## 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.

Here’s the next assumption: that the actual probability of a clean sweep is 1 in 6000. How do we know it’s that? If we have some number *N* of episodes, and a probability *p _{s}* of a clean sweep in any one episode, and no number of clean sweeps makes another clean sweep more or less likely, then the number of clean sweeps to expect is their product,

*N p*. We have about 6,000 episodes; we saw 1 clean sweep; therefore, if we expect only one episode the probability of a clean sweep is 1/6000.

_{s}This is the shakiest assumption, the one most difficult to justify. We are supposing that what we observed was the most probable outcome. It does not need to be. We could work out the probability of just one clean sweep where the probability of a sweep in any one episode is assumed to be 1/2000, or 1/1500, or 1/3000, or any number. For most of those numbers, the chance of seeing one sweep in 6,000 shows is high enough that we aren’t forced to suppose it’s anything but chance. Just because something happened does not mean it was the most likely thing to happen. It does not even mean it was at all likely to happen.

On the other hand, what better assumption can we make? We can extract only so much information from one piece of data, and this is about it.

So let us assume there is a 1/6000 probability of a clean sweep in any one episode. Someone wins the first item up for bids; we make no guesses about which, no assumptions about which. I will call whichever seat wins that first item the “first seat” so we have a label for it. There is a probability — let me call it *p _{1}* — that the second item up for bids is won by the same, first seat. Calling this probability

*p*is not an assumption; there is

_{1}*some*number which describes that probability and I am just giving it a name we can use.

What is the probability that the third item up for bids is won by the same seat that got the first item, though? Here we make another assumption: that this probability is the same number, that same *p _{1}*, as the probability the second item was won by the same seat. This could easily be false. After all, the other three bidders are on their third round of bids and are probably getting a better sense of the particular episode’s flow, and of which members of the audience know what they’re shouting, while the first bidder is on her or his first bid ever. But if we don’t suppose the probability of the third item up for bids being won by the first-seat is the same as the probability for the second item up for bids, we’re stopped. I don’t know offhand how to get to an estimate otherwise.

Similarly we are making the assumption that on every round of items up for bidding the probability of the first seat winning is the same.

Given that constancy of the winning seat being the same as the first seat, then the probability of a clean sweep *p _{s}* is

*p*, this number

_{1}^{5}*p*multiplied by itself five times over. We made the assumption that

_{1}*p*was 1/6000, approximately 0.000 1667. (Oh, hush; that

_{s}*is*approximate. It only looks like a lot of digits because of all those zeroes up front.) Therefore,

*p*is the fifth root of that, or about 0.1755,

_{1}*if*all the assumptions we made hold true. And, therefore, if our assumptions are correct, going first reduces one’s probability of winning by about one-third what it would be if every seat had an equal chance (that is, if the first bidder had a probability of 0.25 of winning).

Are these assumptions right, though? Can we check any of them?

## Chiaroscuro 5:40 am

onMonday, 13 February, 2012 Permalink |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.

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## nebusresearch 4:42 am

onWednesday, 15 February, 2012 Permalink |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.

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## Chiaroscuro 4:22 am

onThursday, 16 February, 2012 Permalink |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.

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## Joseph Nebus 7:33 am

onMonday, 20 February, 2012 Permalink |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.

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## What Can One Week Prove? | nebusresearch 7:13 am

onMonday, 20 February, 2012 Permalink |[…] With One Mon…Proving Something Wi… on What Can One Week Prove?Chiaroscuro on Figuring Out The Penalty Of Go…What Can One Week Pr… on The Significance of the Item U…nebusresearch on Figuring Out […]

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## An Overused Intermediacy | nebusresearch 2:09 am

onWednesday, 29 February, 2012 Permalink |[…] but does have some wonderful results that depend on it. My context was in explaining just what Chiaroscuro had done when he figured out the fifth root of 1/6000th by guessing at it. I mean, he figured he was guessing at it, but […]

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## The Intermediacy That Was Overused | nebusresearch 4:08 am

onSaturday, 3 March, 2012 Permalink |[…] I may sulk, Chiaroscuro did show off a use of the Intermediate Value Theorem that I wanted to talk about because normally the Intermediate […]

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## About Chances of Winning on The Price Is Right | nebusresearch 12:53 am

onSaturday, 21 April, 2012 Permalink |[…] Figuring Out The Penalty Of Going First: There is one explanation for this deficiency which feels quite plausible, though. That’s to suppose that some bidders on Contestants’ Row are at a relative advantage or, maybe more important, at a disadvantage. Watching The Price Is Right suggests that being the last contestant to bid is probably the choice spot. Being the first contestant to bid is probably the worst. If we assume that one clean sweep in six thousand episodes is the most likely outcome, this lets us say at least how big a penalty being the first bidder probably carries. […]

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