The host simply reveals a door that doesn't contain the prize after the contestant chooses a door that s/he believes contains the prize.

Your scenario #4 looks like the same scenario I put for my scenario #2 in my previous comment and shouldn't exist.

There is no scenario #4....

There are only 3 possible scenarios if the prize is behind door #2.

1. Scenario 1:

You pick door 1 and switch and win

2. Scenario 2:

You pick door 2 and switch and lose

3. Scenario 3:

You pick door 3 and switch and win

You can't pick more than three doors..

Hence only 3 scenarios if the prize is behind door #2.

So, you win twice, once in scenario 1 and again in scenario 3 if you chose to switch. So this means you would win 2 out of 3 times or 2/3rd of the time, which is about 66.66%.

The math is sound on this one bkuehlhorn, this does not give the contestant only a 50% chance of winning, this gives the contestant of this game a 2/3=66.66% chance of winning by switching.

Also notice the graph in the article, you would expect the percentage of wins to be much closer to 0.5 than 0.66 if the chance of winning was truly 50% . Instead, we see that the chances of winning are much closer to 0.66 from a larger sample size (a.k.a. the 2000 test games played).

Any ways, I hope this helps. If you are still not convinced, then I suggest reading “The Monty Hall Problem: The Remarkable Story of Math’s Most Contentious Brain Teaser” or you can use Bayes Theorem to prove the statistics.