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The Monty Hall Problem

LostInTyme

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From Wikipedia

The Monty Hall problem is a brain teaser, in the form of a probability puzzle, loosely based on the American television game show Let's Make a Deal and named after its original host, Monty Hall. The problem was originally posed (and solved) in a letter by Steve Selvin to the American Statistician in 1975. It became famous as a question from reader Craig F. Whitaker's letter quoted in Marilyn Vos Savant's "Ask Marilyn" column in Parade magazine in 1990:

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?

Assumptions: The role of the host as follows:
  1. The host must always open a door that was not picked by the contestant.
  2. The host must always open a door to reveal a goat and never the car.
  3. The host must always offer the chance to switch between the originally chosen door and the remaining closed door.
Vos Savant's response was that the contestant should switch to the other door. Under the standard assumptions, the switching strategy has a 2/3 probability of winning the car, while the strategy that remains with the initial choice has only a 1/3 probability.

For the record, many PhDs and Nobel laureates disagreed with Vos Savant's conclusion. In her favor is her 228 IQ, the highest of any woman.

What is your conclusion?
 

KILO NOVEMBER

Practically Family
Messages
969
Location
Hurricane Coast Florida
From Wikipedia

The Monty Hall problem is a brain teaser, in the form of a probability puzzle, loosely based on the American television game show Let's Make a Deal and named after its original host, Monty Hall. The problem was originally posed (and solved) in a letter by Steve Selvin to the American Statistician in 1975. It became famous as a question from reader Craig F. Whitaker's letter quoted in Marilyn Vos Savant's "Ask Marilyn" column in Parade magazine in 1990:

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?

Assumptions: The role of the host as follows:
  1. The host must always open a door that was not picked by the contestant.
  2. The host must always open a door to reveal a goat and never the car.
  3. The host must always offer the chance to switch between the originally chosen door and the remaining closed door.
Vos Savant's response was that the contestant should switch to the other door. Under the standard assumptions, the switching strategy has a 2/3 probability of winning the car, while the strategy that remains with the initial choice has only a 1/3 probability.

For the record, many PhDs and Nobel laureates disagreed with Vos Savant's conclusion. In her favor is her 228 IQ, the highest of any woman.

What is your conclusion?
As many times as I've read or heard this problem, I still can't see why each of the remaining doors has anything other than a 50% probability of hiding the car.

You start off with a 1/3 probability of choosing the correct door.

Using an approach called Bayesian updating (for historical reasons) you should update your estimate of having picked the correct door.

OK, fine.

I now know that I didn't pick the wrong door.
I now know that there are two doors remaining.
I now know that one or the other has the car.

I don't know why the door I picked has any less chance of hiding the car than the one I didn't pick. After all, there are two doors and the car is behind one of them.
 

GHT

I'll Lock Up
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New Forest
Many appear to have trouble with the better odds of switching doors, myself included, until I realised a simple fact: the odds are better if you switch because Monty curates the remaining choices.

Let’s say you played the game where Monty doesn’t know the location of the car. It wouldn’t make any difference if you switch or not (your odds would be 50% no matter what). But this isn’t what happens. The Monty Hall problem has a very specific clause: Monty knows where the car is. He never chooses the door with the car. And by curating the remaining doors for you, he raises the odds that switching is always a good bet.

Another of the reason why it's so hard to wrap your head around the Monty Hall problem is the small numbers. Let’s look at the exact same problem with 100 doors instead of 3. You pick a random door. Instead of one door, Monty eliminates 98 doors. These are doors that he knows do not have the prize! This leaves two doors. The one you picked, and one that was left after Monty eliminated the others.


Do you switch doors now? You should, because when you first picked, you only had a 1/100 chance of getting the right door. Furthermore, it was sheer guesswork. Now you’re being presented with a filtered choice, curated by Monty Hall himself. It should be clear that now your odds are much better if you switch.

Still don’t quite get the Monty Hall problem? Try your own experiment. Put a toy car under one of three boxes and play the game a hundred times yourself, noting your results. But with all of those PhD mathematicians being wrong, don’t feel bad if you’re still stumped.
 

Fifty150

One Too Many
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1,489
Location
The Barbary Coast
You picked door number 1.

Door number 3 has goats.

Door number 1 and door number 2 are unknown.

The game show host asks if you would like to switch from door number 1, to door number 2. Without the option of simply taking door number 3, and keeping the goat.

Then Monty Hall will want to kiss and fondle you. Or was that Richard Dawson?


 

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