The role of probability in solving the Monty Hall problem

[Jenny A]

Wrong.  Monty is required to show you a goat you didn't choose. Those are the rules. 

I never said otherwise and don't understand how this answers my argument. Who are you arguing with here?

At the start of the game one of two things will happen.  One,  you will choose the winning door.  There is a one third chance of that happening.  In that case Monty will show you one of the goats because that is all he can show you. If you switch you will get the other goat and lose.  But chances are two out of three that you will choose a goat to begin with and  then Monty must show you the other goat because those are the rules.  In that case, if you switch, the other door is the prize. So, if you switch you win two out of three times.

Ok, this obfuscatory way of putting it doesn't get you an additional third viable option, which you don't have, since he is going to eliminate one regardless of the choice you make first. You're still going to end up with only two options no matter what you do. The one you already chose and the one left after he revealed the third one. None of this switching and revealing business tells you which one of the two is more likely to hold the desired prize. You can only argue this inane point so long before you realise how wrong you are. I'm hoping the same thing will happen here.

If you don't switch,  your chances remain one in three because that is your chance of picking correctly in the first place.

You can switch or not switch all you like. There are still only two doors to choose from, not three. It is paramount that no matter how much you change the version of the game the moderator always eliminate all but an option from the ones you didn't choose after you make your initial choice. This makes it so that there will always be only two options left. Therefore 50/50, not 66/33.

You are really not thinking.  Get out three cards, one winner and two losers: Ace, 2 of spades, two of diamonds.. Pretend you are Monty and lay them out face up so you can see them.  Walk through what happens if the contestants picks each of the cards and switches after you reveal a 2 he did not pick..

There equal possibilities.  He wins two out of three of them if he switches:

First possibility, contestant chooses the ace.  You show him either of the 2s and he switches to the other 2.  He loses whichever 2 you show him.. 

Second possibility,  he chooses the 2 of spades. You show him the 2 of diamonds. He switches to the ace and wins. 

Third possibility, he chooses the 2 of diamonds. You show him the 2 of spades.  He switches to the ace and wins.


If he does not switch he wins by keeping the ace if he chooses the ace. But he loses if he chooses either of the 2s.  Thus if he doesn't switch he only wins one in three times.  

There is no 50% chance of his winning under any possibility whether he holds or switches.