The role of probability in solving the Monty Hall problem

[Jenny A]

There are no patterns to consider. One door always has the prize, and the others don't.  Choose one door and you have a one third chance of winning.  Choose two doors  and you have a two thirds chance of winning.  After Monty's reveal switching gives you effectively two rather than one doors.  Staying with the first door leaves you with  a one third chance.


EDIT:  Look around it this way. Suppose you had the choice of choosing one door or choosing two doors initially  with the understanding that Monty would show you a goat behind one of the two doors you choose and you'd get the remaining door of your two door choice.  That's what's actually happening if you switch.

Then the choice has been 50/50 all along, and it doesn't change in the least. You're just either in on it or not, that's all that changes. The chances of you being right, however, are the same throughout.

If you think the odds are ever 50/50, then you simply don't understand the math. The odds are one in three if you don't switch and two in three if you do switch after Monty's reveals that one of the doors you didn't choose is a goat..  The only way  you'd get to 50/50 is if Monty showed you a goat first, and then you choose either of the remaining two doors.  But that would be a different and not very interesting puzzle.