The Monty Hall Problem is a notorious puzzle in probability. I will assume you are familiar with it. The well known conclusion is that you should switch, which will give you a 2/3 probability of winning.

However, with different assumptions, the answer turns out differently (see e.g. David MacKay’s two analyses in his book [he calls the problem “three doors”]). The “switch” answer depends crucially on the assumption of a zero prior probability for the host revealing a car when he opens a door. If you think it’s possible that the host will reveal the car, there is no advantage to switching.

The purpose of the host opening a door is presumably to create drama and make the show more exciting. If it is known in advance that the host will not reveal the car, there is no drama, so the hosts’s behavior is pointless! The premise that “the host is now going to open a door to build suspense” implies we think he may reveal the car, which implies the 2/3 answer is wrong.

I am a senior lecturer in the Department of Statistics at The University of Auckland. Any opinions expressed here are mine and are not endorsed by my employer.
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### 4 Responses to Monty Hall Contradicts Itself

1. Cusp says:

In the Monty Hall problem, the host *does* know the location of the car.

• Yes, he does. However, the question is about whether the contestant should switch, which depends on the state of knowledge of the contestant, not the host.

2. Cusp says:

The contestant knows as well – he would have watched a bunch of episodes before hand and would have seen that the host never reveals the car – I feel a new blog post coming on….

• Haha. It’s feeling a bit like groundhog day.

“The contestant knows as well – he would have watched a bunch of episodes before hand and would have seen that the host never reveals the car”

If that’s the case, then the 2/3 answer is correct, but the show is boring.