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.

[Excited Penguin]

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.

The odds are not one in three since there were only ever going to be two doors to choose from in the end, not three, since he was going to eliminate one all along. You can't both have your cake and eat it. When he shows you the goat is irrelevant, all that matters is that he does. And the fact that he only lets you choose between two possible doors doesn't affect the chances in the least. They are still two doors you have no knowledge of, the third one you do get information about after your first decision, regardless of whether you know this will happen or not.


As I said, just a trick. How deluded can people be about it is particularly interesting though. Even explaining it to you on your own terms doesn't seem to work. Fascinating.

[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.

At the start of the game, running the original experiment, from your perspective you have a 1/3 chance of getting right. From the moderators' perspective and the one that is actually right since he has all the information you have a 50/50 chance. So this is not so much a puzzle as it is a trick. No mystery whatsoever here, the chances stay the same from the beginning, only your understanding of them improves, not so the actual chances.

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

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.

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

[Excited Penguin]

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.

At the start of the game, running the original experiment, from your perspective you have a 1/3 chance of getting right. From the moderators' perspective and the one that is actually right since he has all the information you have a 50/50 chance. So this is not so much a puzzle as it is a trick. No mystery whatsoever here, the chances stay the same from the beginning, only your understanding of them improves, not so the actual chances.

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.

[Excited Penguin]

It is two out of three to get a goat first, but after he reveals a goat the pool of possibilities shrinks. Your chances do get better than you thought they were but it doesn't tell you which of the two remaining options are more likely to be true. This is where the illusion comes in.

[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.

[Excited Penguin]

I'm sorry, but I'm not a cards person. You'll have to use a different analogy.

I'm ready to accept a good explanation, but so far there have been none that I've encountered. I might be wrong, but I can't possibly see how in this case.

[Jenny A]

Use three shoe boxes. Put a dollar in one and a penny in another and a nickel in the third. The dollar is the grand prize the coins, duds. Close the lids.

Presuming the contestant switches here are the three equal possibilites.

1. He chooses the box with the penny. You peek in the other two boxes and open the one with the nickle. He switches to the box you did not reveal and gets the dollar.
2. He chooses the box with the nickle. Your peek in the other two boxes and reveal the penny.. He switches to the box you did not reveal and gets the dollar.
3. He chooses the box with the dollar. You peek in the other two boxes and reveal either the penny or the nickle. Whichever you reveal he switches and gets the other coin. He loses.

Notice that by switching he gets the dollar two out of three choices.

Suppose he doesn't switch. Your actions remain the same, but he only gets the dollar if he chooses it to begin with. So if he doesn't switch he has only has a one in three chance of winning.

[Excited Penguin]

After giving it some further thought I understand how you're better off switching if there are more than three options total, but I still can't get my head around why this is so when there are only two false options and one right one to begin with. I guess I'll have to trust that this is so regardless.

[Excited Penguin]

Use three shoe boxes. Put a dollar in one and a penny in another and a nickel in the third. The dollar is the grand prize the coins, duds.  Close the lids.

Presuming the contestants switches here are the three equal possibilites.  

1.  He chooses the box with the penny.  You peek in the other two boxes and open the one with the nickle. He switches to the box you did not reveal and gets the dollar.
2. He chooses the box with the nickle. Your peek in the other two boxes and reveal the penny.. He switches to the box you did not reveal and gets the dollar.
3. He chooses the box with the dollar. You peek in the other two boxes and reveal either the penny or the nickle.  Whichever you reveal he switches and gets the other coin.  He loses.

Notice that by switching he gets the dollar two out of three choices.

Suppose he doesn't switch. Your actions remain the same, but he only gets the dollar if he chooses it to begin with. So if he doesn't switch he has only has a one in three chance of winning.

Ok, I guess that works. Thanks.