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Good old Monty Hall!
http://wmbriggs.com/blog/?p=438 ^ | William Briggs

Posted on 03/18/2009 4:04:29 AM PDT by mattstat

It’s about time we did the Monty Hall problem.

Many of you will already know the answer, but read on anyway because it turns out to be an excellent example to demonstrate fundamental ideas in probability.

Incidentally, I just did this yesterday to a group of surgical residents: you might be happy to know that none of them got the right answer. One even insisted—for a while—that I was wrong.

Here’s the problem.

Setup: Monty Hall shows you three doors, behind one of which is a prize, behind the other two is nothing. Monty knows which door hides the prize. You want that prize.

Rule: You pick a door that Monty will open. You are free to change your mind as often as you like.

The Pick: I’ll suppose, since I can’t quite hear you, that you picked A (the resident yesterday took that one) and did not change your mind.

Question...

(Excerpt) Read more at wmbriggs.com ...


TOPICS: Science
KEYWORDS: letsmakeadeal; montyhall
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To: Nateman
You are correct as far as what my chances were prior to having any door opened (and allowing me, at every oppurtunity to select another).

However, once that first door is opened, my chance of picking the correct door (either remain with mine, or choosing a new one) is now 1 to @-1. This continues until we get down to the last two doors - mine and an unknown. Regardless of how many doors Monty has opened - my chance of choosing the correct door (either the one I have, or the one offered) is 1 in 2.

So, I agree that, before Monty starts opening his doors, my chance of having picked the correct on, is 1-@. I also agree that, once the game is over, my chance of having picked it right, from the beginning is 1-@. However, my chance, on the final decision (stay with Door #1, or take Curtain #3) is 1 in 2.

This is not that different than the old coin toss question ...

Or maybe I'm still missing a key piece of the puzzle?

I'm being honest - if I'm missing something I'd like to understand it!

21 posted on 03/18/2009 9:10:49 AM PDT by An.American.Expatriate (Here's my strategy on the War against Terrorism: We win, they lose. - with apologies to R.R.)
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To: Lancey Howard

Please pardon my “rude ping” - but you seem to understand and agree with the original premise.

If so, would you mind explaining it a bit ?


22 posted on 03/18/2009 9:16:34 AM PDT by An.American.Expatriate (Here's my strategy on the War against Terrorism: We win, they lose. - with apologies to R.R.)
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To: An.American.Expatriate
"...Why is there a 2/3 chance his choice is constrained?

If I made the correct (first) choice, he is not constrained, If I made the wrong (first) choice, he is constrained

From “my” perspective there is a 1 in 2 chance, AFTER my selection, that Monty was “forced” to open a specific door. BEFORE my selection there is a 2 in 3 chance of “forcing” Monty - which is simply the inverse of my probablity of having chosen correctly..."


Please allow me to explain:

The PROBABILITY of an event occuring is the fraction of:

(possible desired outcomes)
---------------------------
(all possible outcomes)


Now, we want to know the PROBABILITY that Monty's choice is CONSTRAINED. Given that there are 3 doors, and the prize is behind only one of them, we get this fraction:

(I can pick 2 doors that have NO Prize)
-----------------------------------------
(there are 3 doors total)


that is 2/3.

Now, the way the game is played, Monty will ALWAYS open an EMPTY door. If he opened the WINNING door, the game would not be "fun".

Given this ADDITIONAL bit of information, you are STATISTICALLY BETTER OFF switching your choice. Monty CANNOT open a door at RANDOM, and THAT is the addional information that changes the probability computations.
23 posted on 03/18/2009 9:18:08 AM PDT by Rebel_Ace (Tags?!? Tags?!? We don' neeeed no stinkin' Tags!)
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To: An.American.Expatriate

You are hung up on the idea that the odds for a choice already made change upon learning more information. What’s done is done. Suppose you pick an apple from a barrel of apples and that apple has a worm in it. Does that worm go away because you are now given the choice of much more reliable apples?


24 posted on 03/18/2009 9:39:29 AM PDT by Nateman (You know you are doing the right thing when liberals scream.)
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To: Rebel_Ace

I think I see it - help me just a bit more here.

If I understand you correctly, I need to be playing the game “in reverse” - i.e. betting that I constrained Monty on my first pick as I have better odds (2/3) of doing so. And, if I did constrain him, my first pick is obviously false, thus I must switch.

So, did I find the Big Deal of the Day?


25 posted on 03/18/2009 9:44:25 AM PDT by An.American.Expatriate (Here's my strategy on the War against Terrorism: We win, they lose. - with apologies to R.R.)
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To: Nateman
You are hung up on the idea that the odds for a choice already made change upon learning more information.

Not really hung up, just trying to see the correct angle.

Suppose you pick an apple from a barrel of apples and that apple has a worm in it. Does that worm go away because you are now given the choice of much more reliable apples?

Not really what we are talking about. Once I know that the aplle has a worm - it is there. We are discussing an unknown condition.

A better example is the coin toss. My odds of flipping "heads" is 1/2. My odds of doing that 4 times in a row is 1/2*1/2*1/2*1/2 or 1/16. However, once I have flipped the first heads (1/2 chance), the chance that the next one is heads is 1/2, and that of continueing the series to the end is 1/2*1/2*1/2 or 1/8. As you said, what is done is done.

After I have done this 1000 times - the chance that the 1001st one will be heads is 1/2. The chance foresseing that this would happen however is extremely small.

26 posted on 03/18/2009 9:53:42 AM PDT by An.American.Expatriate (Here's my strategy on the War against Terrorism: We win, they lose. - with apologies to R.R.)
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To: An.American.Expatriate

Let me try a different approach. Your chance of getting it wrong originally was 2 out of 3. Monty removes 1 of the 3. If you do nothing it’s like you never saw anything Monty showed you. What is the only possible way to use the information Monty just gave you? You have to switch. Then your odds of losing become 2/3 -1/3 = 1/3 (Hence your odds of winning become 2/3)


27 posted on 03/18/2009 10:13:27 AM PDT by Nateman (You know you are doing the right thing when liberals scream.)
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To: An.American.Expatriate
"...I need to be playing the game “in reverse” - i.e. betting that I constrained Monty on my first pick as I have better odds (2/3) of doing so..."

BINGO! That is it precisely.

BTW, I staggered into this thread because my name is "Monty".
28 posted on 03/18/2009 11:31:58 AM PDT by Rebel_Ace (Tags?!? Tags?!? We don' neeeed no stinkin' Tags!)
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To: An.American.Expatriate
Here is an easy min-math way of viewing the problem.

When you make your original choice, ie., select one of three doors, you chance of selecting the door with the prize is 1 in 3 (33%). That is mathematically irrefutable.

Now, AFTER Monty opens an empty door, there are only two doors remaining that could be the winning door and so your "choice" is now between TWO doors, not THREE, making your chances of selecting the winning door 50%, rather than only 33%. If you stay with your original selection, then you are staying with a selection you made when your chance of selecting the winner was 1 in 3. But if you switch, then you are making a new selection where your chance of winning is 1 in 2. So simply by virtue of switching you are improving your odds.

Again, it has been supported with hard empyrical data that when you switch you will have a 2 in 3 chance of winning. If you have an hour or so to kill, grab a friend and go ahead and run your own test. (Forget about all that "constraint" stuff.)

Why is your chance of winning 2 out of 3 (66%) when you switch? That's easy... think backwards.

When you made your original selection the chance that the winner was one of the OTHER two doors was 2 out of 3, right? Now, "Monty" has done you a big favor by eliminating one of those two doors! AND THAT ONE REMAINING DOOR - - the one you switched to after Monty eliminated one - - STILL has that same 2 in 3 chance of being the winner, only now it's the ONLY door between those other two doors you originally failed to select that could be the winner because "Monty" helped you out (big time!) by eliminating the definite loser between the two!

(I hope you followed this rambling explanation.)

FRegards,
LH

29 posted on 03/18/2009 3:08:43 PM PDT by Lancey Howard
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To: Lancey Howard; Rebel_Ace; Nateman

Thanks to each of you for taking the time to explain.

After figuring it out back at 25 - I did some more thinking about it, i.e. “the odds”.

Lets see if this thinking is correct:

Betting on “I picked the winner” involves two choices

1/3 and 1/2 - or odds of 1/6

Betting on “I picked the loser” involves two choices

2/3 and 1/2 or 2/6 = 1/3

Thus, by playing the game that way, may total chance is 1 in 3 of taking home the prize.


30 posted on 03/19/2009 12:40:14 AM PDT by An.American.Expatriate (Here's my strategy on the War against Terrorism: We win, they lose. - with apologies to R.R.)
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To: An.American.Expatriate
Thus, by playing the game that way, may total chance is 1 in 3 of taking home the prize.

Well, yeah, naturally you're chance is 1 in 3 if you pick one of three doors and keep your fingers crossed. But, of course, once you get the additional information ("Monty" revealing a losing door) and are allowed to switch your choice to the one remaining door (which is what you clearly should do), then your "total chance" is 2 in 3 of taking home the prize.

FRegards,
LH

31 posted on 03/19/2009 1:16:46 AM PDT by Lancey Howard
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To: Lancey Howard; Nateman

I think you misunderstood (or - more likely I failed to explain well!!).

Let me try it again ...

When I make my initial pick, I have a 1/3 chance of picking the winner and a 2/3 chance of picking the loser.

Since I get a second pick after Monty kindly eliminates one of the choices, my chance of picking the winner *in this choice* is 1/2.

Now, asumming I stay with my first pick, my chance of picking it right on the first choice were 1/3 and then 1/2 - thus my chance *from the beginning* is 1/6.

OTOH, If I switch (because I assume I chose wrong), my
chance is 2/3 and 1/2 or 2/6 (1/3).

IOW, If I start the game with “I will stick with my first pick” - I have a 1 in 6 chance of taking home the prize.
If I play as “I will switch” I have a 1 in 3 (total) chance.

Whether the “math” is correct, or not - I now understand the principle - either way my chances are far greater by switching.

BTW Nateman - this validates your argument from earlier - my problem was the “angle of attack” on the problem.

By assuming from the beginning that I made the wrong choice - statistically MORE probable than assuming I made the correct one - my probability of being correct one the last choice (switch) is far greater.

Again - thanks to all.


32 posted on 03/19/2009 2:07:43 AM PDT by An.American.Expatriate (Here's my strategy on the War against Terrorism: We win, they lose. - with apologies to R.R.)
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To: An.American.Expatriate
IOW, If I start the game with “I will stick with my first pick” - I have a 1 in 6 chance of taking home the prize.

Huh? There's 3 choices and you pick 1. If you stick with that choice that's a 1 in 3 chance no matter what, and that's mathematical reality.

If I play as “I will switch” I have a 1 in 3 (total) chance.

Sorry, but you can see right away that that makes no sense either. Why would your (total) chance remain 1 in 3 (ie., exactly the same as it was at the very start) even after you are given additional information and switch choices?

No. If you play the game as described, including getting additional information AFTER you have made your first choice (ie., a definite loser door is revealed to you) and then you switch your choice to the last door, your total chance of taking home the prize is 2 in 3.

33 posted on 03/19/2009 2:21:59 AM PDT by Lancey Howard
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To: Lancey Howard

My second choice has “odds” as well, and IIRC, those odds must also be calculated into the overall odds.


34 posted on 03/19/2009 2:32:24 AM PDT by An.American.Expatriate (Here's my strategy on the War against Terrorism: We win, they lose. - with apologies to R.R.)
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To: mattstat

Idiotic, but good to generate hits.


35 posted on 03/19/2009 2:38:21 AM PDT by Petronski (For the next few years, Gethsemane will not be marginal. We will know that garden. -- Cdl. Stafford)
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