# The Monty hall problem, Quantum Games

### The original problem

Three doors, a prize behind only one door (say, a new car). Other two doors are each hiding a goat.

Contestant has to pick a door – and hope that she wins the car.  At just this stage, the chance of her winning the car is 1/3 – that part if simple.Now, add a twist.

After the contestant has picked her door (but not opened it), the game host (Monty Hall), OPENS another door – to reveal a goat. He ALWAYS reveals a goat when he opens his door.Now, he re-asks the contestant – Would you like to stay with your original door – or change to pick the other unopened door.

You may think , it doesn’t matter what the contestant picks at this point. But you would be wrong. The contestant is ALWAYS better off switching her answer!!

The point is, though we know in advance that the host will open a door and reveal a goat, we do not know which door he will open.  If the host chooses uniformly at random between doors hiding a goat (as is the case in the standard interpretation), this probability indeed remains unchanged, but if the host can choose non-randomly between such doors, then the specific door that the host opens reveals additional information.

### Consider the 2 un-chosen doors together

Another way to understand the solution is to consider the two original unchosen doors together. As Cecil Adams puts it (Adams 1990), “Monty is saying in effect: you can keep your one door or you can have the other two doors.” The 2/3 chance of finding the car has not been changed by the opening of one of these doors because Monty, knowing the location of the car, is certain to reveal a goat. So the player’s choice after the host opens a door is no different than if the host offered the player the option to switch from the original chosen door to the set of both remaining doors. The switch in this case clearly gives the player a 2/3 probability of choosing the car.

### Treating non-random information as if it were random (Phillip Martin’s Probability Trap)

Host knowingly opens a door – is different from a host RANDOMLY opening a door.

It is based on the deeply rooted intuition that revealing information that is already known does not affect probabilities. But, knowing that the host can open one of the two unchosen doors to show a goat does not mean that opening a specific door would not affect the probability that the car is behind the initially chosen door. The point is, though we know in advance that the host will open a door and reveal a goat, we do not know which door he will open. If the host chooses uniformly at random between doors hiding a goat (as is the case in the standard interpretation), this probability indeed remains unchanged, but if the host can choose non-randomly between such doors, then the specific door that the host opens reveals additional information.

### Not 3 doors, but N doors

D. L. Ferguson (1975 in a letter to Selvin cited in (Selvin 1975b)) suggests an N-door generalization of the original problem in which the host opens p losing doors and then offers the player the opportunity to switch;

If the host opens even a single door, the player is better off switching, but, if the host opens only one door, the advantage approaches zero as N grows large (Granberg 1996:188). At the other extreme, if the host opens all losing doors but one (p = N − 2) the advantage increases as N grows large (the probability of winning by switching is
N − 1/N  , which approaches 1 as N grows very large).

### Summary

To this day, this problem continues to baffle one’s common sense. However, simulations do not lie – repeated simulations have shown that this is indeed a probability trap – and that the contestant is always better off switching.

Check out also – Problems In Advanced Math and Physics and Rare Finds in Special and General Relativity

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