## Something To Read: Galton Boards

I do need to take another light week of writing I’m afraid. There’ll be the Theorem Thursday post and all that. But today I’d like to point over to Gaurish4Math’s WordPress Blog, and a discussion of the Galton Board. I’m not familiar with it by that name, but it is a very familiar concept. You see it as Plinko boards on The Price Is Right and as a Boardwalk or amusement-park game. Set an array of pins on a vertical board and drop a ball or a round chip or something that can spin around freely on it. Where will it fall?

It’s random luck, it seems. At least it is incredibly hard to predict where, underneath all the pins, the ball will come to rest. Some of that is ignorance: we just don’t know the weight distribution of the ball, the exact way it’s dropped, the precise spacing of pins well enough to predict it all. We don’t care enough to do that. But some of it is real randomness. Ideally we make the ball bounce so many times that however well we estimated its drop, the tiny discrepancy between where the ball is and where we predict it is, and where it is going and where we predict it is going, will grow larger than the Plinko board and our prediction will be meaningless.

(I am not sure that this literally happens. It is possible, though. It seems more likely the more rows of pins there are on the board. But I don’t know how tall a board really needs to be to be a chaotic system, deterministic but unpredictable.)

But here is the wonder. We cannot predict what any ball will do. But we can predict something about what every ball will do, if we have enormously many of them. Gaurish writes some about the logic of why that is, and the theorems in probability that tell us why that should be so.

• #### gaurish 6:11 pm on Tuesday, 26 July, 2016 Permalink | Reply

Thanks for pointing to my blog post. I would like to quote Tim Gowers (A very short introduction to Mathematics, pp. 6) regarding the classical die throwing experiment (“the model”) of probability theory:
“One might object to this model on the grounds that the dice, when rolled, are obeying Newton’s laws, at least to a very high degree of precision, so the way they land is anything but random…”

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• #### Joseph Nebus 6:02 am on Wednesday, 27 July, 2016 Permalink | Reply

Quite welcome. I’m happy to pass along interesting writing.

Granted that falling dice, or balls in a Plinko board like this, are moving deterministically. I do wonder if we get to chaotic behavior, in which the toss is nevertheless random. I’m not well-versed enough in the mechanics of this sort of problem to be really sure about my answer. For the balls falling off pins I would imagine that something like twenty rebounds, on either pin or other balls, would be enough to effectively randomize the result.

(If each rebound doubles the discrepancy between the direction of the ball’s actual velocity and our representation of its direction, then after twenty rebounds the error is about a million times what it started as, and it seems hard to know the direction of a ball’s travel to within a millionth of two-pi radians. But that’s a very rough argument, supposing that randomizing the direction of travel is all we need to have a random ball drop. And maybe two-pi-over-a-million radians is a reasonable precision; maybe we need thirty rebounds, or forty, to be quite sure.)

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## JH van ‘t Hoff and the Gaseous Theory of Solutions; also, Pricing Games

Do you ever think about why stuff dissolves? Like, why a spoon of sugar in a glass of water should seem to disappear instead of turning into a slight change in the water’s clarity? Well, sure, in those moods when you look at the world as a child does, not accepting that life is just like that and instead can imagine it being otherwise. Take that sort of question and put it to adult inquiry and you get great science.

Peter Mander of the Carnot Cycle blog this month writes a tale about Jacobus Henricus van ‘t Hoff, the first winner of a Nobel Prize for Chemistry. In 1883, on hearing of an interesting experiment with semipermeable membranes, van ‘t Hoff had a brilliant insight about why things go into solution, and how. The insight had only one little problem. It makes for fine reading about the history of chemistry and of its mathematical study.

In other, television-related news, the United States edition of The Price Is Right included a mention of “square root day” yesterday, 4/4/16. It was in the game “Cover-Up”, in which the contestant tries making successively better guesses at the price of a car. This they do by covering up wrong digits with new guesses. For the start of the game, before the contestant’s made any guesses, they need something irrelevant to the game to be on the board. So, they put up mock calendar pages for 1/1/2001, 2/2/2004, 3/3/2009, 4/4/2016, and finally a card reading $\sqrt{DAY}$. The game show also had a round devoted to Pi Day a few weeks back. So I suppose they’re trying to reach out to people into pop mathematics. It’s cute.

• #### Marta Frant 5:27 am on Thursday, 7 April, 2016 Permalink | Reply

Questions, questions, questions… The constant ‘why’ is what makes the world go around.

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• #### Joseph Nebus 2:07 am on Saturday, 9 April, 2016 Permalink | Reply

‘Why’ is indeed one of the big questions. ‘What’ and ‘The Heck?’ are also pretty important.

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## Reading the Comics, March 14, 2016: Pi Day Comics Event

Comic Strip Master Command had the regular pace of mathematically-themed comic strips the last few days. But it remembered what the 14th would be. You’ll see that when we get there.

Ray Billingsley’s Curtis for the 11th of March is a student-resists-the-word-problem joke. But it’s a more interesting word problem than usual. It’s your classic problem of two trains meeting, but rather than ask when they’ll meet it asks where. It’s just an extra little step once the time of meeting is made, but that’s all right by me. Anything to freshen the scenario up.

Ray Billingsley’s Curtis for the 11th of March, 2016. I am curious what the path of the rail line is.

Tony Carrillo’s F Minus for the 11th was apparently our Venn Diagram joke for the week. I’m amused.

Mason Mastroianni, Mick Mastroianni, and Perri Hart’s B.C. for the 12th of March name-drops statisticians. Statisticians are almost expected to produce interesting pictures of their results. It is the field that gave us bar charts, pie charts, scatter plots, and many more. Statistics is, in part, about understanding a complicated set of data with a few numbers. It’s also about turning those numbers into recognizable pictures, all in the hope of finding meaning in a confusing world (ours).

Brian Anderson’s Dog Eat Doug for the 13th of March uses walls full of mathematical scrawl as signifier for “stuff thought deeply about’. I don’t recognize any of the symbols specifically, although some of them look plausibly like calculus. I would not be surprised if Anderson had copied equations from a book on string theory. I’d do it to tell this joke.

And then came the 14th of March. That gave us a bounty of Pi Day comics. Among them:

John Hambrock’s The Brilliant Mind of Edison Lee for the 14th of March, 2016. The strip is like this a lot.

John Hambrock’s The Brilliant Mind of Edison Lee trusts that the name of the day is wordplay enough.

Scott Hilburn’s The Argyle Sweater is also a wordplay joke, although it’s a bit more advanced.

Tim Rickard’s Brewster Rockit fuses the pun with one of its running, or at least rolling, gags.

Bill Whitehead’s Free Range makes an urban legend out of the obsessive calculation of digits of π.

And Missy Meyer’s informational panel cartoon Holiday Doodles mentions that besides “National” Pi Day it was also “National” Potato Chip Day, “National” Children’s Craft Day, and “International” Ask A Question Day. My question: for the first three days, which nation?

Edited To Add: And I forgot to mention, after noting to myself that I ought to mention it. The Price Is Right (the United States edition) hopped onto the Pi Day fuss. It used the day as a thematic link for its Showcase prize packages, noting how you could work out π from the circumference of your new bicycles, or how π was a letter from your vacation destination of Greece, and if you think there weren’t brand-new cars in both Showcases you don’t know the game show well. Did anyone learn anything mathematical from this? I am skeptical. Do people come away thinking mathematics is more fun after this? … Conceivably. At least it was a day fairly free of people declaring they Hate Math and Can Never Do It.

• #### ivasallay 11:18 pm on Wednesday, 16 March, 2016 Permalink | Reply

I loved the pi piper. Thanks for bringing these to my attention.

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• #### Joseph Nebus 2:39 am on Thursday, 24 March, 2016 Permalink | Reply

Very happy to be of help. Thank you.

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## November 2013’s Statistics

Hi again. I was hesitant to look at this month’s statistics, as I pretty much fell off the face of the earth for a week there, but I didn’t have the chance to do the serious thinking that’s needed for mathematics writing. The result’s almost exactly the dropoff in readership I might have predicted: from 440 views in October down to 308, and from 220 unique visitors down to 158. That’s almost an unchanged number of views per visitor, 2.00 dropping to 1.95, so at least the people still interested in me are sticking around.

The countries sending me the most viewers were as ever the United States, then Austria (hi, Elke, and thank you), the United Kingdom and then Canada. Sending me a single visitor each were Bulgaria, Cyprus, Czech Republic, Ethiopia, France, Jordan, Lebanon, Nepal, New Zealand, Russia, Singapore, Slovenia, Switzerland, and Thailand. This is also a drop in the number of single-viewer countries, although stalwarts Finland and the Netherlands are off the list. Slovenia’s the only country making a repeat appearance from last month, in fact.

The most popular articles the past month were:

And I apologize for not having produced many essays the past couple weeks, and can only fault myself for being more fascinated by some problems in my day job that’ve been taking up time and mental energy and waking me in the middle of the night with stuff I should try. I’ll be back to normal soon, I’m sure. Don’t tell my boss.

• #### elkement 6:54 pm on Monday, 2 December, 2013 Permalink | Reply

Thanks! As I said, I am playing with anti-spying software but I have excluded your post :-)
I see you have snow on your blog, too! It’s barely visible because of the white background – but it is snowing on your trapezoids.

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• #### Joseph Nebus 4:43 am on Wednesday, 4 December, 2013 Permalink | Reply

And I’d forgot the snow! It’s one of the lesser display options in WordPress’s menu somewhere; I just remember they turned it on one year. I was surprised not to see more people complaining about it.

It’d be really slick to have snow accumulate on the edges of pictures or the like, but that’s probably impossible to do in any reasonable way.

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## The Mathematics Of A Pricing Game

There was a new pricing game that debuted on The Price Is Right for the start of its 42nd season, with a name that’s designed to get my attention: it’s called “Do The Math”. This seems like a dangerous thing to challenge contestants to do since the evidence is that pricing games which depend on doing some arithmetic tend to be challenging (“Grocery Game”, “Bullseye”), or confusing (“The Check Game”), or outright disasters (“Add Em Up”). This one looks likely to be more successful, though.

The setup is this: The contestant is shown two prizes. In the first (and, so far, only) playing of the game this was a 3-D HDTV and a motorcycle. The names of those prizes are put on either side of a monitor made up to look like a green chalkboard. The difference in prize values is shown; in this case, it was $1160, and that’s drawn in the middle of the monitor in Schoolboard Extra-Large font. The contestant has to answer whether the price of the prize listed on the left (here, the 3-D HDTV) plus the cash ($1160) is the price of the prize on the right (the motorcycle), or whether the price of the prize on the left minus the cash is the price of the prize on the right. The contestant makes her or his guess and, if right, wins both prizes and the money.

There’s not really much mathematics involved here. The game is really just a two-prize version of “Most Expensive” (in which the contestant has to say which of three prizes and then it’s right there on the label). I think there’s maybe a bit of educational value in it, though, in that by representing the prices of the two prizes — which are fixed quantities, at least for the duration of taping, and may or may not be known to the contestant — with abstractions it might make people more comfortable with the mathematical use of symbols. x and all the other letters of the English (and Greek) alphabets get called into place to represent quantities that might be fixed, or might not be; and that might be known, or might be unknown; and that we might actually wish to know or might not really care about but need to reference somehow.

That conceptual leap often confuses people, as see any joke about how high school algebra teachers can’t come up with a consistent answer about what x is. This pricing game is a bit away from mathematics classes, but it might yet be a way people could see that the abstraction idea is not as abstract or complicated as they fear.

I suspect, getting away from my flimsy mathematics link, that this should be a successful pricing game, since it looks to be quick and probably not too difficult for players to get. I’m sorry the producers went with a computer monitor for the game’s props, rather than — say — having a model actually write plus or minus, or some other physical prop. Computer screens are boring television; real objects that move are interesting. There are some engagingly apocalyptic reviews of the season premiere over at golden-road.net, a great fan site for The Price Is Right.

• #### elkement 5:38 pm on Wednesday, 9 October, 2013 Permalink | Reply

I have read an interesting book on our abilities to assess simple math – called The Science of Fear (as concerned with the assessment of risk and probabilities). I would not be surprised if the simple calculation involved in this show would make a difference. In this books psychological evidence was given that, for example, 3 of 100 ias perceived intuitively in a different way than 3% although we know these are the same.

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• #### Joseph Nebus 2:53 am on Friday, 18 October, 2013 Permalink | Reply

I’m interested and glad to see that book’s in the library. I figure to borrow it next time I’m there.

I’ve been fascinated informally with how framing a problem makes people more or less likely to solve it ever since long ago I noted that nobody in my family had any problem setting the VCR to record stuff, and someone else pointed out that we talked about “setting” the machine instead of “programming” it to record, the way most people did. Whether our general ability followed from thinking of it as an easy thing to do or vice-versa I couldn’t answer but the correlation interested me.

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• #### elkement 7:23 am on Friday, 18 October, 2013 Permalink | Reply

Thanks for sharing this anecdote. So probably if you consider ‘programming’ as something very interesting (as I do) you might intimidate readers by using such ‘geeky’ / ‘technical’ language although just wanted to share your enthusiasm. I will try to take that into account when writing about something abstract the next time.

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• #### Joseph Nebus 11:30 pm on Tuesday, 22 October, 2013 Permalink | Reply

I don’t actually know there’s a connection, but it does feel intuitively like there’s probably a link between terms that sound like jargon and people feeling they can’t follow it. I suppose that’s similar to the book-publishing lore that every equation cuts book sales in half. That lore seems far too pat to be literally true but it does seem qualitatively to be on to something.

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## My June 2013 Statistics

I don’t understand why, but an awful lot of the advice I see about blogging says that it’s important not just to keep track of how your blog is doing, but also to share it, so that … numbers will like you more? I don’t know. But I can give it a try, anyway.

For June 2013, according to WordPress, I had some 713 page views, out of 246 unique visitors. That’s the second-highest number of page views I’ve had in any month this year (January had 831 views), and the third-highest I’ve had for all time (there were 790 in March 2012). The number of unique visitors isn’t so impressive; since WordPress started giving me that information in December 2012, I’ve had more unique visitors … actually, in every month but May 2013. On the other hand, the pages-per-viewer count of 2.90 is the best I’ve had; the implication seems to be that I’m engaging my audience.

The most popular posts for the past month were Counting From 52 to 11,108, which I believe reflects it getting picked for a class assignment somehow; A Cedar Point Follow-Up, which hasn’t got much mathematics in it but has got pretty pictures of an amusement park, and Solving The Price Is Right’s “Any Number” Game, which has got some original mathematics but also a pretty picture.

My all-time most popular posts are from the series about Trapezoids — working out how to find their area, and how many kinds of trapezoids there are — with such catchy titles as How Many Trapezoids I Can Draw, or How Do You Make A Trapezoid Right?, or Setting Out To Trap A Zoid, which should be recognized as a Dave Barry reference.

My most frequent commenters, “recent”, whatever that means, are Chiaroscuro and BunnyHugger (virtually tied), with fluffy, elkelement, MJ Howard, and Geoffrey Brent rounding out the top six.

The most common source of page clicks the past month was from the United States (468), with Brazil (51) and Canada (23) taking silver and bronze. And WordPress recorded one click each from Portugal, Serbia, Hungary, Macedonia (the Former Yugoslav Republic), Indonesia, Argentina, Poland, Slovenia, and Viet Nam. I’ve been to just one of those countries.

## Solving The Price Is Right’s “Any Number” Game

A friend who’s also into The Price Is Right claimed to have noticed something peculiar about the “Any Number” game. Let me give context before the peculiarity.

This pricing game is the show’s oldest — it was actually the first one played when the current series began in 1972, and also the first pricing game won — and it’s got a wonderful simplicity: four digits from the price of a car (the first digit, nearly invariably a 1 or a 2, is given to the contestant and not part of the game), three digits from the price of a decent but mid-range prize, and three digits from a “piggy bank” worth up to $9.87 are concealed. The contestant guesses digits from zero through nine inclusive, and they’re revealed in the three prices. The contestant wins whichever prize has its price fully revealed first. This is a steadily popular game, and one of the rare Price games which guarantees the contestant wins something. A couple things probably stand out. The first is that if you’re very lucky (or unlucky) you can win with as few as three digits called, although it might be the piggy bank for a measly twelve cents. (Past producers have said they’d never let the piggy bank hold less than$1.02, which still qualifies as “technically something”.) The other is that no matter how bad you are, you can’t take more than eight digits to win something, though it might still be the piggy bank.

What my friend claimed to notice was that these “Any Number” games went on to the last possible digit “all the time”, and he wanted to know, why?

My first reaction was: “all” the time? Well, at least it happened an awful lot of the time. But I couldn’t think of a particular reason that they should so often take the full eight digits needed, or whether they actually did; it’s extremely easy to fool yourself about how often events happen when there’s a complicated possibile set of events. But stipulating that eight digits were often needed, then, why should they be needed? (For that matter, trusting the game not to be rigged — and United States televised game shows are by legend extremely sensitive to charges of rigging — how could they be needed?) Could I explain why this happened? And he asked again, enough times that I got curious myself.

## About Chances of Winning on The Price Is Right

Putting together links to all my essays about trapezoid areas made me realize I also had a string of articles examining that problem of The Price Is Right, with Drew Carey’s claim that only once in the show’s history had all six contestants winning the Item Up For Bids come from the same seat in Contestants’ Row. As with the trapezoid pieces they form a more or less coherent whole, so, let me make it easy for people searching the web for the likelihood of clean sweeps or of perfect games on The Price Is Right to find my thoughts.

• #### Joe Fix It 2:21 pm on Saturday, 21 April, 2012 Permalink | Reply

Did you watch all 6000 episodes yet???

Ya gotta put this stuff into an e-book!!!

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• #### Joseph Nebus 6:47 pm on Saturday, 21 April, 2012 Permalink | Reply

Not even nearly six thousand episodes. I’d be surprised if I’ve managed two thousand episodes lifetime, and that’d be impossible without the modern conveniences of Tivo and online streaming.

I have thought about e-books, but certainly haven’t got enough material for it yet.

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## 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). • #### Joseph Nebus 6:05 pm on Sunday, 19 February, 2012 Permalink | Reply Tags: binomial distribution ( 11 ), clean sweeps ( 7 ), expectation value ( 9 ), game show ( 13 ), Matlab ( 8 ), Octave ( 5 ), probability ( 73 ), significance level ( 3 ), significance testing ( 5 ), television ( 7 ), The Price Is Right, tv ( 14 ) ## 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 Like • #### 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. Like • #### Joseph Nebus 4:16 am on Thursday, 16 February, 2012 Permalink | Reply Tags: binomial distribution ( 11 ), clean sweeps ( 7 ), Drew Carey ( 10 ), expectation value ( 9 ), game show ( 13 ), Matlab ( 8 ), Octave ( 5 ), probability ( 73 ), significance testing ( 5 ), television ( 7 ), The Price Is Right, tv ( 14 ) ## 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? • #### Joseph Nebus 8:24 pm on Sunday, 12 February, 2012 Permalink | Reply Tags: ansatz ( 2 ), assumption, assumptions ( 3 ), clean sweeps ( 7 ), Drew Carey ( 10 ), expectation value ( 9 ), game show ( 13 ), penalty, probability ( 73 ), television ( 7 ), The Price Is Right, tv ( 14 ) ## Figuring Out The Penalty Of Going First Let’s accept the conclusion that the small number of clean sweeps of Contestants Row is statistically significant, that all six winning contestants on a single episode of The Price Is Right come from the same seat less often than we would expect from chance alone, and that the reason for this is that whichever seat won the last item up for bids is less likely to win the next. It seems natural to suppose the seat which won last time — and which is therefore bidding first this next time — is at a disadvantage. The irresistible question, to me anyway, is: how big is that disadvantage? If no seats had any advantage, the first, second, third, and fourth bidders would be expected to have a probability of 1/4 of winning any particular item. How much less a chance does the first bidder need to have to get the one clean sweep in 6,000 episodes reported? Chiaroscuro came to an estimate that the first bidder had a probability of about 17.6 percent of winning the item up for bids, and I agree with that, at least if we make a couple of assumptions which I’m confident we are making together. But it’s worth saying what those assumptions are because if the assumptions do not hold, the answers come out different. The first assumption was made explicitly in the first paragraph here: that the low number of clean sweeps is because the chance of a clean sweep is less than the 1 in 1000 (or to be exact, 1 in 1024) chance which supposes every seat has an equal probability of winning. After all, the probability that we saw so few clean sweeps for chance alone was only a bit under two percent; that’s unlikely but hardly unthinkable. We’re supposing there is something to explain. • #### Chiaroscuro 5:40 am on Monday, 13 February, 2012 Permalink | Reply A much nicer explanation of the sort of thing I just did with a fair amount of the [1/x] button in the windows XP calculator and some messing around. Indeed, it’s some very rough assumptions made; but we’ve got to start somewhere, and this is a good place to start. Like • #### nebusresearch 4:42 am on Wednesday, 15 February, 2012 Permalink | Reply Oh, ew, you worked it out from trying out different percentages until you found one that matched? Actually, that’s a respectable numerical-solution technique, called “regula falsi”, that I should probably explain since it’s powerful, simple, and works. I’ll make a note of that. Like • #### Chiaroscuro 4:22 am on Thursday, 16 February, 2012 Permalink | Reply Oh goodness no. I *estimated* a few times to get to the proper neighborhood, then figured a way to reverse what I was doing. “So 23% yields… and how about 20%.. hmmm. lower. How about 16.6%.. too low. okay, then we do this in reverse and start with 1/6000…” It is a wonderful method for ‘ballpark figures’, quite true. Like • #### Joseph Nebus 7:33 am on Monday, 20 February, 2012 Permalink | Reply Ah, OK, I follow now. I think I can tie this in to something I’d wanted to talk about anyway, too, so I appreciate the hook. Like • #### Joseph Nebus 3:05 am on Wednesday, 8 February, 2012 Permalink | Reply Tags: binomial distribution ( 11 ), Bob Barker ( 2 ), Chuck Woolery, clean sweeps ( 7 ), Drew Carey ( 10 ), probability ( 73 ), significance testing ( 5 ), television ( 7 ), The Price Is Right, tv ( 14 ), wheel of fortune ## 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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## Finding, and Starting to Understand, the Answer

If the probability of having one or fewer clean sweep episodes of The Price Is Right out of 6,000 aired shows is a little over one and a half percent — and it is — and we consider outcomes whose probability is less than five percent to be so unlikely that we can rule them out as happening by chance — and, last time, we did — then there are improbably few episodes where all six contestants came from the same seat in Contestants Row, and we can usefully start looking for possible explanations as to why there are so few clean sweeps. At least, that’s the conclusion at our significance level, that five percent.

But there’s no law dictating that we pick that five percent significance level. If we picked a one percent significance level, which is still common enough and not too stringent, then we would say this might be fewer clean sweeps than we expected, but it isn’t so drastically few as to raise our eyebrows yet. And we would be correct to do so. Depending on the significance level, what we saw is either so few clean sweeps as to be suspicious, or it’s not. This is why it’s better form to choose the significance level before we know the outcome; it feels like drawing the bullseye after shooting the arrow the other way around.

• #### BunnyHugger 4:38 am on Sunday, 5 February, 2012 Permalink | Reply

Good explanation.

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• #### nebusresearch 4:46 am on Monday, 6 February, 2012 Permalink | Reply

Aw, thank you kindly.

Don’t worry. I can drag it out to the point nobody cares what I was explaining anymore.

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• #### snidelytoo 8:23 pm on Sunday, 18 March, 2012 Permalink | Reply

“declaring as the result of something suspicious what is actually only chance” … that’s not as transparent a phrase as I’d like to see in its position. Hmmm, “declaring something to be a suspicious result when it is actually only chance”? I’m not sure I’ve come up with anything better, but I hope you can see my problem.

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• #### Joseph Nebus 1:00 am on Tuesday, 20 March, 2012 Permalink | Reply

I agree with your assessment, that it’s not a very clear phrase. Unfortunately it’s hard to get the concept just right.

The classic example is probably this: imagine flipping a fair coin over and over and over and over, and it comes up tails every single time, from now to the end of time. It’s impossible that this should happen if the coin is fair and the toss isn’t rigged in any way. Yet, it’s really not impossible, since there’s no reason a fairly flipped coin shouldn’t come up tails, and that it’s come up tails the last million times doesn’t mean it shouldn’t come up tails again.

It can be put reasonably precisely in the form of theorems and proofs, but that just changes the kind of difficult-to-read statements that one looks at.

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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.

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