Monty Hall Problem, as discussed by a 5 year old Illuminati

October 1, 2014

Monty Hall Problem, as discussed by a 5 year old Illuminati

Filed under: Uncategorized — rrtucci @ 9:52 pm

A friend just sent me a link to the following NYT article:

The Odds, Continually Updated by Flim D. Flam (would I lie to you?) Sept. 29, 2014, New York Times

Even though the article doesn’t mention the obvious fact that quantum computers will someday soon revolutionize the Bayesian field (because they will be able to do Bayesian calculations much faster than classical computers), it’s not such a hopelessly outdated, clueless piece of reporting. Really. The article doth sing the praises of Bayesian techniques, so it ain’t that utterly bad.

The article makes a big deal about the Monty Hall problem. It’s a cute problem, for a five years old. Correction: A five year old that has been taught Bayesian networks. We B net advocates teach our children Monty Hall on day one, of pre-kindergarten. It’s quite conceivable that an intelligent 50 year old kid who has been taught only the frequentist canon might spend hours on such a problem, and ultimately give up.

Here is how a 5 year old illuminati who is conversant with the language of B nets would discuss such a piddling problem:

Let $\theta({\cal S})$ stand for the “truth function”. It equals 1 if statement ${\cal S}$ is true and 0 otherwise. For example, $\theta(a=b)=\delta_a^b$ is the Kronecker delta function.
monty

The Monty Hall problem can be modelled by the B net shown, where

$c=$ the door behind which the car actually is.
$y=$ the door opened by you (the contestant), on your first selection.
$m=$ the door opened by Monty (game host)

If we label the doors 1,2,3, then $m,c,y\in \{1,2,3\}$ and

$P(c)=\frac{1}{3}$

$P(y)=\frac{1}{3}$

$P(m|c,y)=\frac{1}{2}\theta(m\neq c)\theta(y=c)\;\;\;+\;\;\;\theta(m\neq y,c)\theta(y\neq c)$

It’s easy to show that the above node probabilities imply that

$P(c=1|m=2,y=1)=\frac{1}{3}$

$P(c=3|m=2,y=1)=\frac{2}{3}$

So you are twice as likely to win if you switch your final selection to be the door which is neither your first choice nor Monty’s choice.

Comments (2)

2 Comments »

Hello Bob: This problem was stated in a book by the University of Toronto mathematician, Jeffrey S. Rosenthal, titled “STRUCK BY LIGHTNING-The curious World of Probabilities” On page 212, he relates the story of a gifted woman by the name of Marilyn vos Savant who had written an article in Parade magazine recommending that contestants have twice the chance of winning the car if they switched. He writes: “Unexpectedly, vos Savant’s column ignited a storm of controversy, and thousands of letters were received. Mathematicians from such institutions as George Mason University, the University of Florida, the University of Michigan, and Georgetown University wrote in to complain that vos Savant’s answer was incorrect, saying that she “blew it big,” that her “logic is in error,” and that she is “utterly incorrect.” One intemperate academic even wrote that vos Savant herself was the goat!” My question is this: How did these mathematicians get hired by their respective universities? I find that very odd for such a simple problem!.

Comment by Sol Warda — October 2, 2014 @ 12:55 am
That’s a great story Sol. I didn’t know it. That reminds me of one of the reasons why I like B nets so much; they are a nice way of getting the kinks out of faulty probabilistic reasoning

Comment by rrtucci — October 2, 2014 @ 2:26 am

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October 1, 2014