Re:Seeing Theory: A Visual Introduction to Probability and S...
by jonsen on Wednesday September 30, 2020 @09:52AM
(#24638687)
It does not clearly say that it is Monty’s procedure to
- allways open a door
- allways open a door with a goat
- and not open a door at random
Often discussion of the solution reveals that this is not clear.
Re:Seeing Theory: A Visual Introduction to Probability and S...
by 542354234235 on Wednesday September 30, 2020 @11:02AM
(#24639439)
There are three doors. You have picked one, leaving two other doors. It absolutely explicitly says he opens one door. The only options are a goat or a car. If it was a car, you would have lost already and so there is no problem. If it was random, you still get the same information (what is behind one of the unpicked doors).
Re:Seeing Theory: A Visual Introduction to Probability and S...
by jonsen on Wednesday September 30, 2020 @12:29PM
(#24640528)
> If it was random, you still get the same information
No, if both you and Monty pick a door at random, there’s 1/3 chance of a car behind each door. If Monty’s door reveals a goat, it’s 50/50 for the remaining two doors. It’s mandatory to specify Monty’s procedure precisely.
Re:Seeing Theory: A Visual Introduction to Probability and S...
by jibal on Wednesday September 30, 2020 @09:58AM
(#24638742)
This is goalpost moving that has nothing to do with the original point. If people misunderstand the conditions of the problem, that has nothing to do with intuitions about probability.
I won't respond further.
That has been part of the explicit problem ever since it was first presented back in 1975.
"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?"