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?

That’s found by the same binomial distribution we used for the clean sweeps, although with different numbers. In this case, the trial is a single item being up for bids. The successful outcome is the first seat winning the item up for bids. If we suppose every seat has an equal chance, the probability p of success on any one trial is 1/4, or 0.25. In one week, the number of items up for bids, the number of trials N, is 30. We can use that binomial distribution referred to a while ago to evaluate the chance of the first bidder having 0 wins, 1 win, 2 wins, 3 wins, and so on. The evaluation is a little tedious, but made a lot less so through a numerical calculations package like Matlab or its open-source clone, Octave. In Matlab and in Octave the function binopdf will calculate, for a given number N of trials and a probability p of success on any one trial, the chance of seeing a certain number x of successes.

The chance of the first bidder getting 5 or fewer wins out of 30 is the chance of getting 0 wins plus the chance of 1 win plus the chance of 2 wins et cetera up to the chance of 5 wins. If we worked out the chance of each individual number of wins, we can add that up easily. Or we can let Matlab do even more of the hard work; its binocdf function works out the probability of at most a specified number of successes, out of a set number of trials and a given probability of a success on each trial. I strongly recommend letting Matlab, or Octave, or something similar do the work. It’s too much arithmetic for the math involved otherwise.

The probability, if the first bidder has a 25 percent chance of winning any one item up for bids, and if there are 30 items up for bids, of the first bidder winning 5 or fewer times is just a bit over 20 percent (20.26 to be slightly more exact). That’s not very likely, I suppose, but it’s common enough that one week wouldn’t prove much of anything to anyone.

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Author: Joseph Nebus

I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there.

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