Did the famous "Monty Hall problem" actually ever appear in a game show hosted by Monty Hall?

Question

The problem stated and answered here is quite well known: http://en.wikipedia.org/wiki/Monty_Hall_problem

Did this situation ever appear in one of Monty Hall's game shows? If so, were contestants more likely to switch than not?

Answer 1

Yes, all the time. It wasn't always a goat I don't think, and I'm not sure if it was ever a car. I don't remember how often they switched, probably they more often kept to their first choice.

Answer 2

It's not clear to me what happens when B happens to choose the grand prize. Does the reshuffle of the prizes count as one of the players retries? If not, is the player aware of this reshuffle? I'm not sure it affects the player's strategy, though. I don't see how anything beats: The player should reroll the game, if allowed, any time a door is opened. If no door is opened, he should stay with his latest choice. If a door is opened on the final reroll, he should switch. Assuming an invisible reshuffle that doesn't count as a retry, then B is indistinguishable from A, provided that A doesn't have a known bias such as picking the lower numbered door when two zonkers are available to show. All that's known is that a random door, known to be a zonker by the opener, is opened. At that point, the prize is behind the unpicked, unshown door with probability 2/3, as in the standard version of the problem. Only C can refuse to open a door, and that's a certainty that the chosen door is the winner. That happens with probability 1/9 on any trial. So, with k total tries (k-1 rerolls maximum) the player sees a door opened on the final try with probability (8/9)^k and loses only on 1/3 of those cases. The overall probability of a win is then 1 - (1/3)*(8/9)^k. Even without a retry (k=1), the player wins more than in the classic game, just because of C. Edit: After a reread, it sounds like the odds might be better for the player. 1/9 of the time, (1/8 of the cases where the player didn't win outright by C failing to open), the player will know who B is and can avoid him for the rest of the game. That improves the odds of winning outright to 1/6 for any future games. Further, he should switch hosts after a reveal, since that host is less likely to be C. I don't have the energy to work out the exact probabilities tonight, but it's do-able...if messy...on this basis for specific values of k. I don't know if a readable closed form in terms of k is there.

Answer 3

it happened all the time.
A lot of people just stuck with the 1st choice because now they knew it was a 50/50 chance

Answer 4

most stuck with curtain number 1