Re: [FRIAM] the Monty Hall problem

[Stephen Guerin]

On Wed, Aug 9, 2023 at 9:19 PM Eric Smith  wrote:

> Wow.  Old terminator.  New terminator
>

Which one works for Skynet?  The fight scene:
https://youtu.be/TVA4-SNxErc?t=41
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Re: [FRIAM] the Monty Hall problem

[Eric Smith]

Wow.  Old terminator.  New terminator.

> On Aug 10, 2023, at 12:15 PM, Stephen Guerin  
> 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 
> CEO, https://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  > 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]  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]  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 

Re: [FRIAM] the Monty Hall problem

[Stephen Guerin]

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 
CEO, https://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 
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/
>
-. --- - / ...- .- .-.. .. -.. / -- --- .-. ... . / 

[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.

--

[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.”
-. --- - / ...- .- .-.. .. -.. / -- --- .-. ... . / -.-. --- -.. .
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
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