Well the title of this post is a bit inflammatory. So I won’t be arguing that it “refutes” your religion, but will be arguing more that it’s weak Bayesian evidence against your religion.

So. The Monty Hall problem is an illustration of how our intuitions of probability don’t always match up with reality. In its original formulation, you’re given a choice between three doors. One door has a prize, the other two do not. If you choose one of the doors, then another door that doesn’t have a prize is shown to you. You then have the option of staying with the door you chose or switching doors.

Most people think that it either doesn’t matter whether you switch or that switching lowers your probability of winning. Neither of those is true!

Your initial probability of winning the prize is 1 out of 3. Once one of the doors is opened, the probability that you had picked the correct door stays at 1 out of 3 whereas the other non-picked door now contains the remaining probability of 2 out of 3. Because you have to do a Bayesian update once new information — in this case, the one door revealed to not have the prize — is introduced.

I’ve gone over this before. Yet, I want to add an additional wrinkle to the problem to make intuition fall more in line with Bayesian reasoning.

If, instead of picking one door out of three to win the prize, what if it were one door out of 100? And once you’ve made your selection, 98 other doors are opened up to show that they have no prize, leaving only your choice and one other unknown door? In this case it seems more obvious that something is suspicious about the only other door that wasn’t opened up. And this intuition lines up with a Bayesian update using the same scenario:

P(H): 1 out of 100 or .01

P(~H): 99 out of 100, or .99

P(E | H): Probability of all other doors besides yours and one other being opened to reveal no prize given that you’ve picked the correct door: 100%.

P(E | ~H): Probability of all other doors besides yours and one other being opened to reveal no prize given that you’ve picked the *incorrect door *is 100%.

This is an easy Bayesian update to do. Both conditional probabilities, P(E | H) and P(E | ~H) are both 100%. Meaning the likelihood ratio is 1, and your posterior probability is the same as your prior probability. But now your selection is still 1 out of 100 and the only other remaining door has a probability of 99 out of 100 of having a prize! So in this case, both Bayesian reasoning and intuition line up: There is something suspicious about the only other door that wasn’t opened.

How does this relate to religion? Specifically, the religion that you grew up with?

Using Willy Wonka’s logic in the meme above, the chance that you just happened to grow up with the correct religion is pretty low. Instead of the chance of picking the correct door out of 3, or out of 100, you’ve picked a door out of thousands of religions; many of which no longer exist. They are “opened doors” revealing no prize in the analogy.

So a Bayesian update will work the same way as it did with picking one door out of 100. Meaning, your religion is probably wrong. And you should probably switch religions. The only reason I say this is weak Bayesian evidence is because there are still a few religions to choose from. But their joint probability of being correct is yet higher than the single chance that your family religion is the correct one.

Analogously, it would be like if, say, you had a choice between choosing one door out of 10,000, and after your choice all but 10 of the doors are closed. Your initial chance of having chosen the correct door is still 1 out of 10,000, but the 10 doors that remained open after closing the rest have a joint probability of 9,999 out of 10,000 of being the correct door: Those 10 other doors individually have (approximately) 10% chance of being the correct door. As opposed to your original selection’s probability of 1 out of 10,000.

So the Monty Hall problem is weak Bayesian evidence against your religion.

peter cresswell

March 7, 2018 at 3:43 am

I agree with your logic. There is a better chance now in going with the other non-picked door, as your initial chance was only in three. But surely, with two doors remaining unopened and one of these having a prize, the now-better chance is one in two rather than two in three?

Marcel Kincaid

April 28, 2018 at 8:34 pm

“But surely, with two doors remaining unopened and one of these having a prize, the now-better chance is one in two rather than two in three?”

Um, no, and way to go completely ignoring everything written in this article. Your nearly explicit fallacy here is the baseless belief that, given two options, they are equally likely. But the two doors are not at all the same, due to how Monty Hall deals with them.

Again: Suppose that there are 100 doors. You pick one. MH then opens 98 doors showing you 98 goats, as you *knew* he would because of the conditions of the puzzle. The probability that the car is behind your door is 1/100, *as it always was*, because you have received no new information. And the probability that the car is behind one of the 99 other doors is 99/100, *as it always was*, because you have received no new information. But 98 of those doors stand open, so you know the car isn’t behind any of those. So the entire 99/100 accrues to the other door.

Marcel Kincaid

April 28, 2018 at 8:43 pm

P.S. Critical to this answer is the condition that Monty Hall does not open doors randomly … that he purposefully avoids showing you the car. If he does open doors randomly, and he happens to open 98 doors without revealing a car (the probability of that happening is 1/50), then indeed the chances that the car is behind the two remaining doors is each 1/2. In this case, each time he opens a door with a goat behind it, that provides information that increases the probability that the car is behind your door. (The first door opened increases it from 1/100 to 1/99, the second door opened increases it from 1/99 to 1/98, and the 98th door opened increases it from 1/3 to 1/2.)

Marcel Kincaid

April 28, 2018 at 9:02 pm

P.P.S. One might wonder, given these two different probability outcomes for two slightly different problem statements, how the Monty Hall problem might be any sort of evidence against one’s religion … the claim seems *suspicious*. Indeed, the claim is completely bogus — it’s a category mistake; the Monty Hall problem isn’t the sort of thing that can be Bayesian evidence for anything. *Understanding* and *accepting* the Bayesian analysis of the Monty Hall problem might give someone an appreciation for the validity of Bayesian analysis and thus new information, but few people do that and even then they could come to the opposite conclusion … If, say, there were some reason to think that more popular gods are more likely to exist (see, e.g., Terry Pratchett’s “Small Gods”), then some gods and religions are more likely to be correct than others. Or if, say, the holy texts of some religions contain predictions that seem implausible without foreknowledge, that would make those religions more likely to be correct than others. Or if the holy books of some religions seem to display consilience — the same claims deriving from disparate sources — that would make those religions more likely to be correct than others. And there are people who believe each of those things. (FWIW, I’m not one of them.)