I had big issues understanding the Monty Hall problem. Someone even went as far as explaining it with a hundred doors, and 98 were opened. It didn't dawn on me until I was walking home later and I had an aha-moment.
I've just given up on the Monty Hall problem. I get it, but I just don't get it. Sort of like a lot of physics concepts. I get that certain things happen and I get the reasons, but I couldn't work out on a board exactly why.
The point lies in what Monty does when you choose a door. He deliberately opens a door that doesn't have anything behind it. Thus making sure that the 2/3-probability still lies with the door(s) you haven't chosen. I had my students try it out on eachother and it pays off to switch in two thirds of the cases.
I had no idea what the Monty Hall problem was. Upon looking it up, I think I can explain it:
Monty introduced NEW INFORMATION into the system once he exposed the door that was a zonk (he told the contestant which door NOT to choose). The contestant had NO new information on the door he/she chose, but new (and beneficial) information on the doors he/she didn't choose. To benefit from that new info, the logical choice was to switch.
If that still doesn't make sense, i can try to explain it differently.
Do you agree that when you choose your initial door, the probability of being right is 1/3? Baring a switcheroo behind the door, that probability cannot change. The car won't jump to or away from the door.
The other doors also have 1/3 chances. Until Monty opens one, now that door has 0/3 chances of being right. You door is still 1/3, we already established it can't change. There's now 1/3 unaccounted for. Where did it go? To the remaining door which brings it to 2/3.
I almost lost a friend trying to explain this to them. I usually turn to a pack of cards and the ace of spades. It's amazing the stubbornness of our minds!
I know that a lot of people have big issues with the Monty Hall problem, and I believe it is because it is usually explained in a misleading way. In order for switching doors to be advantageous, we must assume
1. That the host always opens a door that the contestant did not open, and
2. That the door which the host opens always contains a goat.
To me, these assumptions seem counterintuitive.
If he is a game show host, couldn't he be trying to trick you, and only open a door when you first pick the door with the car? If this is the case, then the host opening a door means there is a 100% certainly that the original door you picked has the car.
If he is trying to build suspense, wouldn't he open the door with the car behind it 1/3 of the time? If this is the case, then him opening a door with a goat means the remaining doors have equal chances.
Maybe the host has some other set of rules that he follows that we don't know. Maybe he opens the door with the car if you have red hair and the door with a goat if you have blond. I just don't know why we should assume that he always opens a door and that his door always has a goat.
If I were playing the Monte Hall game, I would flip a coin and on heads I would change doors, on tails I would not. This way, the host's strategy is irrelevant; I will always have a 1/2 chance of winning.
The results are very counterintuitive. For the three-door problem, if the contestant is correct on the first choice, then he or she will be incorrect after a switch. Or, if a contestant was incorrect on the first choice, he or she will be correct after a switch. Since the probability of being incorrect on the first choice is 0.667, then the probability of being correct after a switch is 0.6667.
Yes, I understand that if you make the assumptions that I laid out, then it is advantagous to switch. I'm not confused or bafffled by the problem. The simulation you linked to is predicated on those assupmtions.
But I don't think those assuptions are intuitive based on how the problem is usually presented. At least they are not intuitive for me. Here is the famous Ask Marylin column that popularize the Monty Hall problem;
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 #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?
When I read this, I think that the host is potentially a trickster, or possibly he was waiting to see what I chose before he gave me the choice. I don't view him as a mechanical device that must always open a door that I did not open and that has a goat.
If you think that it is intuitive to say that he always opens a door that the contestant did not open and that the door he opens must contains a goat, then I suppose the result would seem really cool. I don't share that intuition, and so the result seems cheap and misleading.
I had big issues understanding the Monty Hall problem. Someone even went as far as explaining it with a hundred doors, and 98 were opened. It didn't dawn on me until I was walking home later and I had an aha-moment.
I've just given up on the Monty Hall problem. I get it, but I just don't get it. Sort of like a lot of physics concepts. I get that certain things happen and I get the reasons, but I couldn't work out on a board exactly why.
For those wondering, this is a great video highlighting the concept of the Monty Hall problem. https://www.youtube.com/watch?v=4Lb-6rxZxx0
The point lies in what Monty does when you choose a door. He deliberately opens a door that doesn't have anything behind it. Thus making sure that the 2/3-probability still lies with the door(s) you haven't chosen. I had my students try it out on eachother and it pays off to switch in two thirds of the cases.
I had no idea what the Monty Hall problem was. Upon looking it up, I think I can explain it:
Monty introduced NEW INFORMATION into the system once he exposed the door that was a zonk (he told the contestant which door NOT to choose). The contestant had NO new information on the door he/she chose, but new (and beneficial) information on the doors he/she didn't choose. To benefit from that new info, the logical choice was to switch.
If that still doesn't make sense, i can try to explain it differently.
Why isn't the choice to stay or switch independent from the first choice to choose one of the 3?
Do you agree that when you choose your initial door, the probability of being right is 1/3? Baring a switcheroo behind the door, that probability cannot change. The car won't jump to or away from the door.
The other doors also have 1/3 chances. Until Monty opens one, now that door has 0/3 chances of being right. You door is still 1/3, we already established it can't change. There's now 1/3 unaccounted for. Where did it go? To the remaining door which brings it to 2/3.
Ohhhhhhhhhhhhhhhhhh! I got it now. Thanks!
I almost lost a friend trying to explain this to them. I usually turn to a pack of cards and the ace of spades. It's amazing the stubbornness of our minds!
Is a friend truly a friend if he can't understand the Monty Hall problem? :-P
A dear acquaintance then ;)
For those who aren't familiar with it: the Monty Hall problem.
I know that a lot of people have big issues with the Monty Hall problem, and I believe it is because it is usually explained in a misleading way. In order for switching doors to be advantageous, we must assume
1. That the host always opens a door that the contestant did not open, and
2. That the door which the host opens always contains a goat.
To me, these assumptions seem counterintuitive.
If he is a game show host, couldn't he be trying to trick you, and only open a door when you first pick the door with the car? If this is the case, then the host opening a door means there is a 100% certainly that the original door you picked has the car.
If he is trying to build suspense, wouldn't he open the door with the car behind it 1/3 of the time? If this is the case, then him opening a door with a goat means the remaining doors have equal chances.
Maybe the host has some other set of rules that he follows that we don't know. Maybe he opens the door with the car if you have red hair and the door with a goat if you have blond. I just don't know why we should assume that he always opens a door and that his door always has a goat.
If I were playing the Monte Hall game, I would flip a coin and on heads I would change doors, on tails I would not. This way, the host's strategy is irrelevant; I will always have a 1/2 chance of winning.
We need data.
Try it our for yourself! http://onlinestatbook.com/2/probability/monty_hall_demo.html
This quote is what got me to understand it.
Yes, I understand that if you make the assumptions that I laid out, then it is advantagous to switch. I'm not confused or bafffled by the problem. The simulation you linked to is predicated on those assupmtions.
But I don't think those assuptions are intuitive based on how the problem is usually presented. At least they are not intuitive for me. Here is the famous Ask Marylin column that popularize the Monty Hall problem;
When I read this, I think that the host is potentially a trickster, or possibly he was waiting to see what I chose before he gave me the choice. I don't view him as a mechanical device that must always open a door that I did not open and that has a goat.
If you think that it is intuitive to say that he always opens a door that the contestant did not open and that the door he opens must contains a goat, then I suppose the result would seem really cool. I don't share that intuition, and so the result seems cheap and misleading.