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### A Bayes' Solution to Monty Hall

For any problem involving conditional probabilities one of your greatest allies is Bayes' Theorem. Bayes' Theorem says that for two events A and B, the probability of A given B is related to the probability of B given A in a specific way.

Standard notation:

probability of A given B is written $$\Pr(A \mid B)$$
probability of B is written $$\Pr(B)$$

Bayes' Theorem:

Using the notation above, Bayes' Theorem can be written: $\Pr(A \mid B) = \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)}$Let's apply Bayes' Theorem to the Monty Hall problem. If you recall, we're told that behind three doors there are two goats and one car, all randomly placed. We initially choose a door, and then Monty, who knows what's behind the doors, always shows us a goat behind one of the remaining doors. He can always do this as there are two goats; if we chose the car initially, Monty picks one of the two doors with a goat behind it at random.

Assume we pick Door 1 and then Monty shows us a goat behind Door 2. Now let A be the event that the car is behind Door 1 and B be the event that Monty shows us a goat behind Door 2. Then
\begin{aligned}
\Pr (A \mid B) &= \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)} \\
&= \frac{1/2\times 1/3}{1/3\times 1/2+1/3\times 0+1/3\times 1} \\
&= 1/3.
\end{aligned}The tricky calculation is $$\Pr(B)$$. Remember, we are assuming we initially chose Door 1. It follows that if the car is behind Door 1, Monty will show us a goat behind Door 2 half the time. If the car is behind Door 2, Monty never shows us a goat behind Door 2. Finally, if the car is behind Door 3, Monty shows us a goat behind Door 2 every time. Thus, $\Pr(B) = 1/3\times 1/2+1/3\times 0+1/3\times 1 = 1/2.$The car is either behind Door 1 or Door 3, and since the probability that it's behind Door 1 is 1/3 and the sum of the two probabilities must equal 1, the probability the car is behind Door 3 is $$1-1/3 = 2/3$$. You could also apply Bayes' Theorem directly, but this is simpler.

So Bayes says we should switch, as our probability of winning the car jumps from 1/3 to 2/3.

### Comments

1. This comment has been removed by the author.

2. Doesn't Pr(B) = 1/2?

[Pr(B) = 1/3 x 1/2 + 1/3 x 0 + 1/3 x 1 = 1/2]

3. Hi John! You're correct, of course. I'll correct these calculations now.

4. How to you get Pr(BIA) = 1/2?

Thx

1. This comment has been removed by the author.

5. @Mare
Pr(B|A) is the probability Monty opens Door2 given the car is behind Door1 (the door you picked). Since Monty has a choice of 2 goat doors to open in this scenario, the probability he opens Door2 is 1/2.

6. Given your interest in the quarks, may I ask if you have a view on Bayesian interpretations of quantum theory, such as QBism (http://en.wikipedia.org/wiki/QBism)? As an interested non-mathematician, I'd love to read an explanation of how QBIsm resolves "spooky" quantum theory features such as action-at-a-distance and many-state-reality, especially in the context of developments like quantum key distribution or quantum computing, which appear to use those exactly those features.

1. Statistical/Math Bayes has nothing in common with QBism except for the appellation.

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9. Your error is here: "and since the probability that it's behind Door 1 is 1/3 ".

This is not true. The probability that it's behind Door 1 changes to 1/2 when a third of the probabilities are eliminated by opening Door 2.

Talk about mass hypnosis.

1. "The probability that it's behind Door 1 changes to 1/2 ..."
I would be interested to see a Bayesian explanation for that.

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12. Hi Christopher,

If we define C as Contestant selects Door #1

Isn't the value you calculated as Pr(B) in actuality Pr(B|C)? Shouldn't we use Pr(B) without condition C?

Regards,
Tom

13. This is a rather strange and unnecessary use of Bayes' theorem.
You really didn't need Bayes to conclude that the probability of the car being behind the door you (initially) picked is 1/3. This is given! And in fact you use this same probability as "P(A)".

The fact that Monty does "B" doesn't matter to event "A" (there's either a car behind door 1 or not, no matter what Monty does afterwards (!) with the remaining doors), so "P(A|B)" and "P(A)" must always be equal here.

So your explanation of the Monty Hall problem really just starts being interesting/correct and addressing the problem at
"The car is either behind Door 1 or Door 3, and since ..."

14. quite an easy puzzle

15. quite an easy puzzle

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