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Something to blow your mind
11 years ago
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The Monty Hall Problem
Say you are on a game show, the Monty Hall show in fact. Part of the game is you are given a selection of 3 doors. Behind one door is a brand new sports car :D, behind the other two doors there are goats.
You can select the door A, B, or C with no idea which door the car lies behind?
Say you select A. Monty (the host) will then remove one of the remaining doors, always one with a goat behind it. You are then given the option to switch doors, do you?
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So the Monty Hall problem is a logical paradox. It seems like the remaining doors will be 50/50 chance for the car.
In reality if you switch doors you have a 2/3 chance of getting the car than if you stayed with your first door. Wait, you say, that makes no sense. It should be 50/50 there's 2 doors.
So, when you start there is a 1/3 chance that the car is behind any door. A 2/3 chance that it is not behind the door you choose. When a door is removed from the system it is still a 1/3 chance that the door you picked first is right, and still a 2/3 chance it is not that door, but since there is only one door left you can pick it and have a 2/3 chance of a car.
The fact of the matter is, our brains hate this sort of probability stuff, and it's always going to seem wrong. I personally wouldn't believe it until I did the problem over and over and over.
check out an interactive version here http://stayorswitch.com/ you can click the explanation tab to get a more in depth explanation
Say you are on a game show, the Monty Hall show in fact. Part of the game is you are given a selection of 3 doors. Behind one door is a brand new sports car :D, behind the other two doors there are goats.
You can select the door A, B, or C with no idea which door the car lies behind?
Say you select A. Monty (the host) will then remove one of the remaining doors, always one with a goat behind it. You are then given the option to switch doors, do you?
-
--
---
--
-
So the Monty Hall problem is a logical paradox. It seems like the remaining doors will be 50/50 chance for the car.
In reality if you switch doors you have a 2/3 chance of getting the car than if you stayed with your first door. Wait, you say, that makes no sense. It should be 50/50 there's 2 doors.
So, when you start there is a 1/3 chance that the car is behind any door. A 2/3 chance that it is not behind the door you choose. When a door is removed from the system it is still a 1/3 chance that the door you picked first is right, and still a 2/3 chance it is not that door, but since there is only one door left you can pick it and have a 2/3 chance of a car.
The fact of the matter is, our brains hate this sort of probability stuff, and it's always going to seem wrong. I personally wouldn't believe it until I did the problem over and over and over.
check out an interactive version here http://stayorswitch.com/ you can click the explanation tab to get a more in depth explanation
Goats have alot of uses as well :3
When you're asked to stay or switch, the revealed goat is irrelevant. One door has a car. One door has a goat. Regardless of what the third door may have had, you are given a new question: Stay or switch? Which is essentially the same question as "Do you pick door number one, and stay? Or do you pick door number two, and switch?" So you're still picking between two doors.
You can't pick the third door because it's out of the question.
It's not "Do you stay OR switch OR pick the obvious goat?" It's "Stay or Switch?" "One or two?"
At the point in which Monty intervenes, the problem is changed to a new problem.
The probability of you picking the right door the first time is 0. Monty steps in and doesn't let you know if you were correct. There are no correct doors when you make your first selection because the car is not revealed when you make your first selection.
The illusion is that it seems like a logical progression of events when it's actually two entirely separate events. One event with a zero probability. One event with a 50/50 probability.
Or at least that's how I see it.
Goat goat car. If you pick A he removes B. If you pick B he removes A. If you pick C he will take A or B it doesn't matter.
Its not predestined because it depends entirely on you picking first. You can't pick montys goat because Monty won't pick until you do and will pick the goat you do not.
You seem to be wanting to make this more than it is.
I think that explanation makes sense, but I'll have to let you know after I have had some coffee.
The real problem is that the odds of winning the car is 33% for those who don't understand the problem yet 67% for those who do.
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The Monty Hall Problem has a solution that balances the odds to 50% whether contestants game the original problem or not.
Monty flips a coin to decide ahead of time whether he'll let the contestant switch doors or not. Then, in the game, the contestant chooses a door.
If the contestant chose the car, Monty always opens one of the goat doors and offers the contestant the choice to switch, even if the coin flip told him not to.
If the contestant chose a goat and Monty chose to let the contestant switch, he'll open the other goat door and offer the choice.
If the contestant chose a goat and Monty chose not to let the contestant switch, however, then Monty reveals the goat as the contestant's prize without offering a chance to switch.