17 percent. My next question is how we could test this? The chance for the first bidder to win might be higher than 17 percent — around 1/6, which is near enough and easier to work with — or lower than 25 percent — exactly 1/4 — or conceivably even be outside that range.

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?

## Interpreting Drew Carey

If we’ve decided that at the significance level we find comfortable there are too few clean sweeps of any position in Contestants Row, the natural question is why there are so few. We estimated there should have been six clean sweeps, based on modelling clean-sweep occurrences as a binomial distribution. Something in the model went wrong. Let’s try to reason out what it was.

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?

• #### Chiaroscuro 5:56 am on Wednesday, 8 February, 2012 Permalink | Reply

And I didn’t want to interrupt.

Of course, we might estimate how much more likely the advantage given to the last bidder is- let’s say it’s roughly 31% over time (Figuring it could raise to a rough 1/3, but there’s exact bids, and \$1 bids which are sometimes too low, which might lower it from that). Making the big assumption that the other three bidders have a roughly equal chance as well (Which is also a big assumption, given that the second bidder knows the first bid, the third knows the first and second, but it’s pretty good for me) Then we’ve got the last seat at 31%, and the other three seats at 23%.. From THERE we’d have to solve for the odds of a clean sweep in one show, and see where that lies, and I suspect that’s enough to throw the odds just towards harder enough to make it a bit more towards 1 in 6,000.

But I’m not going to do that math, offhand.

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• #### nebusresearch 2:54 am on Thursday, 9 February, 2012 Permalink | Reply

I’m not sure that, with the given information, there’s really a way to say how much of an advantage the last bidder gets over the ones who go first. We can come up with decent reasons to think it’s one thing or another, but I think the limits of calculation on this data, the one clean sweep in 6,000 shows, are estimating how big a disadvantage the person going first has.

Of course, someone tracking every episode to see which number bidder — first, second, third or fourth — as opposed to which seat could probably make a pretty good estimate within a couple of months.

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• #### Chiaroscuro 6:08 am on Thursday, 9 February, 2012 Permalink | Reply

A very rough calculation would lead to one in six thousand being the expected result of each round, the person winning being the first bidder is 17.6%. (A 20% chance would yield one in 3,125.) You are correct though in that it does not matter what breaks a clean sweep for this- if the second, third, or fourth bidder wins the item up for bid, that breaks the clean sweep. It doesn’t matter which has the higher advantages, just how low the first bidder’s is.

(To note, my original estimate of 23% put the odds at one in 1,555.)

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• #### nebusresearch 12:26 am on Sunday, 12 February, 2012 Permalink | Reply

You’ve got just the right calculation, yes, and I make it out to be the same estimate.

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• #### BunnyHugger 8:47 pm on Wednesday, 8 February, 2012 Permalink | Reply

I miss when they used to shop for prizes. Everyone remembers the dalmatian statue that was almost always in the prize showcase, but there was an end table shaped like an elephant that recurred often and which I liked as a kid.

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• #### nebusresearch 2:57 am on Thursday, 9 February, 2012 Permalink | Reply

I miss the shopping for prizes too, even if they did bring the flow of gameplay to a stop. (I suppose I might search YouTube for ancient episodes to see what I think of the pacing now.) Between all the special spots on the wheel and the toss-up puzzles it can be hard spotting the clean lines of the original game underneath all the cruft now.

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## The Significance of the Item Up For Bids

The last important idea missing before we can judge this problem about The Price Is Right clean sweeps of Contestants Row is the significance level. Whenever an experiment is run — whether it’s the classic probability class problems of flipping coins or rolling dice, or whether it’s watching 6,000 episodes of a game show to see whether any seat produces the most winners, or whether it’s counting the number of red traffic lights one gets during the commute — there are some outcomes which are reasonably likely, some which are unlikely, and some which are vanishingly improbable.

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.

Or …

## The First Tail

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.

## Significance Intrudes on Contestants Row

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?

## A Simple Demonstration Which Does Not Clarify

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.

## Off By A Factor Of 720 (Or More)

To work out the task of figuring out whether it was plausible that there had been only one “clean sweep”, of all six contestants winning the Item Up For Bid on The Price Is Right coming from the same seat, we had started a little into the binomial distribution. The key ideas included that we have “Bernoulli trials”, a number of independent chances for some condition to happen — in this case, we had about 6,000 such trials, the number of hourlong episodes of The Price Is Right — and a probability p of successfully seeing some event occur on any one episode. We worked that out to be somewhere about p = 1/1000, if every seat is equally likely to win every time. There is also a probability of 1 – p or 999/1000 of the event failing to see this event, that is, that one or more contestants comes from a different seat.

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.

• #### fluffy 8:10 pm on Friday, 13 January, 2012 Permalink | Reply

Just out of curiosity, how can one expect any sort of predictable probabilistic distribution on a problem where by nature of the action every choice made affects the future choices?

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• #### nebusresearch 3:23 am on Saturday, 14 January, 2012 Permalink | Reply

Probabilistic distributions don’t require that outcomes be independent of earlier ones; for example, you could model the likelihood that a particular word is being typed in, with the outcome depending on what’s been typed already (whether just by what letters were in already, or by what could possibly make grammatical or semantic sense in the sentence).

But that’s more work than I want to do right here. I’m working as far as I can with the assumption that who wins the Item Up For Bid is independent of who won the last time, and seeing if that assumption forces me to accept something absurd.

After all, while who gets the first and who gets the final bid is dependent on the previous winner, the first person to have a perfect bid can win anytime, whatever position she or he’s in. It looks like the final bidder probably has an advantage, in being able to do a dollar-over the best-looking bid or bidding a dollar if they all seem high, but just because it looks like there must be an advantage doesn’t mean there necessarily is. Part of the results of this little investigation should be learning, albeit indirectly, whether there is an advantage in any of the bidding spots.

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• #### fluffy 3:35 am on Saturday, 14 January, 2012 Permalink | Reply

I have a sneaking suspicion that this is both 1) beyond my feeble capabilities as a statistics dilettante and 2) something that could possibly be modeled fairly well using Markov chains (but see 1).

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• #### fluffy 5:01 am on Saturday, 14 January, 2012 Permalink | Reply

also I guess I wasn’t quite clear but I didn’t mean that probabilities can’t depend on previous cases (because obviously they can, I mean Markov and Bayes theory are based entirely on that) but that the bidding on items is a strategy and not a purely stochastic process. It seems like the fourth player to go has the best chance of winning because they have the other peoples’ guesses to build on, and none of the guesses are particularly random to begin with (despite sometimes seeming such for people who are well-versed in the product areas that they are bidding on).

I also wonder what percentage of fourth players would have won by simply doing a +1 bid on a previous bid (and which previous bid is best to increment on).

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• #### nebusresearch 4:05 am on Monday, 16 January, 2012 Permalink | Reply

Well, bidding is obviously a case where strategy, or at least outside information, helps. If you go in with a good idea of the prices of typical prizes, or if you figure out who in the audience knows — and there was the perfect bid on the Showcase, where the constant heard the perfect bid from the guy in the audience who had memorized the prices of all the prizes that kept coming up — and listen to them, you can assuredly win.

However, in practice, contestants usually have only a vague idea what the prizes should cost. I’m not sure their bidding practices are appreciably better than random guesses. Another article or two on this thread, though, and we should reach a point where we can infer whether we can treat their bids as random events.

I’m curious too how many contestants have won, or would win, with dollar-over bids, but I haven’t got such records. I remember in the glory days of alt.tv.game-shows some people encapsulating episodes with that information, but it would take a Google-like service that worked on Usenet to find them.

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## From Drew Carey To An Imaginary Baseball Player

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.

## Ted Baxter and the Binomial Distribution

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 —

• #### Chiaroscuro 7:44 am on Saturday, 10 December, 2011 Permalink | Reply

So the chance here of one die failing to come up heads is 1 – 1/6 or 5/6

I think the chance of most dice failing to come up heads is quite a bit higher.

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• #### nebusresearch 4:29 pm on Saturday, 10 December, 2011 Permalink | Reply

Gr. Yeah.

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• #### Joe Fix It 2:18 am on Monday, 12 December, 2011 Permalink | Reply

Ted would have liked it

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• #### nebusresearch 11:09 am on Monday, 12 December, 2011 Permalink | Reply

Liking is one thing. Calculating right another. When they can get together at once, I’m reasonably content.

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