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The original Parade problem is fairly ambiguous and incomplete. Therefor there is no "correct solution". Other formulations typically include the conditions tha
by em500 4y ago
The original Parade problem is fairly ambiguous and incomplete. Therefor there is no "correct solution". Other formulations typically include the conditions that the host will always open a door, and that door will always contain a goat as explicit assumptions. In that case the math and logic are pretty straightforward.
But in the setting of an actual repeated game show, I think the more realistic case to keep the show interesting is that the host will mix up his strategy, i.e. sometimes he opens another door, sometimes he doesn't. If it's a long running daily or weekly show, if he always opens a door with a goat, the public will work out pretty quickly that it's advantageous to always switch.
- mcphage 4y agoThe original formulation was probably written under the assumption that the readers were familiar with the game show it was drawn from.
- em500 4y agoAs TFA notes the Parade magazine formulation is only "loosely based" on the actual show Lets Make a Deal" hosted by Monty Hall. The real Lets Make a Deal* is a bit more complicated[1], and AFAICT there is no easy analysis or solution for an optimal strategy. The game described in Parade magazine is different from the TV show game, and the assumptions (about the host's behavior) are incomplete/ambiguous enough that switching is not necesarrily the optimal strategy. This, like almost everything else you can think about the Monty Hall Problem, is also described in TFA[2]. [1] https://en.wikipedia.org/wiki/Let%27s_Make_a_Deal#Format https://en.wikipedia.org/wiki/Let%27s_Make_a_Deal#Format [2] https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_behaviors https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_...