[FRIAM] the Monty Hall problem

Stephen Guerin stephen.guerin at simtable.com
Wed Aug 9 23:15:03 EDT 2023


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.Guerin at Simtable.com <stephen.guerin at simtable.com>
CEO, https://www.simtable.com <http://www.simtable.com/>
1600 Lena St #D1, Santa Fe, NM 87505
office: (505)995-0206 mobile: (505)577-5828


On Wed, Aug 9, 2023 at 8:46 PM Nicholas Thompson <thompnickson2 at 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] <#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] <#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.”
> -. --- - / ...- .- .-.. .. -.. / -- --- .-. ... . / -.-. --- -.. .
> FRIAM Applied Complexity Group listserv
> Fridays 9a-12p Friday St. Johns Cafe   /   Thursdays 9a-12p Zoom
> https://bit.ly/virtualfriam
> to (un)subscribe http://redfish.com/mailman/listinfo/friam_redfish.com
> FRIAM-COMIC http://friam-comic.blogspot.com/
> archives:  5/2017 thru present
> https://redfish.com/pipermail/friam_redfish.com/
>   1/2003 thru 6/2021  http://friam.383.s1.nabble.com/
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://redfish.com/pipermail/friam_redfish.com/attachments/20230809/f87ea275/attachment.html>


More information about the Friam mailing list