# Thread: Simple puzzles to stump people

1. Originally Posted by athenian200
No matter which door you choose originally, one of the irrelevant ones is eliminated, meaning it's now it's a 50/50 chance between the car door and the goat door. This other answer makes absolutely no sense to me. I think it's beyond my comprehension.
(Okay, this I can respond to in my sleep. )

Sometimes it helps to think about it this way: Instead of only three doors, picture a thousand doors. You pick one, the host closes 998 doors and there is then one left apart from the one you chose. Does it still seem right to you that because of the closing of 998 irrelevant doors the odds become fifty-fifty that you picked the right door at first - or are you more inclined now to zero in on that one door the host didn't open?

2. Originally Posted by Jennifer
That's because you are "readjusting the situation" in your head once Door #2 has been eliminated. Don't do that.

You would only have to readjust the situation/odds IF the item was moved AFTER the second door was eliminated. Since the item is not moved and stays in the same spot, the host has simply done you the favor of telling you which door it's not behind.
If the door I originally chose had the goat, and he eliminated the other door containing a goat, then I would have a choice between keeping the original choice of the goat, or switching to the car. If the door I originally chose had the car, and he eliminated one of the goats, then I would also have a choice between switching to the goat or keeping the car. No matter what, he eliminates one goat. I can't see why the door I didn't originally choose has a higher likelihood of having been the door with the car. Since neither of the doors remaining was eliminated, they both have the potential to be the correct one. I'm sure you're right, but my brain doesn't bend that way.

Economica: Yes, it still seems right, because it wasn't only one door he didn't open. He left two doors closed, one of which contain the car. You could just as easily have chosen the one with the car as not. I don't see why it's less likely.

3. Originally Posted by athenian200
I understand what the results were, I just can't comprehend why. I don't see how it become 1/3 in the first place. I see it as 1/2. No matter which door you choose originally, one of the irrelevant ones is eliminated, meaning it's now it's a 50/50 chance between the car door and the goat door. This other answer makes absolutely no sense to me. I think it's beyond my comprehension.
The fact that Monty eliminates a goat door doesn't alter the fact that you had a 1/3 chance of choosing correctly at the time you chose. Your odds of choosing correctly were one in three, and your odds of having chosen the right door remain one in three even after Monty's goat door is opened.

4. My example had the benefit of a pretty picture... but Economica's example was better at showing the absurdity of recalculating the odds after 998 doors are closed.

5. There is also the cab problem:

A cab was involved in a hit and run accident at night. Two cab companies, the Green and the Blue, operate in the city. 85&#37; of the cabs in the city are Green and 15% are Blue.
A witness identified the cab as Blue. The court tested the reliability of the witness under the same circumstances that existed on the night of the accident and concluded that the witness correctly identified each one of the two colors 80% of the time and failed 20% of the time.
What is the probability that the cab involved in the accident was Blue rather than Green?

6. What if there's a car behind one door, a goat behind another door, and Jennifer's behind the third door?

Monty Hall opens a door you didn't pick, revealing the goat.

So do you stick with the door you chose, or switch to the remaining door?

7. Monty Hall I have heard of before, so didn't bother with that one.

Wason card I first though A and 4, but after a bit realized that the 4 didn't matter to the problem, so figured out that it was just A that was needed.

If the door I originally chose had the goat, and he eliminated the other door containing a goat, then I would have a choice between keeping the original choice of the goat, or switching to the car. If the door I originally chose had the car, and he eliminated one of the goats, then I would also have a choice between switching to the goat or keeping the car. No matter what, he eliminates one goat. I can't see why the door I didn't originally choose has a higher likelihood of having been the door with the car. I'm sure you're right, but my brain doesn't bend that way.
The host is not making a random choice of which door to show you, he (or she) deliberately picks the goat door to show you. If the host had randomly chosen a non-picked door to show you, and that door happened to be a goat, it would be a 1/2 chance. (though in that case he might open up the prize car door instead.)

Here's an example of the three possibilities. Say you pick door A

If the car is in door A (1/3 chance): Host can now pick door B or C to open (both these give the equivalent situation, though the total probability of opening B plus opening C is still 1/3). Now the unopened door has a goat, your picked door has the car.

If the car is in door B (1/3 chance): Host must open door C, because that is the only non-picked door to have a goat. Now, your picked door has a goat, and the unopened door has a car.

If the car is in door C (1/3 chance): Host must open door B, and you get the equivalent situation to above, with doors B and C switched. This still has a 1/3 total probability of happening, as there was a 1/3 probability of the car being in each door.

All of these work the same if you pick doors B or C. Because there is a 2/3 probability of the car being behind the unopened or unpicked door for each of your possible choices of original door, the total probability works out to be 2/3 of you getting the car if switching.

(For comparison, I didn't figure this out when originally hearing about it either, and it took some convincing also.)

8. Re: Eco's "Cab" puzzle:
I'm going to guess there was a 12&#37; chance that the cab was Blue... but I'm pretty sure that must be wrong, since otherwise it would be too easy.

And I will also assume that none of the Blue cab company's cars were painted green, and none of the Green company's were painted blue.

Originally Posted by oberon67
What if there's a car behind one door, a goat behind another door, and Jennifer's behind the third door? Monty Hall opens a door you didn't pick, revealing the goat. So do you stick with the door you chose, or switch to the remaining door?
Most people who were hungry might just take Oberon (the goat) and go home.

9. Originally Posted by athenian200
Economica: Yes, it still seems right, because it wasn't only one door he didn't open. He left two doors closed, one of which contain the car. You could just as easily have chosen the one with the car as not. I don't see why it's less likely.
If it's any consolation, there are Ph.D.'s in math and statistics who don't get it either, cf. the responses to Marilyn vos Savant's column (which incidentally constitute a very fun read for those who have no illusions left about human rationality).

Originally Posted by Jennifer
Re: Eco's "Cab" puzzle:
I'm going to guess there was a 12&#37; chance that the cab was Blue... but I'm pretty sure that must be wrong, since otherwise it would be too easy.
Once again, if it's any consolation, not a single student in my class of 30 got it right either. And we all knew Bayes' Theorem by heart. :blushing:

10. Originally Posted by Zergling
Wason card I first though A and 4, but after a bit realized that the 4 didn't matter to the problem, so figured out that it was just A that was needed.
No, you need to flip the 7 as well. If the 7 has a vowel on the back, you've disproven the statement, which was the assignment.

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