EDIT: Let’s cool it with the downvotes, dudes. We’re not out to cut funding to your black hole detection chamber or revoke the degrees of chiropractors just because a couple of us don’t believe in it, okay? Chill out, participate with the prompt and continue with having a nice day. I’m sure almost everybody has something to add.
The Monty Hall problem.
You are given a choice of three doors, let’s call them 1, 2, and 3.
Behind one of the doors is a fabulous prize. Behind the other two are joke prizes worth nothing.
You are asked to pick a door. It doesn’t matter which one you choose, because it’s not opened inmediately.
Instead, the host opens one of the doors you did not pick to reveal the gag gift.
He then asks you if you want to change your choice.
What are the chances of winning? Should you choose a different door, or keep your existing choice?
The math says, your chance of winning if you stay with your choice is 1/3. Revealing the contents of one door does not change that, it’s still 1/3.
Switching to the other door gives you a 2/3 chance of winning. Not 1/2 or 1/3.
https://behavioralscientist.org/steven-pinker-rationality-why-you-should-always-switch-the-monty-hall-problem-finally-explained/
“If the car is behind Door 1, you lose. If the car is behind Door 2, Monty would have opened Door 3, so you would switch to Door 2 and win. If the car is behind Door 3, he would have opened Door 2, so you would switch to Door 3 and win. The odds of winning with the “Switch” strategy are two in three, double the odds of staying.”
Are you saying you don’t believe it? Because you explained why it works pretty well. When the host opens the door, they will always open a non-winning door, so it doesn’t affect the odds at all. There is still a 1:3 chance it’s the door you picked, and a 2:3 that it’s one of the other 2. All the host did is showed you which one it wasn’t behind, and that means the odds of that remaining door is 2:3
Thanks to your explanation I think I can get my head around it.
I can explain it, it doesn’t mean I believe it. ;)
This problem doesn’t make any sense.
If one wrong door is always opened, your chance was never 1/3 to begin with, so you are thinking about this problem with the wrong premise, making it hard to grasp. You were just assuming it was 1/3 because you didn’t know one door would be taken away.
As soon as the wrong door is opened, your odds are never 1/3 nor 2/3. It’s 1/2 because there’s only two doors. What did you think the number after / stood for?
EDIT: Now I’ve tried to look through the examples in the article, and it honestly just makes it worse.
The example about picking a door at 1/1000, and then Monty removing 998 of the doors, leaving two doors, therefore making it more likely you should pick the one Monty left open, is also stupid - because it’s not comparable.
The above example is true. The likelihood of Monty being right is much higher.
But your pick is never 1/1000 when there’s only 3 doors, making the example not compatible with the other. The 1000 door example is not wrong - you just can’t compare them.
And now to explain why it’s different:
In the 3 door example, your “pick power” is 1. Means you can pick 1 door. Montys “pick power” is also 1, making you both equally strong.
This means that you picking a door gives as much intel as Monty picking a door does. No matter what, you will always be left with 1 door not being picked.
Now you look at the 2 doors. The one you picked, and the one nobody did. Now this problem suggests that Monty has given you new information because he removed a door, but he didn’t give you that, and here’s why:
The problem suggests that Monty gives you intel by removing a door in a 1/3 scenario. But he doesn’t. That’s an illusion.
From Montys perspective, he only has 2 doors to pick from, because he can NEVER remove yours, no matter what you picked.
Now Monty has made his choice, and this is where we turn the game around making it clear it was a 1/2 choice all along.
Because the thing you are picking between is not the doors anymore. It was never about the doors.
You are picking between if Monty is bluffing or not.
Let’s say you always pick door 1 as your first option. Monty will always remove 2 or 3. Either Monty removes door 2 or 3 because he helps you, or he’s doing it because he’s bluffing.
If you didn’t get any more help, this WOULD’VE been a 1/3. You’d have to choose between if Monty bluffed at door 2 or he bluffed at door 3, or he bluffed at both, because it was your door.
But then Monty goes ahead and removes a door, let’s say 3 (or 2 if you want, it doesn’t matter). He tells you it’s not that one. Now you have to choose if he’s bluffing at door 2 or he’s bluffing at your door.
You now have a 1/2 to call his bluff.
Monty was the enemy all along - not the doors.
You would think, right? Try it out yourself:
https://www.mathwarehouse.com/monty-hall-simulation-online/
The Monty Hall problem has always bothered me when considering it on the basis of 3 doors. However when the concept is extended to 100 doors, and 98 are opened, it starts to click for me that of course the odds arent 50/50. It’s much more obvious that the prize was in the field (and the odds shift to reflect that)
That’s the thing though, according to the explanation, it’s never 50/50.
If there are 3 doors, and 1 is opened, you have a 2/3rds chance in winning by picking the other door.
I meant are not
Typo :)
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You’d THINK, right?
Try it out here!
https://www.mathwarehouse.com/monty-hall-simulation-online/