A even simpler explanation is that your initial pick has a 1/3 chance of 
winning. Nothing has changed as far as that door is concerned. Thus, with host 
removing one of the other two doors, the probability of winning must be 2/3. 

These word problems which involve a priori vs a posteriori probabilities can be 
very tricky. In Contact Bridge, one important variant is known as the principle 
of restricted choice. Suppose you have 9 cards in a suit but are missing the Q 
and the J. Suppose you have the A and K in your hand. You play the A and the 
player on your left drops the Q or J. What do you do next? You can either play 
the K hoping the other Q or J will drop or you can finesse hoping the player on 
your right has the missing Q or J. Most people think the two strategies are 
(approximately) equally probably to succeed but in fact finessing is twice as 
probable to succeed. It’s really frustrating to be playing a competition and 
have the opponents play the losing strategy and luck out.

Ed
__________

Ed Angel

Founding Director, Art, Research, Technology and Science Laboratory (ARTS Lab)
Professor Emeritus of Computer Science, University of New Mexico

1017 Sierra Pinon
Santa Fe, NM 87501
505-984-0136 (home)                     edward.an...@gmail.com
505-453-4944 (cell)                             http://www.cs.unm.edu/~angel

> On Aug 9, 2023, at 9:15 PM, Stephen Guerin <stephen.gue...@simtable.com> 
> wrote:
> 
> I think this might be a more concise explanation:
> 
> Switching wins if you initially pick a goat (2/3 chance) and loses if you 
> pick the car (1/3 chance), so the win probability with switching is 2/3.
> 
> _______________________________________________________________________
> stephen.gue...@simtable.com <mailto:stephen.gue...@simtable.com>
> CEO, https://www.simtable.com <http://www.simtable.com/>
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> office: (505)995-0206 mobile: (505)577-5828
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> 
> On Wed, Aug 9, 2023 at 8:46 PM Nicholas Thompson <thompnicks...@gmail.com 
> <mailto:thompnicks...@gmail.com>> wrote:
>> In a  moment of supreme indolence [and no small amount of arrogance] I took 
>> on the rhetorical challenge of explaining the correct solution of the Monty 
>> Hall problem (switch).   I worked at it for several days and now I think it 
>> is perfect.  
>> 
>> The Best Explanation of the Solution of the Monty Hall Problem
>> 
>> Here is the standard version of the Monty Hall Problem, as laid out in 
>> Wikipedia:
>> 
>> Suppose you're on a game show, and you're given the choice of three doors: 
>> Behind one door is a car; behind the others, goats. You pick a door, say No. 
>> 1, and the host, who knows what's behind the doors, opens another door, say 
>> No. 3, which has a goat. He then says to you, "Do you want to pick door No. 
>> 2?" Is it to your advantage to switch your choice?
>> 
>> This standard presentation of the problem contains some sly “intuition 
>> traps”,[1] <x-msg://68/#m_3313630866437708646__ftn1> so put aside goats and 
>> cars for a moment. Let’s talk about thimbles and peas.  I ask you to close 
>> your eyes, and then I put before you three thimbles, one of which hides a 
>> pea.  If you choose the one hiding a pea, you get all the gold in China.  
>> Call the three thimbles, 1, 2, and 3.
>> 
>> 1.        I ask you to choose one of the thimbles.  You choose 1.  What is 
>> the probability that you choose the pea.   ANS: 1/3.
>> 2.       Now, I group the thimbles as follows.  I slide thimble 2 a bit 
>> closer to thimble 3 (in a matter that would not dislodge a pea) and I 
>> declare that thimble 1 forms one group, A, and thimble 2 and 3 another 
>> group, B.
>> 3.       I ask you to choose whether to choose from Group A or Group B: i.e, 
>> I am asking you to make your choice of thimble in two stages, first deciding 
>> on a group, and then deciding which member of the group to pick. Which group 
>> should you choose from?  ANS: It doesn’t matter.   If the pea is in Group A 
>> and you choose from it, you have only one option to choose, so the 
>> probability is 1 x 1/3.  If the pea is in Group B and you choose from it, 
>> the pea has 2/3 chance of being in the group, but you must choose only one 
>> of the two members of the group, so your chance is again, 1/3:  2/3 x ½ = 
>> 1/3. 
>> 4.       Now, I offer to guarantee you that, if the pea is in group B, and 
>> you choose from group B, you will choose the thimble with the pea. (Perhaps 
>> I promise to slide the pea under whichever Group B thimble you choose, if 
>> you pick from Group B.)  Should you choose from Group A or Group B?   ANS:   
>> Group B.  If you chose from Group A, and the pea is there, only one choice 
>> is possible, so the probability is still 1 x 1/3=1/3.   Now, however, if you 
>> chose from group B, and the pea is there, since you are guaranteed to make 
>> the right choice, the probability of getting the pea is 1 x 2/3=2/3.
>> 5.       The effect of Monty Hall’s statement of the problem is to sort the 
>> doors into two groups, the Selected Group containing one door and the 
>> Unselected Group, containing two doors.   When he then shows you which door 
>> in the unselected group does not contain the car, your choice now boils down 
>> to choosing between Group A and Group B, which, as we have known all along, 
>> is a choice between a 1/3 and a 2/3 chance of choosing the group that 
>> contains the pea. 
>> 
>> 
>> [1] <x-msg://68/#m_3313630866437708646__ftnref1> The intuition trap has 
>> something to do with the fact that doors, goats, and cars are difficult to 
>> group.  So, it’s harder to see that by asking you to select one door at the 
>> beginning of the procedure, Monty has gotten you the group the doors and 
>> take the problem in two steps.  This doesn’t change the outcome, but it does 
>> require us to keep the conditional probabilities firmly in mind. “IF the car 
>> is in the unselected group, AND I choose from the unselected group, and I 
>> have been guaranteed to get the car if I choose from the unselected group, 
>> THEN, choosing from the unselected group is the better option.”
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