Subject: The monty hall problem
Date: Wed, 18 Apr 2001 12:23:06 +0200
From: Peter Stikker

Dear madam/sir,

As a fourth year student at the Educational faculty of Amsterdam (The Netherlands), studying math-teacher, I found your website very usefull over the years.

I recently took some interest in the Monty Hall problem (a.k.a. the goat problem) and found no satisfactory explenations of the answer. Sure they are all correct but "non-mathematical" friends of mine weren't really convinced by these "proofs". So I started to try to proof it myself. I went back into my very elemantery books on probability and decided to draw a probability tree well the results are attached as a word-2000 document.

If you find this usefull enough to put on your website please feel free to use it and correct my English (I am dutch).

Hope to hear from you soon.

Your sincerely,
P.H. Stikker

The Monty Hall Problem

When I read about this problem for the first time, I could not believe it (like many of us). I started looking for some proofs and finally found some on the internet and in some books, but none where really easy or not satisfactory (? Sorry for the bad English, I'm Dutch). As I am studying to become a math-teacher I wanted an easy to understand proof and started making a probability tree. This gave the following result:

But what happened? If I add up the "Wins" and add up the "Looses" they are both 6. Then I remembered that I forgot to put in the chances. So I added those with the following result:

Now is the chance of loosing: L = 1/3 · 1/3 · 1/2 · 1 · 1 = 1/18

And I got 6 L's so: Total chance for loosing is: 6· 1/18 = 1/3

For winning: W = 1/3 · 1/3 ·1 · 1 · 1 = 1/9

And I got 6 W's so: Total chance for winning is: 6 · 1/9 = 2/3. (check: 2/3 + 1/3 = 1 (OK))

I persuaded lots of people with this solution and hope it also helped you.

Your sincerely,
P.H. Stikker

Monty Phister found a simplified diagram even more illuminating:

  simplified tree diagram

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