# Freedonia Monty Hall Puzzle

**URL:** <https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566>\
**Category:** The Game Room\
**Created:** [October 24, 2016, 4:34am UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566 "2016-10-24T04:34:36Z")\
**Posts on this page:** 2\
**Page:** 2

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**Author:** ![newme](https://avatars.discourse-cdn.com/v4/letter/n/e0b2c6/32.png) [@newme](https://boards.straightdope.com/u/newme)\
**Post date:** [October 26, 2016, 12:37am UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/21 "2016-10-26T00:37:18Z")

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> [@Some\_Call\_Me.Tim](#):
>
> Several things: The original scenario never specified how it was randomized. It might be that “Sometimes it’s all cars. Sometimes it a mix, 2 goats and a car, or two cars and a goat” means that those four options are all a 1 in 4 chance, which seems to be how you interpreted it. It might mean that a fair coin is flipped for each door, and thus there’s a 1 in 8 chance of all goats or all cars, and a 3 in 8 chance of the other two. The problem doesn’t specify.
> 
> The rules do specify that if he is able to Monty **must** show a goat behind a contestant’s door. Therefore scenarios 3 and 5 don’t exist; Monty has no choice if only one of the contestant’s doors has a goat behind it.
> 
> Also, scenario 2 is half as likely to occur as scenario 6: the two goats scenario is equally likely to put the car behind door 3 as door 1. However, half the time the car is behind door 3 Monty will show player 1 his goat and he won’t get to choose.

Yes, I am making the assumption that,before anyone chooses a door, you flip a coin for each and every door to decide if a goat or a car is behind that door.

I agree my wording is awkward, but it was my way of trying to show some scenarios will show up twice as often as others. If you scratch scenarios 3 and 5, then scenarios 4 and 6 are twice as likely to occur as 1 and 2, and you get the same result.

Stay - Win 2/3  
Switch - Win 1/2

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**Author:** ![Isosleepy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/isosleepy/32/479_2.png) [@Isosleepy](https://boards.straightdope.com/u/Isosleepy)\
**Post date:** [October 26, 2016, 3:51pm UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/22 "2016-10-26T15:51:49Z")

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> [@newme](#):
>
> Yes, I am making the assumption that,before anyone chooses a door, you flip a coin for each and every door to decide if a goat or a car is behind that door.
> 
> I agree my wording is awkward, but it was my way of trying to show some scenarios will show up twice as often as others. If you scratch scenarios 3 and 5, then scenarios 4 and 6 are twice as likely to occur as 1 and 2, and you get the same result.
> 
> Stay - Win 2/3  
> Switch - Win 1/2

We don’t know the odds of cars and goats. My earlier post also partially ignored this. We do know that if player 2 is shown to have a goat, you are more likely to not have a goat than box 3. So we don’t know what the winning percentage is, just that in this scenario staying is better than switching.

If we knew the distribution, say goats show up 9 out of 10 times, we could calculate the exact odds (by we I mean someone else, because lazy). But it doesn’t change that staying is better than switching, because of the extra info of p2’s goat, _and the rules of its reveal._

[Previous page](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566.md?page=1)