Right. That’s what I was saying in post #3 - that the Monty Hall game rules are exactly equivalent to this. So I think this is the best way to gain intuition that switching doubles your chance of winning. For me, the “100 doors” scenario does not grant the same intuition.
I’ve discussed this with a lot of friends, and many of them get hung up on the “I chose Door 1 and I’m sticking with MY door, right or wrong!” Even when they come to accept the math they still won’t take the 2 out of 3 chance.
Try getting a 3rd party involved.
“Hey random stranger - pick a number between 1 and 3.”
“2.”
“Great. OK friend, you can have Door #2 or Doors #1 and 3.”
It sort of helps. Sometimes. Some people will simply never accept it.
Here’s my take on it.
First assume you play the strategy of never switch. Then you have one chance in three of winning. Clear?
Now play the strategy of always switch. Now whenever you choose the wrong door, you always win. What are your odds of choosing a wrong door? Two chances in three. Clear?
The most mysterious thing about this is that even Paul Erdös could not be convinced of it. And Brian Hayes who at the time wrote a regular math/cs column for the American Scientist didn’t believe it until he ran a Monte Carlo simulation.
Mathematically it’s the same. But I feel that describing it in ways where doors are being opened obscures the issue. People still see the final choice as being between two doors and don’t see why that isn’t a fifty/fifty choice.
Explaining it in terms of choosing what’s behind one door versus choosing what’s behind two doors simplifies the issue down to an immediate level. You don’t have to follow any trail of math or logic that leads you to the conclusion that one door is a better choice than the other. You can just see that two chances are better than one.
Aaaahhhh.
That’s the only one that makes any sense to me at all.
Increasing the number of doors doesn’t make the problem and its claimed solution make any more sense to me than it does with three doors. Neither does the information that the person who opened a door knows what’s behind it. But being told that one’s getting to pick two out of three –
only appears to help very briefly, until I stop and think about it. Because I’m still picking two out of three if I stick with the door that I started with.
Suppose you start the problem up from scratch at that point: there are three doors, you know something you want is behind one of them, you know that it isn’t behind another, you don’t know which of the two remaining doors it’s behind. I understand that there’s a 50/50 chance of picking the right door out of two. I understand that if I don’t know what’s behind any of three doors there’s a 1 in 3 chance of picking the right door out of three. I understand that 50/50 is better odds than 1 in 3.
What I don’t understand is why it’s not supposed to still be 50/50 between the two remaining doors, giving no advantage to either switching or staying with the earlier choice, just because of a choice made previously. If it would be 50/50 if the problem started there, why is it not still 50/50?
But the choice isn’t between Door 2 and Doors 1 and 3. Being told that Door 3 doesn’t have what you want makes it a choice between Door 2 and 3 or Door 1 and 3.
Forget about the fact that Monty opens a door for a moment.
Consider a much simpler game. You pick door A. Now you’re offered the choice: do you want to keep door A, or do you want to take BOTH of doors B and C? Obviously you switch for a 2/3 chance. Now refine that slightly: if you switch to take B & C, you only get the better of the two things behind B & C, if one is a goat and one is a car. Obviously you still switch, you don’t care about giving up the lesser prize, you still get the car if it’s behind either one of the two doors.
Now go back to the actual game.
When Monty opens one of B & C and shows you a goat, he is effectively comparing B & C, and giving you information about them. He is showing you which of out of B & C contains the worse prize, if they are different. So when he offers you the switch to the door he has not opened, he is offering precisely the same as in the scenario above: the better prize out of both B & C if you switch, a 2/3 chance.
The reason sticking with A does not give you the same 2/3 chance is clear: Monty will never open door A. He is not comparing the other doors to door A and giving you any new information about door A. He is only comparing door B to door C, and giving you information about which of B and C contains the worse prize.
This sounds awesome! You are then guaranteed to get at least one goat. And in the best case scenario you get two goats and don’t have to worry about what to do with a stupid, old car.
Why is he only comparing door B to door C, and not either of them to door A?
He only opens one door. He’s not telling me any more about one of the doors that he doesn’t open than he is about the other.
Again: back up, and presume we’re just starting the problem at that point, and there was no earlier choice. There are three doors, and I know I don’t want what’s behind one of them (I don’t see that it matters whether you call that one A, B, or C.) I can pick between the other two. Is that not a 50/50 chance?
And if it is a 50/50 chance, what does anything previous have to do with it? Is the chance that a given roll of dice will give a specific result changed by the outcome of earlier rolls?
Because those are the rules of the game. He compares B & C and eliminates one of them as an option by opening it. He is not allowed to eliminate door A that you have picked as your initial choice.
Ah. But I still don’t see that he’s giving me any more information about one of the doors that he doesn’t open than he is about the other.
And I still don’t see how its being a 50/50 chance is affected by anything that happened earlier. Again, is an individual roll of the dice affected by the results of earlier rolls?
There are only 3 possible scenarios, so it may be simplest just to spell them out and make it obvious.
Assume that your initial choice is always door A, obviously it’s symmetrical, and consider the car being in each of the 3 possible positions.
(1) Car behind door A.
Monty may open either B or C, showing a goat. You switch. => LOSE
(2) Car behind door B.
Monty must open door C, showing a goat. You switch to B => WIN
(3) Car behind door C.
Monty must open door B, showing a goat. You switch to C => WIN
So you win 2/3 times by switching, because Monty is COMPARING door B to door C, showing you which contains the worse prize, and allowing you to switch to the BETTER of doors B and C.
Because he’s not comparing anything to A. Under the rules of the game, he’s not allowed to open and eliminate door A, the door that you initially pick. So he’s comparing only B to C, and giving you information about which of B and C is worse.
If he opens door B, he’s telling you that it’s worse than (or equal to) C. He’s telling you nothing about whether it’s better or worse than A.
If he opens door C, he’s telling you that it’s worse than (or equal to) B. He’s telling you nothing about whether it’s better or worse than A.
OK. I think what’s going on in my head is that, in the example, he’s already opened a door. So the situation I’m dealing with when I’m supposed to choose again isn’t ‘he’s going to open a door’, but ‘he has opened a door’; and at that point I’m still seeing a choose-one-out-of-two option.
If I consider it as a one out of three option, then yes, switching makes sense. But once that door is open it’s not a one out of three any longer.
OK. That almost makes sense to the back of my head.
That . . . sort of makes sense. For a minute. But it still seems to me that (presuming he opens B) what I then know is that B is worse than or equal to C – and also that B is worse than or equal to A. He doesn’t have to open A to give me that information, any more than he has to open C.
Brilliant!!
And if you get two goat of differing sex then you can have all the little goatlets your heart desires. It’s the gift that keeps on giving.
I hate to rain on your parade but
- I’ve read this before and
- Trust me; people will still think it’s a 50-50 shot to switch.
The Monty Hall problem is incredibly counterintuitive. (It’s also often incorrectly explained; you MUST know Monty always deliberately opens a goat door.)
Yes, that’s a fair criticism of the way I presented it.
The difference is that there is no option under the rules for Monty to open door A, so you have no new information about door A. Since there are two goats, when you initially pick door A, you already know that there must be at least one goat among B and C. So the fact that Monty opens one out of B and C and reveals a goat tells you nothing new about door A. It DOES tell you something you did not know before about which out of B and C is better.
But I actually do have new information about door A, and it’s the same new information that I have about whichever of B and C Monty didn’t open: I have the information that both unopened doors are better or equal to the opened one; and neither of them is worse than the opened one.
That’s why at that point I’ve got a 50/50 chance instead of a 1 in 3 chance: because I have more information about one of the three doors. That information about the door that was opened, however, tells me that both unopened doors are better or equal to the open door. It doesn’t tell me that only one of them is.
No, you don’t. When you initially picked door A, you already knew that there was at least one goat among B and C. Knowing which of B and C contains that goat tells you nothing new about A.
No. You have to remember that there is information not just in what actually happened, but in the range of things that are ALLOWED to happen under the rules, including things that were allowed but which do not actually occur.
If you pick A initially, Monty is NOT ALLOWED to open A under the rules, simply because you chose it, not because of what it contains.
But door C is ALLOWED to be opened under the rules, unless it contains the car. The fact that Monty actually opens door B, when he was ALLOWED to open door C but did not do so, gives you information about C that you do not have about A.
If you’re having a conversation with someone about this in person, I think a deck of cards is the way to go.
Pull out 3 cards, one of them being the ace of spades (the winning card), and go through the game a couple times, where after they pick you look at the remaining two cards and flip over one of the losers. Maybe they catch on at this point, maybe they don’t.
If they don’t catch on yet, do it with the full deck. Have them pick a card, any card. Now fan the remaining 51 cards (not showing them) and pull out the ace spades, but keep it covered. Now flip over the 50 loser cards, then say “You can stick with your first choice, or switch to this covered card here.”
Do this a few times and they will start to see that it is not, in fact, a 50/50 split between the final two cards. (The one they originally picked and the ace of spades that you pulled out of the remaining cards.)
The stumbling block people have is getting stuck on “But at the end there are only 2 choices, so it has to be 50/50!” Except that at the end of the full deck version it’s also only 2 choices but very clearly not 50/50.
Try it this way.
There’s a 1/3 chance the car is behind door A when you pick it. You always know that there is at least one goat among B and C, so if someone who knows what’s behind all the doors DELIBERATELY (not just by picking at random) shows you a goat among one of B and C, you have no new information. You still have a 1/3 chance of winning if you stick with A.