**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.

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ReplyDeleteDoesn't Pr(B) = 1/2?

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

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

ReplyDeleteHow to you get Pr(BIA) = 1/2?

ReplyDeleteThx

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Delete@Mare

ReplyDeletePr(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.

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.

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteThis comment has been removed by the author.

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

ReplyDeleteThis 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.

"The probability that it's behind Door 1 changes to 1/2 ..."

DeleteI would be interested to see a Bayesian explanation for that.

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DeleteActually, that wasn't his error. p(A) is the a priori probability of the car being behind door 1, and equals 1/3.

DeleteHis error is p(A|B) which actually equals 1/2. What he's really seeking is

p(A|B & P), where P is the event that the player chooses door 1; then the math adds up.

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ReplyDeleteThis comment has been removed by the author.

ReplyDeleteHi Christopher,

ReplyDeleteIf 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

This is a rather strange and unnecessary use of Bayes' theorem.

ReplyDeleteYou 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 ..."

quite an easy puzzle

ReplyDeletequite an easy puzzle

ReplyDelete