🚀 go-pugleaf

>Leon ter Beek (leon@unixg.ubc.ca) writes:
>
>>I've made up another problem, similar to the Monty Hall problem. The
>>similarity is very obvious, but it does give it a different twist,
>>though.
>>
>>Disclaimer: First, I have been reading this group for just two days, so
>>excuse me if a similar extension has already been posted.
>>Second, I don't claim this to be a smart problem (what do you expect ;-)).
>>
>>There are five doors, behind two of them a car, and behind each of the
>>other three doors a goat. You pick a door and the quizmaster (who knows
>>what's behind the doors) opens a door with a goat behind it.
>>1) Should you change doors?
>>2) Why is the sum of the probability of changing and not changing not
>>equal to one?
>
The simple reason that the two probalities do not add up to one is that, unlike
the "classic" problem you can lose either way; i.e. even if you know for
certain that the door you chose originally was wrong, there are still 3 other
doors to chose from and only 2 winners...