User:Rick Block/MH solution

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[edit] Solution

The overall probability of winning by switching is determined by the location of the car. Assuming the problem statement above and that the player initially picks Door 1:

  • The player originally picked the door hiding the car. The game host must open one of the two remaining doors randomly.
  • The car is behind Door 2 and the host must open Door 3.
  • The car is behind Door 3 and the host must open Door 2.


Player picks Door 1
Car behind Door 1 Car behind Door 2 Car behind Door 3
Player has picked Door 1 and the car is behind it Player has picked Door 1 and the car is behind Door 2 Player has picked Door 1 and the car is behind Door 3
Host opens either goat door Host must open Door 3 Host must open Door 2
Host opens Door 2 half the time if the player picks Door 1 and the car is behind it Host must open Door 2 if the player picks Door 1 and the car is behind Door 3
Host opens Door 3 half the time if the player picks Door 1 and the car is behind it Host must open Door 3 if the player picks Door 1 and the car is behind Door 2
Switching loses with probability 1/6 Switching loses with probability 1/6 Switching wins with probability 1/3 Switching wins with probability 1/3

Players who choose to switch win if the car is behind either of the two unchosen doors. In two cases with 1/3 probability switching wins, so the overall probability of winning by switching is 2/3.

The reasoning above applies to all players at the start of the game without regard to which door the host opens, specifically before the host opens a particular door and gives the player the option to switch doors (Morgan et al. 1991). This means if a large number of players randomly choose whether to stay or switch, then approximately 1/3 of those choosing to stay with the initial selection and 2/3 of those choosing to switch would win the car. This result has been verified experimentally using computer and other simulation techniques (see Simulation below).

A subtly different question is which strategy is best for an individual player after being shown a particular open door. Answering this question requires determining the conditional probability of winning by switching, given which door the host opens. This probability may differ from the overall probability of winning depending on the exact formulation of the problem (see Sources of confusion, below).

In the figure above, the cases where the host opens each door are shown on separate rows. For example, if the host opens Door 2 switching loses in a 1/6 case where the player initially picked the car and otherwise wins in a 1/3 case. Regardless of which door the host opens switching wins twice as often as staying, so the conditional probability of winning by switching given either door the host opens is 2/3 — the same as the overall probability. A formal proof of this fact using Bayes' theorem is presented below (see Bayesian analysis).