monty hall game theory

[21frogman]

The prior thread on this topic in the Subscription Forum, Advanced Strategies, Theory & Math. It is titled "Bayes Theorem-The Monte Hall Problem". RCJH posted a very simple proof (post #10) showing that the contestant is better off changing his guess after a goat is found behind the opened door.
As noted above, the reason is Bayes Theorem. The switch improves the contestant's chances from 1 in 3 to 2 in 3.
If someone with computer skills can take a picture of the proof by RCJH & post it here, this thread won't have to go to 100 posts.

[Three]

The logic to the 2/3rds chance of winning if you switch is predicated on the door being opened never having a goat behind it. Monty never showed a goat so on the show the logic is sound. But if the door where chosen randomly instead. 1/3rd of the time a goat would be revealed and the other 2/3rds the goat would be behind either door with equal probability, it would be 50%-50% as to which of the remaining doors the big prize was behind. While the OP didn't specify that the choice was not random, he did call his OP "Monty Hall game theory" which would indicate that the host never shows a goat. If you are thinking it is 50-50 after the goat is revealed just think about the difference between the Monty Hall scenario where a goat is always revealed and a scenario where the door is chosen randomly between the remaining doors and reveals a goat 1/3rd of the time. To sum up the link that was requested. 1/3rd of the time you chose right and a goat is revealed and 2/3rds of the time you chose wrong and a goat is revealed. So switching doubles your chances of winning the big prize if you know the big prize is never revealed. But without that knowledge the odds become 50%-50% when a goat is revealed

[Three]

I would switch no matter what if I did't know the door revealed had to be a goat. There are two possibilities. The first is the door had to be a goat and you double your chances of winning the big prize by switching. The other is the door was a random selection that happened to reveal the goat and would reveal a goat 1/3rd of the time, in which case your odds of winning the big prize doesn't change if you switch. Given those two possibilities switching becomes a no brainer.

[angle_sh00ter]

I would switch no matter what if I did't know the door revealed had to be a goat. There are two possibilities. The first is the door had to be a goat and you double your chances of winning the big prize by switching. The other is the door was a random selection that happened to reveal the goat and would reveal a goat 1/3rd of the time, in which case your odds of winning the big prize doesn't change if you switch. Given those two possibilities switching becomes a no brainer.

Lol I would assume that anyone in the business of giving away either a car or a goat would prefer to give away less cars than goats. Therefore with no information about his process if I was shown a goat and given the choice to change my guess I would assume it's a trick and that the option to change is only given if you guess correctly. If you guess wrong initially the game is over.

Of course if I knew that the host had to show me a goat regardless of whether I picked right or wrong then I would always switch.

[Freightman]

Monty Hall and the Zohan!

You dont mess with the Zohan - goat scene

https://youtu.be/8DkbFJ3uz54

[DSchles]

The Monty H. Schlesinger problem:

Suppose we have a thread with 100 posts. If we eliminate all the irrelevant ones, how many will be left?

Three or four. Max!

Don

[SteinMeister]

Okay so having a discussion with employees. You are on a game show and have three choices to pick a winning door. you have 33% chance of picking the winning door. You pick one door #2. Host the host comes in a eliminates a losing door #3. He asks if you want to switch to door #1 or keep you original choice of #2. Supposedly if you switch your choice you have 66% chance of winning, but to me it becomes a 50/50 chance....

Reminiscent of a statistical class I took many years ago.

The professor walks into the waiting class, and quietly writes the following on the board: “BRD-149” (or something equivalent). And then exclaims “I saw this license plate on the way to campus today…. What are the odds?”

So we all start calculating, thinking this is a pop-quiz. 1/26 * 1/26 * 1/26 * 1/10 * 1/10 * 1/10 = ….freaking small, right?

His point was this: Yes, those WERE the odds of seeing that particular license plate BEFORE he left his house… but AFTER the fact (i.e. in our classroom) those odds were irrelevant.


Seems to me the OP is correct is saying it’s a 50/50 chance at that time.


(sorry Don to perpetuate this beaten horse)

[Dbs6582]

Monty Hall answers this problem:

https://www.nytimes.com/1991/07/21/u...nd-answer.html

It depends on how this problem is worded.

[Three]

Monty Hall answers this problem:

https://www.nytimes.com/1991/07/21/u...nd-answer.html

It depends on how this problem is worded.

I think I heard someone say that a few times in this thread.

[Dbs6582]

I think I heard someone say that a few times in this thread.

You were one of them. I think I saw you make at least three posts on this thread staying pretty much the same thing. So what’s new?

[Three]

You were one of them. I think I saw you make at least three posts on this thread staying pretty much the same thing. So what’s new?

Anglesh00ter pointed out the possibility of host manipulation. But the logic problem always denies this and usually states a door will always be opened with a goat and you will be allowed the option to switch.

[Dbs6582]

Anglesh00ter pointed out the possibility of host manipulation. But the logic problem always denies this and usually states a door will always be opened with a goat and you will be allowed the option to switch.

Point taken and I get that. But I did think it was interesting that in the real game, Monty Hall manipulated it so it wasn’t an easy probability problem. I found this article interesting and thought some would like reading it.