[FRIAM] the Monty Hall problem

[Nicholas Thompson]

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] <#_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.

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[1] <#_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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