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Monty Hall Problem Extended

The Monty Hall Problem is quite famous: Behind one of three closed doors is a new car, the other two doors have goats behind them. You pick one of the doors. Monty opens one of the other doors and it reveals a goat. You are given the choice of sticking with your choice or switching to the other unopened door. Should you switch? What are your chances of winning the car if you do?

If you have never played with this problem, stop and do so.

Once you have mastered the correct logic underlying the above classic problem, try your hand at this extension, called the Four-Door Monty Hall Problem: Behind one of four closed doors is a new car. You pick door #1. Monty opens the door #4 and it reveals a goat. You switch your choice to door #2. Now Monty opens door #3 and it is also reveals a goat. Given the choice, should you switch back to door #1, and what are your chances of winning the car if you do?

 


Hint: Take out three playing cards (one black, two red)...grab a friend...and simulate the problem by trying both options--staying and switching. Record the number of successes in each case over a large number of games. Now, can you explain the experimental data using a theoretical model?

If you are unable to find cards, this web site provides a visual simulation of the three door case.

 


Solution Commentary: If you want to check your work, consider this web site (and the source of my graphic) for a great explanation and opportunity to run simulations of the problem. Or, try web site which suggests that it has a solution...and explores other variants on the problem.

Micahel Naylor (WWU) adds these fascinating comments on 11/6: I did Monty Hall for 10 doors... each time he shows a goat and each time you switch, until you finally switch for the remaining door.

The fraction is beautiful...
with 3 doors: 1 - (1 - 1/3)/2 = 2/3
with 4: 1 - (1 - (1 - 1/4)/3/2 = 5/8
with 5: 1 - (1 - (1 - (1 - 1/5)/4)/3)/2 = 19/30
the pattern continues:
91/144, 177/280, 3641/5760, 28673/45260, 28319/44800

As decimals, the pattern is:
3 doors = 0.666
4 doors = 0.625
5 doors = 0.633
6 doors = 0.6319
7 doors = 0.63214
8 doors = 0.632118
9 doors = 0.6321208
10 doors = 0.6321205...
I believe this is converging to 1 - 1/e, but I don't know enough yet (about the problem) to prove that.

Notice how the decimal hops up and down, like Fibo ratios converging on phi. Curiously, the four door problem gives you the worst probability of any number of doors greater than 2. (2 doors is 0.5)

And in another e-mail from Michael (11/7): I have confirmation that the Monty Hall problem converges on 1-1/e.

It is a beautiful series of calculations that boils down to the form: 1 - 1/2! + 1/3! - 1/4! + 1/5! - ...

And in yet another e-mail from Michael (11/10):
Here's a little bit showing how the series develops... don't know if you want to try to slip this graphic in but it sure is pretty.

If you write it neatly:





Dividing through, starting from the bottom, we get: 1 - 1/2 - -1/(3*2) - - -1/(4*3*2) - - - -1/(5*4*3*2) - - - - -1/ ...= 1 - 1/2! + 1/3! - 1/4! + 1/5! - ...

This series also arises in the "Hat Check" problem: it's the probability that if a large number of people are given their hats at random, at least one person gets their own hat. But how is the Monty Hall problem like the Hats problem? . . .