His brain was full of two-base hits and curves of every kind

[austin_dern]

On Friday's The Price Is Right Drew Carey gave a little mention to ``the people keeping track of spins at home'' by noting that consecutive spins of the big wheel for the Showcase Showdown, for consecutive contestants, were both 25 cents. Either he's trusting in the boundless limits of fandom or he's aware that among the comment threads for the online episodes are occasionally people keeping track of how often Secret X or the Check Game have been won or lost this season. They aren't quite covering things in baseball game-like detail, but they're working on it.

And I'm not sure how to admit this but I've quietly joined the weird league of people taking statistical measure of The Price Is Right. I got into it with a fair spirit of inquiry: I wondered if, at the Showcase Showdown (in which three contestants spin the big wheel trying to get as close as possible to $1.00 without going over) whether spinning second or third is more likely to win. Spinning order is determined by ascending order of how much you won in the item-up-for-bid and your pricing games, but it seemed to me that the second person may have an edge: she has to do better than the first contestant to win the Showdown, yes, but the third person has to do better than that. And nonscientific tracking seemed to indicate a possible preference for the second position. The only thing to do was start keeping notes.

Well. From the 10th through 30th of November, ten times the first spinner won the Showcase Showdown, nine times the second did, and eleven times the third did. This suggests little preference. For the shows from 1 through 26 December (the last week of the year was reruns, not that this matters), the first contestant won nine times, the second thirteen, and the third eighteen times. For 5 through 30 January, the first and second contestants won nine times each, and the third twenty. And for February, the first contestant won eleven times, the second fourteen, and the third fifteen. So from November through February the first contestant won 39 times, the second 45, and the third 64. This would seem to suggest that the third contestant is the one to be, but who knows if I'll ever convince myself adequately of that fact to stop counting. I've made some interesting studies of the Showcase, too.

Trivia: Baseball published-statistics pioneer John Lawres began collecting the records of every baseball player after fans questioned his high opinion of the 1889 New York Giants. His collection grew to a full-sized ledger book and from 1912, Who's Who In Baseball. Source: The Numbers Game: Baseball's Lifelong Fascination With Statistics, Alan Schwarz.

Currently Reading: Flash Of Genius And Other True Stories Of Invention, John Seabrook. Inspiration for the Greg Kinnear movie that you didn't see, although that makes up only the first chapter.

Comments

[bunny-hugger.livejournal.com]

It's good to have hobbies.

[austin-dern.livejournal.com]

I admire your ability to make what I'm doing sound vaguely rational.

[bunny-hugger.livejournal.com]

As a rationalist I find your irrationalities strangely fascinating. (They're more whimsical than my own.)

[austin-dern.livejournal.com]

You are really very kind to my irrationalities. I'm glad they do come across as whimsical. It's hard telling from the inside of them.

I'm sorry to be so long replying; the combination of your and CL's comments made me not realize at a glance there were unanswered mentions. I'm more sharply attuned to notice odd numbers.)

[argonel.livejournal.com]

I would think that there is an optimal strategy to try winning the spins similar to an optimal strategy for winning blackjack with a multideck shoe. It is controlled primarily by luck, but wise choices on the part of the first part can make it much more difficult for contestents 2 and 3.

[chefmongoose.livejournal.com]

Oh, absolutely. The question, of course, being what are the numbers to stop on, ideally, for contestant 1, 2, and.. well, 3 has no strategy, it's just spinning. I would imagine that the break is something about 75 cents for the first contestant and 65 for the second; that is, the odds you'll go over become higher than the odds that in remaining spins you'll be exceeded. I'm certain there's easy statistical determination for exact values, but that's my gut feeling.

[chefmongoose.livejournal.com]

http://www.carthage.edu/faculty/msnavely/sdlnew/vol15/massey.pdf has some information on it, but it's flawed; it assumes only one more contestant. (A second-contestant scenario). It gives the break point at 55 cents to spin again, 60 to stay.

[xolo.livejournal.com]

There's an unexamined assumption there that the distribution is even. I've always kind of wondered (although not enough to collect the data) how well balanced that wheel is, and whether some numbers might be favoured.

[austin-dern.livejournal.com]

I haven't been tracking the spins as they come up, but I'd be surprised if the distribution were quite uniform. There are enough players who try to finesse the wheel --- and they can get up to three spins to try, and there are some regions that are obviously dangerous with a lot of high-value amounts in a row --- that even a perfectly uniform wheel would give not quite uniform results. I'd think it has to be close, though, since people don't, practically speaking, get more than one chance to feel how heavy the wheel is, and many people play to the audience and spin as fast as they're able.

I don't think the wheel can be terribly unbalanced apart from going longer without having the axle cleaned than is maybe ideal: if it were there'd be sometimes that the weighting would cause a backspin at the end as the weight was on the `forward' side. If the weighting were so light that this backspin was never noticeable it wouldn't affect the prize distribution enough to be worth doing.

[chefmongoose.livejournal.com]

Right. There's some ability to control spins, but really only on your second spin.. the first spin is as good as random, until you have a feel for the wheel.

I think for a hold/stay strategy it's of minimal effect.

[austin-dern.livejournal.com]

And it's probably not so easy to get a feel for the wheel. If it were then Pat Sajak's Final Spin on Wheel of Fortune would never hit anything but $5,000 (for the most exciting shows) or $250 (for the most economical shows).

So what you have, really, is a case where the first spin produces a uniform distribution, and the second spin a Gaussian distribution centered on $1.05, and contestants have a mandatory second spin if they're behind the current leader and a voluntary spin if they're above the current leader; in practice, I think they stop spinning once they've got $0.70 (but haven't really tested this) ... maybe it could be written as algorithm after all.

[chefmongoose.livejournal.com]

The First contestant's stop value would have to be higher than the second contestant's, though; He has to not be outspun by both the second and third contestants.

The second contestant, the link upthread gives the correct strategy; if he/she spins and beats the first contestant on the first spin, then spin again 55 cents, stay on 60 cents or higher.

The third contestant has no real choices, given that if he/she achieves the lead, he/she won't want to spin again under just about any circumstance.

[austin-dern.livejournal.com]

Oh, there's a wealth of conditional probability problems to be gotten from this. You could probably do a whole exam's worth of probability questions based just on the Showcase Showdown. It's marvellous.

I'm convinced that there should be a most efficient possible algorithm for solving the Race Game --- where you have to match four prices to four prizes, and have 45 seconds to swap out any given only the information of how many prices you have right --- but I haven't worked it out.

[chefmongoose.livejournal.com]

Well, for that.. Let's assume you make your best possible guess, fairly promptly at the first You'll have either 0, 1, 2 or 4 (and win). The effective rate of switching that I've seen really gives you two possible 'resets'. Considering 4 is your desired state, you've got to figure out the best strategy for: 0, 1, and 2; then figure a second round possibility set of:

0->0 0->1 0->2
1->0 1->1 1->2
2->0 2->1 2->2

Also keeping in mind these aren't random numbers to determine. But yes, rather complex.

[austin-dern.livejournal.com]

The game's complicated in that there are three serial contestants and that they can have one or two spins. If everyone had to take two spins you could probably make do with a bell curve centered on $1.05, but someone spinning $0.95 the first spin is going to stop unless absolutely required.

I suspect the least awful approach may be just keeping track of each spin, which I'm not doing just yet.

[austin-dern.livejournal.com]

Ah, but that's the thing ... it's easy to say if you have 80 cents then the next contestant has a 20 percent chance of beating you on the next spin. But there being two spins ... And then on top of that there's the devious arrangement of coin values on the wheel: there are some patches which are all high values, some all low values, and quite a few great suspense-building pairs of a high and a low value.

I'm really impressed that the values are so well arranged considering the wheel arrangement appeared (with the axis facing the audience rather than parallel to the cameras, initially) in 1975, and they can't have had its arrangement modelled to heighten suspense. Maybe it could conceivably have been tested on computer but I doubt it. Whoever put the values in did a really good job of value placement.

[austin-dern.livejournal.com]

That is part of what makes this such an exciting question and I suspect one that really can't be handled except empirically. It's very much like blackjack, except that every possible number is infinitely replaceable, and yet there's a dependence for one spin on the prior. I've got exactly the wrong set of mathematical tools to model it correctly.