I Please Explain (actually explain) The Monty Hall Problem

[Ibix]

A friend of mine at work was unconvinced until he started writing a computer program.

I actually initially convinced myself by drawing the probability tree:

Step 1: pick a door to hide the prize (3 equi-probable branches)
Step 2: contestant picks a door (3 equi-probable branches from each of the first three)
Step 3: Monty decides which door to open (2 equi-probable branches from the three branches where the contestant picked a winner, one branch from the other six)

This is, of course, another way of formally writing out the algorithm, but one that doesn't require any knowledge of programming. Laying it out formally leads you to the realisation that Monty sometimes has a choice and sometimes doesn't depending on the contestant's first choice, and that his actions don't change the probability that the contestant guessed right first time. So switching is always the better strategy.

 

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Fact:
He has knowledge of his first chosen door, that it's 1/3 the winning door..

And he has knowledge of the other remaining door that it's also 1/3 the winning door.


So he has no useful knowledge to help him in the second choice. Because he has knowledge that his chosen door is 1/3, and he also has the same knowledge of the other optional door has a 1/3 chance from the previous round of choices. So this knowledge, that both doors previously had 1/3 chance, is immaterial, useless, not really information. Now he knows they both have a 1/2 chance, and that's all the probability ever is.

The proponents state the fact he has knowledge of his chosen door after the first round (1/3), but they imply that he has no similar knowledge of the second remaining door. That implication is false. The only relevant knowledge would be if he knew something about his door that he DIDN'T know about the other door. Then he could use that knowledge. But he has no such knowledge of the DIFFERENCE between the two doors.

Look at it the other way around; what are the probabilities it is a losing door instead of a winning door? Facts:

• On the first round, he has knowledge of his first chosen door, that it's 2/3 a losing door.
• On the second round, he has knowledge of the remaining door, that it's 1/2 a losing door.

2/3 > 1/2, therefore his first chosen door has a bigger probability that it is a losing door; he better select the other one.

Repeating the process with 100 doors and Monty opening 98 doors makes it even clearer:

• On the first round, he has knowledge of his first chosen door, that it's 99/100 a losing door.
• On the second round, he has knowledge of the remaining door, that it's 1/2 a losing door.