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.

Six thousand is a lot to deal with. That’s usually a sign of trouble. It’s typically easier to work with, and much easier to imagine, smaller problems. So instead of picking a couple cases out of 6,000 episodes, let’s imagine picking a couple out of *N = 6* episodes instead. How many ways are there to pick one episode out of six possible choices? That’s obvious, though, and we can answer without thinking: there are six possible ways. Even better than being obvious, the answer is right even when we slow down and think about why it should be that. We can pick the first, or the second, or the third, or the fourth, or the fifth, or the sixth, and there are no other ways to pick no fewer and no more than one out of the six.

Now, how about picking exactly two out of the six? This we can understand as picking any one of the six to start with — there’s six choices there — and then picking any one of the five remaining — any of the six, except the already selected one. For example, we could pick the first and second; or the first and third; or the first and fourth; or the first and fifth; or the first and sixth. Or the second and first, or the second and third, or the second and fourth, and so on. We had six choices for the first episode we pick, and five choices for the second episode we pick, and since we can pick any combination that gives us six times five combinations: 30 possibilities.

… Well. Maybe we have thirty. We have to think about that a bit again, but I’ll get to think about what after this.

Suppose we want to pick exactly three episodes out of the six? We have, still, six choices for the first selection, and again five choices for the second selection, and then there are four remaining episodes for the third selection. That is, there’s the first-second-third, or the first-second-fourth, or the first-second-fifth, or the first-second-sixth; or the second-first-third, or the second-first-fourth, or the second-first-fifth, or the second-first-sixth, and so on. Six choices for the first episode, times five choices for the second episode, times four choices for the third, gives us six times five times four or 120 possibilities.

Following this same reasoning we figure out that if we wanted to pick exactly four episodes out of the six we should find six times five times four times three, or 360, possible groupings. To pick five episodes out of six gives us six times five times four times three times two, or 720, possibilities. And to pick six episodes out of six gives us six times five times four times three times two times one, or 720, possibilities, and at this point we throw the pen and paper we’ve been working this out on — you *have* been following along with pen and paper, haven’t you? — and say that’s ridiculous.

If we’re picking six episodes out of six, there’s just *one* way to do it, and that is to select all six episodes. For that matter, it’s absurd to think there could be 720 ways to select five episodes out of six. Choosing five episodes out of six — say, picking the ones to delete from the video recorder — is the same thing as picking one episode out of six — say, picking which one to *not* delete from the video recorder — so it’s absurd that there should be any difference in the number of ways to do it. Reason tells us there should be exactly 6 ways to select five episodes out of six. And, similarly, there should be exactly the same number of ways to pick four out of six as there are to pick two out of six. Somewhere along the line we are over-counting and quite badly so.

The problem appears right in that first listing of the selecting-two-episodes-of-six, where we came up with 30 possibilities. What’s the difference between selecting the first and second episodes or selecting the second and first episodes? It matters if you’re figuring out what to have on while you’re working out at the WiiFit’s harsh direction (“keep a steady pace for a more aerobic exercise”), but if all you’re interested in is the number of times something happens, you don’t care. We should be counting the selection of first-second as the same as second-first. We had picked episodes in a particular order, but we don’t care about that order and we have to correct for that.

In selecting exactly three episodes out of six, which gave us 120 combinations, we again over-counted. We did worse than doubling the number of combinations: we counted each possibility six times over. The selection of first-second-third episode also appeared as first-third-second, and second-first-third, and second-third-first, and third-first-second, and third-second-first. There are three times two times one or 6 ways to arrange three things in a particular order. So, we had to divide the 120 ways to select three things out of six *in order*. We have 6 ways to put any three things in order. So, there are 120/6 or a less intimidating 20 selections of three episodes out of six.

When we wanted to pick four episodes out of six, we thought we had 360 ways to do it. For any set of four episodes, there are four times three times two times one or 24 ways to arrange them. So there are 360/24 or 15 ways to pick out four episodes out of six if we don’t care what order they’re in. And this is comfortable since picking four episodes out of six ought to be equivalent to picking two episodes out of six, and we knew there were only 15 ways to pick a pair out of six. There were 720 ways to pick five items in different orders out of six; there are 120 ways to order any selection of five things; so there are 720/120 or 6 ways to pick five things out of six, the same number of ways there are to pick one thing out of six.

For six things out of six, we had 720 ways to pick them in order, and there are six times five times four times three times two times one or 720 ways to organize six things. So there are 720/720 ways, or one way, to pick all six things out of six, which again is just as we reasoned there should be.

Picking six items out of six ought to be the same as picking zero things out of six — choosing to include all six is the same as choosing to exclude nothing — and this leads to the conclusion that there may be only one way to pick nothing out of six items (leave them all behind), but there must also be one way to organize nothing. After all, we had been taking the number of ways to pick things in order, divided by the number of ways to order that many things, to give us the number of ways to pick some things. There’s 1 way to pick nothing out of six items where we care about the order, and 1 way to pick nothing out of six items where we don’t care about the order, and if 1 divided by something equals 1 then that something has to equal 1 or we’ve got division very wrong. It’s a bit heady to imagine the number of ways to organize no items, but we’re driven to this conclusion by steps that certainly looked reasonable enough at the time.

We’re getting closer to being able to say something meaningful about how likely it is there was exactly one clean sweep in The Price Is Right history.

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?

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.

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).

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).

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

hadmemorized 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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