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re: Variable Change And The Movie 21

Posted on 6/14/18 at 3:42 pm to
Posted by LNCHBOX
70448
Member since Jun 2009
86782 posts
Posted on 6/14/18 at 3:42 pm to
quote:

I’m never arguing in Monty Hall threads.


Me too. Anyone that doesn't get it or can't figure out how to google it isn't worth the effort.
Posted by hawkster
Member since Aug 2010
6261 posts
Posted on 6/14/18 at 4:05 pm to
quote:

Don’t think about in terms of 3 think about in terms of 100 like my example. The only way you would be wrong by switching doors is if your initial pick was correct which will occur 1% of the time. In other words, there’s a 99% chance it’s the other door. That’s variable change.


Good, simple analysis of the question that anyone should be able to grasp.

And in a real gambling scenario, there is another psychological factor that plays into the house's favor here. For those who don't grasp the variable change, their ego tends to make them keep their first pick, because it would be more disappointing to have given away the winning pick than it would be to keep your first choice and lose that way. That bit of human nature stacks the deck even more for the house.
Posted by The Egg
Houston, TX
Member since Dec 2004
81789 posts
Posted on 6/14/18 at 4:07 pm to
might want to put the scene in the OP

LINK
Posted by Who_Dat_Tiger
Member since Nov 2015
22182 posts
Posted on 6/14/18 at 4:09 pm to
quote:

If you're on The Price is Right on the punch-out game, and they eliminate 98 of the 100 sockets to win the prize (the correct choice is not 1-98; it's either 99 or 100), by this logic you have a 99% chance to win. Of the whole, that's true. But is the whole really still available? You have a 50% chance of picking the correct one as between 99 and 100. I think it's a matter of perspective as much as anything.

This is how I was thinking of it too, in a case of it being 1 out of 100.

So say you have a 1% chance of choosing the one box with a prize in it out of 100 boxes where the other 99 are empty. If you pick a box then remove 98 other boxes which are empty and are left with only 2 boxes. According to the Monty hall problem if given the option at that point to switch boxes you then have a 99% chance of winning the prize rather than 50% like you're suggesting by switching.
This post was edited on 6/14/18 at 4:10 pm
Posted by Merck
Tuscaloosa
Member since Nov 2009
1693 posts
Posted on 6/14/18 at 4:45 pm to
quote:

Assume the person is always selecting door 1 first and then one of the other doors is opened to reveal a goat:

These are the only possibilities for the initial arrangements.
Door 1 - Door 2 - Door 3 - Stay - Change
goat --- goat ----- car -- ---- lose - win
goat --- car ----- goat -- ---- lose - win
car - -- goat ---- - goat ------ Win - lose

Overall there's a 50% chance of winning, but you can see that switching results in a win 2/3. The part that often gets left out is that one of the remaining doors is opened.


This is the most logical explanation I've seen so far and it has 3 down votes. Which of you morons down voted this?
Posted by castorinho
13623 posts
Member since Nov 2010
84390 posts
Posted on 6/14/18 at 4:54 pm to
quote:

66% and 50% are both right given the perspective. We tend to like the odds of 66% better so we choose that option of thinking.
what?

The key is the host KNOWS. That's the main factor with Monty Hall. So yeah you switch every time and increase your odds. Dude did a case by case à page ago, that should clear it up.
Posted by Mo Jeaux
Member since Aug 2008
59752 posts
Posted on 6/14/18 at 5:09 pm to
quote:

He picks a door and they reveal its one of the piles of shite.


This isn't the Monty Hall problem. In that, they do not reveal what is behind the door that you have picked. Rather, they reveal one of the piles of shite behind another door and them ask you if you want to change your choice.
Posted by Hopeful Doc
Member since Sep 2010
15388 posts
Posted on 6/14/18 at 6:37 pm to
Why don't they just buy a house that's already painted?
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