#### Comment Preferences

• ##### My own generalization of the problem . . .(6+ / 0-)

Original with me, though no doubt others have come up with it independently.

In the original problem, switching doors gives you a 2/3 chance of winning. If, on the other hand, Monty opens a door at random, and the door he opens happens not to conceal the prize, then you have a 50-50 chance of winning whether you switch or not. (The diarist alludes to this variant but doesn't explicitly state the resulting probability.)

Now consider the following two further variants. In each, we start with four doors and you pick one as your very own.

Variant A:  After you pick your door, Monty (knowing where the prize is) opens a door that he knows does not contain the prize.  Then he opens one of the remaining two doors at random and the prize does not appear.

Variant B:  After you pick your door, Monty opens one of the other three doors at random and the prize does not appear. Then he opens one of the remaining two doors, one that he knows does not contain the prize.

In each case there are now two doors left. Question: Are the two situations identical? If so, should you switch doors, and what is your chance of winning?  If not, in which variant (if either) should you switch, and what is your chance of winning in each variant?

In general, start with N doors (behind one of which there is a prize) and a bit vector V of length N-2.  You pick a door.  Monty now processes V left to right: Each time he encounters a 0 he opens a door other than yours at random. Each time he encounters a 1 he opens a door other than yours that he knows doesn't conceal the prize. At the end there are two doors unopened. Given that (by chance) the prize has not been revealed, what is the probability that you will win if you switch, as a function of V?

(The original Monty Hall problem is V="1", variant A is V="10", and variant B is V="01".)

For even more generality we can start with N doors concealing k < N prizes and a bit vector of some length less than N-k. I don't think this makes things more interesting.

• ##### Both of those are considered in the book(5+ / 0-)
Recommended by:
Orinoco, kurt, Jimdotz, elropsych, NCrissieB

and there is a proof that the best method, when Monty knows which door, is to switch at the last possible time.

The proof gets into more math than I want to do here, but the intuition is that, once Monty has opened 1 door, your door is the LEAST likely of all the doors, so Monty provides more information by opening OTHER doors than by opening yours.  If you  switch, then he might open your first choice, which would give you the least information.

• ##### Interesting, though not exactly what I said(5+ / 0-)
Recommended by:
Orinoco, plf515, Jimdotz, elropsych, NCrissieB

I'm not offering you the choice of when to switch. I'm asking for a probability computation based on Monty's scheme for opening doors.

• ##### Intuitively(1+ / 0-)
Recommended by:
plf515

I would do better switching in variation B. Either way you have only a 1 in 4 chance of being right the first time but in var. A Monty has a 1 in 2 chance of randomly reveling the car but in var. B he has a 1 in 3 chance of reveling it. The advantage comes in reducing the odds of Monty opening the door you want to pick.

Blackwater is changing its name to Xe.

[ Parent ]

• ##### My first guess(2+ / 0-)
Recommended by:
plf515, NCrissieB

Is that you have a 60% chance if you switch with A, and a 66% chance with B.  I probably missed something though.

Yes, there are progressives in the rural South. 50 States.

[ Parent ]

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