This helps me understand the Monty Hall problem...

Miscellaneous and Personal Stuff I Must Share
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[If you’re not familiar with the “Monty Hall problem,” see below links.]

When I first encountered the Monty Hall problem, I answered incorrectly–like many people, even math professors.

I reasoned: After the “dud prize” door is opened, what’s left? Two doors, one with the prize, the other with a dud. Obviously, a 50-50 probability for each door. Wrong!

This explanation from Cecil Adams’s column helped me understand:

“A friend of mine did suggest another way of thinking about the problem that may help clarify things. Suppose we have the three doors again, one concealing the prize. You pick door #1. Now you’re offered this choice: open door #1, or open door #2 and door #3. In the latter case you keep the prize if it’s behind either door. You’d rather have a two-in-three shot at the prize than one-in-three, wouldn’t you? If you think about it, the original problem offers you basically the same choice. Monty is saying in effect: you can keep your one door or you can have the other two doors, one of which (a non-prize door) I’ll open for you. Still don’t get it? Then at least have the sense to keep quiet about it.”

For me, the below explanation makes it even clearer that by switching, the player in effect gets two doors instead of only one:

Three doors, behind which:
-One has a check for $100,000
-one has a penny
-one has a nickel

Monty Hall knows what’s behind each door, and he always opens either the penny or the nickel door.

Player chooses door #1.

Three possibilities: Door #1 has penny, nickel, or $100k

–Door #1 has penny, Monty opens the nickel door, player switches, wins $100k
–Door #1 has nickel, Monty opens the penny door, player switches, wins $100k
–Door #1 has $100k, Monty opens either penny or nickel door, player switches, loses.

So in two of three cases, the player wins by switching.

Cecil’s column:

http://www.straightdope.com/columns/read/916/on-lets-make-a-deal-you-pick-door-1-monty-opens-door-2-no-prize-do-you-stay-with-door-1-or-switch-to-3

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the best way of explaining it is to first of all accept the the person offering to open the door knows what is behind each of them. That has to be true for any of this to make sense.

Now imagine that instead of three doors there are a hundred. You choose a door and the host begins opening 98 of the other doors, there is nothing behind any of them.
Now he offers you to swap for the remaining unopened door.

Do you swap? yes of course you do.

The three door problem offers a greater chance that you picked correctly yourself but the principle is the same.

3

A lot of the “Math professors got it wrong” stuff comes from people not stating the problem right.

Does Monty know which door has the prize? Does he open a door at random? Does he offer the switch regardless of which door you choose. Etc.

You omit or misstate one of the conditions and you allow different interpretations with different correct answers.

I think the best explanation is a combo of multidoor grouping and 100 doors.

You pick a door. Monty gives you the choice of sticking with that door or getting everything behind the other 99 doors. What do you do?

(Why some people fail to understand this problem with resorting to computer simulations is beyond me.)

4

The fact that the host knows what’s behind each door is key. He doesn’t open a door at random, otherwise, sometimes he would open a door with the big prize behind it. The fact that this NEVER happens is what alters the odds.

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There are 3 doors. One winner 2 losers. You pick #1. Monty knows the winner door. He of course opens a dud door to build suspense. Offers you the chance to switch. You reason that your original pick had a 1/3 chance, if you switch its 1/2. New wrinkle. Monty brings out another contestant who was in the back, unaware of what’s gone on. Monty gives him a choice of the 2 remaining doors doors. He also picks #1. His odds on #1 is 50/50. Your odds on staying with #1 is 1/3. Same door. Different odds. How? Lets do 1000 doors. 999 duds Monty knowingly opens 998 doors that are a dud, You bet I’m switching. However. if Monty opens them randomly, I think its 50/50. Would you bet I could flip 20 heads in a row, probably not. But if I did 19 the 20th is now just 50/50. When its controlled the odds stay the same as the original. When its random the odds change.

6

That happened with the plane on the treadmill one. I read it (or assumed) as where the proponents were saying that it would take off while stationary, which of course was absurd (as in takeoff with no airflow over the wings, which is impossible). A few posts later I realized that the plane was allowed to move forward (but the original writeup never made that clear).

Later on someone raised the issue of friction on the wheels, and in that scenario if the treadmill was capable of speeding up sufficiently quickly to very high speeds it could counteract the friction and prevent takeoff.

A third scenario then was devised where the treadmill itself would as it zipped on by generate a sort of ground effect and allow even stationary takeoff (if the plane was an ultralight or such).

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The trouble with explaining it is getting past the incorrect assumptions people have starting out. Much of that is not understanding the problem.

I can reduce this down pretty well to one question: Would you rather have the prize behind one door, or the grand prize if it’s behind either of the other two doors. From that point the explanation gets a little more complicated. However, it’s pretty intuitive that you have a 1 in 3 chance of the prize being behind the one door you pick, and a 2 in 3 chance of it being behind one of the other two doors, and the unopened door of the other two doors has a 2 in 3 chance of being the grand prize no matter which of the two is opened, so you switch to put the odds in your favor.

8

Yet another person dropping in to say that what I didn’t get at first is that Monty Hall knows what’s behind each door, and always opens one that doesn’t have the prize. Somehow, the way the problem is stated doesn’t make that clear.

9

Math prof here (retired). When I first heard about the problem (in the form of a “look what stupidity Marilyn has pulled now” email from my son) I agreed with my son at first. Fifteen minutes later, I emailed him that in this case she was right.

Here’s my explanation. First off, I assumed that Monty would always choose a door that didn’t contain the prize. I redefined the question as choosing a losing door. Obviously, I have 2 chances in 3 of being successful in that task. Assuming I succeed and always switch, I win. If I didn’t succeed I lose. So by playing always switch I win 2 times in 3. If I never switch, I win 1 out of 3.

Incidentally, if there 100 doors and only one is opened you should still switch. Your don’t improve your odds much, only from 1 in a 100 to 1 in 99, but they do improve. If you pay $1 to play and the payoff is $100 you will win in the long run.

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The problem assumes real-world knowledge. Many Americans were well aware of Monty’s TV show and could observe his door offers (or infer what his policy would be). Others might have treated it as an abstract logic puzzle, with Hall a fictitious character — then the precise wording of the problem becomes important.

I don’t think the real Monty Hall always opened a door; we’d need to know the correlation between that refusal and prize location to make an exact assessment. (And did he ever expose the door with the grand prize?)

Monty Hall himself was interviewed about the Monty Hall problem. Of course he knew contestant should switch!