TaeKahn

10

While the overall probabilities for the game will never change, the contestant’s perception of the current state of the game will cause them to affect their win rate. To elaborate on what I was saying, imagine the following internal monologue of a contestant:

I’ve eliminated one goat. Two doors left. One is a goat, one is a car. No way to tell which is which, so I’ll just randomly pick one.

IMO, this is probably what most contestants believe when faced with the final choice. Obviously, there is a way to have a greater success rate, but this person is evaluating in a vacuum. If contestants were aware of the actual probabilities involved, I think we would see less “agonizing” moments as the contestants decide if they should switch or not. By randomly picking door A or B, irrespective of the entirety game, you’ve lost your marginal advantage and lowered your win rate. That being said, if they still “randomly” pick switch every their win rate will be the expected, actual probability.

Edit: The same behaviour can be seen in Deal or No Deal. If for some insane reason, they go all the way to the final two cases, the correct choice is to switch. I don’t know exactly how many cases you have to choose from, but the odds are greatly against you that you picked the 1k case. If the case is still on the board, the chick is holding it. Yet, people make the choice to switch based entirely on the fact that there are two cases and 1 has 1k and the other has 0.01. They think they have a 50/50 shot, so they make their odds essentially 50/50 by randomly choosing. In other words, they might as well as have flipped a coin to make the decision between the two cases.

-10

I have a small issue with the way you presented the Monty python problem. In my opinion, the setup could be a little clearer. The Bayesian model you presented holds true iff you make an assumption about the door you picked; either goat (better) or car (less wrong). If you pick a door at random with no presuppositions (I believe this is the state most people are in), then you have no basis to decide to switch or not switch, and have a truly 50% chance either way. If instead you introduce the assumption of goat, when the host opens up the other goat, you know you had a 2/3 chance to pick a goat. With both goats known or presumed, the last door must be the car with an error rate of 1/3.

You are making perfect sense; it’s me that is not. I had thought to clarify the issue for people that might still not “get it” after reading the article. Instead, I’ve only muddied the waters.