Behind Monty Hall's Doors: Puzzle, Debate and Answer?

[MikeK]

Now hold on a minute here...I've taught the Monty Hall problem in various math classes for years, and, in all three textbooks that I have used that explained how it works, what Dale wrote is the exact explanation they use.  You can even go more general and show that, even if the probability that Monty opens either of the two doors in the "bad" case is not equal, you will still get the same 2/3 probability winning when switching.

Take it a step further.  Let's say there are 1000 doors.  The contestant picks a door.  Monty opens up 998 other doors, all showing zonks.  The probability the car resides behind the unrevealed door not chosen by the contestant is 999/1000.  The probability the car is behind the the unpicked door is (x-1)/x, where x is the number of doors in play.

Mythbusters tackled The Monty Hall Problem beautifully about 6 months ago:

[dale_grass]

If I were you, I'd keep it a secret.

You know full well you don't get full credit unless you show your work.  Care to explain why my method is incorrect?

[Bobby B.]

I completely understand why it's better to switch, but I've always thought that if I were ever actually on a game show where it was used, I'd be stubbornly apprehensive about switching.  That little thought of "But what if I DID pick the right one the first time?" would make the decision a little tougher.  But that's just how my mind works...lol.

[chris319]

Take it a step further.  Let's say there are 1000 doors.  The contestant picks a door.  Monty opens up 998 other doors

Monty opens only 27 of those 1000 doors, it being a 30-minute show. Monty has plum run out of time. The stagehands are about to go into overtime. We have proven nothing.

[chris319]

If I were you, I'd keep it a secret.

You know full well you don't get full credit unless you show your work.  Care to explain why my method is incorrect?

I already did.

[MikeK]

Take it a step further.  Let's say there are 1000 doors.  The contestant picks a door.  Monty opens up 998 other doors

Monty opens only 27 of those 1000 doors, it being a 30-minute show. Monty has plum run out of time. The stagehands are about to go into overtime. We have proven nothing.

You and your realism.  We're doing a 24-hour marathon.  After every 100 doors are opened, the audience is entertained by naked dancing girls.  There won't be a dry seat in the house.

[dale_grass]

I already did.

You explained why yours was correct.  You didn't explain why mine was incorrect.  As a logician, you should be the first to realize those two situations aren't equivalent.

[davidhammett]

You're double counting the door 2 pick.

Not when you count the pick and reveal as the experiment.  Then there are two outcomes associated with picking the correct door: either one of the two remaining doors can be revealed by the host.  Thus, there are 4 elements in the sample space.  The catch is that the two middle rows in my table each have probability of 1/6: 1/3 you'll pick Door 2 and 1/2 Monty will reveal 1 or 3.

To clarify, you're both right.  Chris correctly mentions that by counting the "pick door 2" event twice the probability appears to be 1/2.  Dale is correct, however, in acknowledging that there are two different events that can happen after picking door 2, each with probability 1/6, as opposed to the probability of 1/3 when picking door 1 or door 3.

[GameShowFan]

Alternate explanation?

Let p = Probability of winning
Let q = Probability of not winning

As I recall from my statistics and probability classes, p + q =1.

If p = 1/3, then 1/3 + q = 1 -> q = 1 - 1/3 -> q = 2/3? In selecting your original door, your change of winning is p. By switching, your probability of winning is now q (or, rather, not p).

Just checking...

[davidhammett]

Alternate explanation?

Let p = Probability of winning
Let q = Probability of not winning

As I recall from my statistics and probability classes, p + q =1.

If p = 1/3, then 1/3 + q = 1 -> q = 1 - 1/3 -> q = 2/3? In selecting your original door, your change of winning is p. By switching, your probability of winning is now q (or, rather, not p).

Just checking...

Perfectly (and simply) accurate.

[chris319]

The contestant has only three options for the original pick. The truth table must reflect that. Each door can be one of two states: CAR or ZONK. Here is the truth table reworked to reflect that, assuming the car is behind door 2:

PICK     REVEAL     REMAINING     SWITCH     NO SWITCH

  1       ZONK     CAR OR ZONK     CAR         ZONK

  2       ZONK     CAR OR ZONK     ZONK        CAR

  3       ZONK     CAR OR ZONK     CAR         ZONK

Note that the "REVEAL" and "REMAINING" columns are the same across the contestant's three possible picks.

Mike will now post the truth table for 1,000 doors which will include naked dancing girls.

[MikeK]

Mike will now post the truth table for 1,000 doors which will include naked dancing girls.

No.  The dancing girls are staying with me.  Giggity.

[chris319]

Mike will now post the truth table for 1,000 doors which will include naked dancing girls.

No.  The dancing girls are staying with me.

You may have your dancing girls provided you can satisfy 10 per night. I know you can do it.

[MyronMMeyer]

From Comics I Don't Understand, "You Can't Fight Monty Hall." Convincingly demonstrating why not switching is a bad idea.

-3M

[Chuck Sutton]

Does anyone know if Wayne's version uses the scenario, or is there any rhyme or reason to the deals there?

In the Big Deal Wayne always reveals one the of the lessor deals first.   One of the players who most likely knew the puzzle asked Wayne "Can I switch now?"  Without hesitation he answered, "No"