[FRIAM] the Monty Hall problem
Angel Edward
edward.angel at gmail.com
Thu Aug 10 11:55:51 EDT 2023
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.angel at gmail.com
505-453-4944 (cell) http://www.cs.unm.edu/~angel
> On Aug 9, 2023, at 9:15 PM, Stephen Guerin <stephen.guerin at 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.
>
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> On Wed, Aug 9, 2023 at 8:46 PM Nicholas Thompson <thompnickson2 at gmail.com <mailto: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] <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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