On the December 15th episode of The Price Is Right, host Drew Carey mentioned as the sixth Item Up For Bids began that so far that show, all the contestants who won their Item Up For Bids (and so got on-stage for the pricing games) had come from the same spot so far, five out of six. He said that only once before on the show had all the contestants come from the same seat in Contestants Row. That seems awfully few, but, how many should there be?

We can say roughly how many “clean sweep” shows we should expect. There’ve been just about 6,000 episodes of The Price Is Right played in the current hour-long format (the show was a half-hour its first few years after being revived in 1972; it was a very different show in previous decades). If we know the probability of all six contestants in one game winning their Item Up For Bids — properly speaking, it’s called the One-Bid, but nobody cares — and multiply the probability of six contestants in one show coming from the same seat by the number of shows, we have the number of shows we should expect to have had such a clean sweep. This product, the chance of something happening times the number of times it could happen, is termed the “expected value” or “expectation value”, or sometimes just the “mean”, as in the average number to be, well, expected.

This makes a couple of assumptions. All probability problems do. For example, it assumes the chance of a clean sweep in one show is unaffected by clean sweeps in other shows. That is, if everyone in the red seat won on Thursday, that wouldn’t make everyone in the blue seat winning Friday more or less likely. That condition is termed “independence”, and it is frequently relied upon to make probability problems work out. Unfortunately, it’s often hard to prove: how do you prove that one thing happening doesn’t affect the other?

In this case, it seems safe to just assume that episodes are probably independent. It’s hard to see how all the contestants from one seat winning in one episode could possibly make the same thing happening again more or less likely, unless the program’s designers decided to tinker with the rules. Even then it’s hard to see what they *could* do. While the show’s producers pick the contestants, they can hardly pick better or worse contestants or assign them seats. (The first four contestants called on down may sit in any seat they like, or at least could the last time I looked this up; subsequent contestants have to sit in the vacant seat.)

We have, roughly, the number of shows made in the hour-long format. It’s actually about 5800 as of the end of December 2011, but 6,000 is near enough and we can wait. The question is what is the probability of all six contestants coming from the same seat?

Well, there are four seats. The first contestant getting on stage comes from one of them. What is the probability that the second contestant comes from the same seat? If we have no reason to suppose that any seat has become more or less likely — particularly, if we suppose the contestants remaining haven’t gotten any better or worse at pricing items for having some experience in the real live show — then there’s one chance in four the second contestant comes from that same seat. So that probability is one out of four, or ^{1}/_{4}.

What about the chance the third contestant comes from the same seat as the first? Well, again, if we suppose the chance of each seat winning hasn’t been affected by the first two Items Up For Bid being won, the probability of the third contestant coming from the same seat as the first is one out of four, or ^{1}/_{4}. But we do have to take a moment to figure out: how do we go from knowing the probability of the first and second contestants coming from the same seat, and the probability of the first and third contestants coming from the same seat, affect the probability of the first, second, and third coming from the same seat?

The rule is that the probability of two independent events happening is equal to the probability of the first event happening times the probability of the second event happening. This is fairly easy to be convinced happens if you use simple events, like coin flips or die rolls, or particularly a coin flip and a die roll together since the chance of a coin coming up tails is different from the chance of a die coming up “four”. If you’re not convinced that this is obvious, we can explain it anyway, but I don’t want to do that today, as it’s longer and harder to do.

Now for a trick: what is the chance the second and third contestants come from the same seat? That should be one in four, also, for a probability of ^{1}/_{4}.

So, ought the probability of the first and second, the first and third, and the second and third contestants all coming from the same seat be ^{1}/_{4} times ^{1}/_{4} times ^{1}/_{4}, that is, ^{1}/_{64}?

No, not in this case. We crash up against independence of events again: if the first and second contestants come from the same seat, and the first and third come from the same seat, then the second and third *must* come from the same seat. If the first and second come from the same seat, and the first and third do not, then the second and third *must not* come from the same seat. Since we don’t have independent events, we don’t have a multiplication of probabilities.

This is nice enough, though. We can work out the probability of each of the five later contestants coming from the same seat as the first, and the probability of a clean sweep is just the probability of each of the five successor contestants coming from the same seat as the first multiplied together. (Equivalently, we could work it out as the probability of the second coming from the same as the first, times the probability of the third contestant coming from the same seat as the second, times the probability of the fourth contestant coming from the same seat as the third, and so on.) This probability is ^{1}/_{4} times ^{1}/_{4} times ^{1}/_{4} times ^{1}/_{4} times ^{1}/_{4}, which works out to be ^{1}/_{1024}.

Based on this, then, we would expect about one out of every thousand shows to turn up a clean sweep. Since there have been about six thousand shows, we would expect to see six clean sweep shows. According to Drew Carey, there had actually been one.

Let’s suppose for a moment that Carey wasn’t simply making small talk about the in any cases remarkable coincidence of five winners from the same seat. We can’t rule that out, by the way: a game show host has to entertain, and puffing up cute coincidences into historic occasions is part of that. Bob Barker liked to proclaim many minor coincidences into “historic occasions”, making for good television and frustrating the insufficiently skeptical would-be archivist watching at home. Where did the missing five clean sweeps go?

There wasn’t a clean sweep that episode either, by the way; someone from another seat won the final Item Up For Bid.

I think I know what’s happened to the missing five clean sweeps; but you may have a follow-up article and I don’t want to spoil things you’ve likely planned for it. I’ll just say that due to The Price Is Right and how Item Up For Bid is structured, the bids are not *precisely* independent events.

..that or Drew Carey could just have meant ‘since I took over on the show’.

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You’re quite right that I have a follow-up in mind, and that is going to look at whether we have to conclude something like that the winning seat isn’t independent of the ones before, or that Carey was wrong to suggest it happened once in six thousand shows.

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