I have one last important thing to discuss before I finish my months spun off an offhand comment from The Price Is Right. There are a couple minor points I can also follow up on, but I don’t think they’re tied tightly enough to the show to deserve explicit mention or rate getting “tv” included as one of my keywords. Here’s my question: what’s the chance of winning an average pricing game, after one has got an Item Up For Bid won?

At first glance this is several dozen questions, since there are quite a few games, some winnable on pure skill — “Clock Game”, particularly, although contestants this season have been rotten at it, and “Hole In One … Or Two”, since a good miniature golfer could beat it — and some that are just never won — “Temptation” particularly — and some for which partial wins are possible — “Money Game” most obviously. For all, skill in pricing things help. For nearly all, there’s an element of luck.

I’m not going to attempt to estimate the chance of winning each of the dozens of pricing games. What I want is some kind of mean chance of winning, based on how contestants actually do. The tool I’ll use for this is the number of perfect episodes, episodes in which the contestant wins all six pricing games, and I’ll leave it to the definers of perfect such questions as what counts as a win for “Pay The Rent” (in which a prize of $100,000 is theoretically possible, but $10,000 is the most that has yet been paid out) or “Plinko” (theoretically paying up to $50,000, but which hasn’t done so in decades of playing).

According to golden-road.net’s impressive FAQ, there have been 80 perfect episodes. There have been somewhere around 6,000 hourlong episodes. I’ll use this the same way I estimated the chance of the first seat winning an Item Up For Bid. I suppose that every episode is independent: today’s episode being perfect does not make tomorrow’s more or less likely to be perfect. I suppose that every episode has the same probability of being perfect. So if there have been N = 6000 episodes, and there is a probability p_{e} of any one episode being perfect, then I expect there to be N times p_{e}, or 6000 times p_{e}, perfect episodes. I’m using the name p_{e}, with the p standing for “probability”, and the subscript “e” for “episode”, because there’ll be another probability to consider later on. I further suppose — and this is the weakest assumption — that the 80 perfect shows is the most probable outcome, the outcome to expect. So 6000 times p_{e} is 80, and therefore, the probability p_{e} of a perfect episode is 80/6000 or a surprisingly high (to me) 1/75.

Next, I suppose that every pricing game has the same probability, p_{g}, of being won. They don’t, of course. Some games are easy; some are near impossible. I am taking a mean, an average, of the chance of winning each game. Specifically, I am taking the geometric mean, which is in your statistics book somewhere just after the arithmetic mean and skipped because the class hasn’t got enough time to cover it. I figure to talk more about that later. But supposing each game is equally likely to be won is practical, and won’t be too far off until we know which game you get to play.

I next assume that each of the six pricing games is independent of the other: whether the first game is won or lost does not affect whether the next is won or lost. This is probably fair enough, although a contestant who knows the last five games were all lost or won might be more thoughtful about the sixth, so as to keep the show from being a perfect failure or to not break the streak. Whether that thoughtfulness helps any I couldn’t guess; let’s suppose not.

If each of the six games is independent, and each has the same probability p_{g} of being won — that’s the other probability, and the subscript g means this p represents the probability of a win for the one game — then the probability that all six games are won will be the probability p_{g} times itself, six times over. That is, it will be p_{g}^{6}.

The chance of winning six games in one episode is p_{g}^{6}. Six pricing games won in one episode is a perfect game, and we know what the probability p_{e} of a perfect game is: it’s 1/75. p_{g}^{6} and 1/75 are therefore two ways of writing the same number; they have to be equal to each other. And this tells us that the probability of winning one game must be the sixth root of 1/75. You can work this out by figuring out how the power button (the x^{y} or y^{x} button) on a calculator works, or by fiddling around with examples until you get close enough. I’ll skip right to the answer.

The probability of winning any one game, p_{g}, on these assumptions, turns out to be 0.487. Considering all the approximations made here, I’d call that a probability of 1/2.

If we take every pricing game as having a probability of 1/2 of being won, we can estimate what fraction of games ought to have zero wins, one win, two wins, and so on. The reciprocal of that is about how many episodes you should expect to watch to see one show with that many wins in it turn up, although it isn’t as though if you do watch 64 episodes you’ll be sure of seeing one perfect game. It might happen after 12 episodes; it might happen after 90 episodes. On average, about 64 episodes, though. The roster of all these possibilities turns out to be:

Won Games | Probability | One In Each (How Many) Games |
---|---|---|

0 | 0.015625 | 64 |

1 | 0.093750 | 10.666667 |

2 | 0.234375 | 4.266667 |

3 | 0.312500 | 3.2 |

4 | 0.234375 | 4.266667 |

5 | 0.093750 | 10.666667 |

6 | 0.015625 | 64 |

In short, if you do get up on stage, and they offer “Temptation”, just take the four prizes they give you and don’t try for the car. You never win and you’ll just spoil the episode.

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