# Thread: Simple puzzles to stump people

1. ## Simple puzzles to stump people

I was inspired by how often the Wason card problem stumped the people I interviewed (save one, and that one got it instantaneously).

So I wanted to make a list of really simple puzzles that seem to stump people.

I will start with two famous examples.

The Wason Card Problem.

The "Monty Hall Problem".

2. I always loved the Monty Hall problem. I remember when Marilyn vos Savant covered it in her column and had to deal for weeks with people writing in and arguing with her about it.

There's also the Shared Birthday problem (i.e., you only need about 35-40 people in one room in order to ensure that at least two of them will have the same birthday... and NOT 365 people as most would think).

3. Originally Posted by ygolo
I was inspired by how often the Wason card problem stumped the people I interviewed (save one, and that one got it instantaneously).

So I wanted to make a list of really simple puzzles that seem to stump people.

I will start with two famous examples.

The Wason Card Problem.

The "Monty Hall Problem".
Okay, with the Wason card problem, you're trying to determine if the following statement is false: "If a card has a vowel on one side, it has an even number on the other side." So to determine if it is false, you have to find just one example of a card that does not follow the rule.

So you turn over the A to see if the number on the reverse is odd, and you turn over the 7 to see if the letter on the reverse is a vowel. If either or both is true, you've accomplished your purpose. The other two cards are irrelevant, because the B does not match the premise, and the presence of the 4 can only confirm, not void, the conclusion.

I confess, I fell victim to confirmation bias before I went back and thought it through.

4. When I saw this:

Let the doors be called X, Y and Z.

Let Cx be the event that the car is behind door X and so on.

Let Hx be the event that the host opens door X and so on.

Supposing that you choose door X, the possibility that you win a car if you then switch your choice is given by the following formula

P(Hz ^ Cy) + P(Hy ^ Cz)

= P(Cy) P(Hz Cy) + P(Cz) P(Hy Cz)

My nose and ears started bleeding.

5. On the card problem, my guess was that if I could only turn over one card, I would choose "A." Was that a good guess or bad one? But I believe that I would actually have to turn over every card besides "B" in order to be completely certain.

I still don't understand the Hall problem. It seems to be based on probability. My understanding is that if you choose any particular door, and then one of the other non-car doors is eliminated, then you have two choices, one of which is the car, and one of which is the goat. No matter which door you choose, the one that you previously chose, or the other, you still have a 1 in 2 chance of winning the car. I don't understand the answers they came up with.

6. Originally Posted by athenian200
I still don't understand the Hall problem. It seems to be based on probability. My understanding is that if you choose any particular door, and then one of the other non-car doors is eliminated, then you have two choices, one of which is the car, and one of which is the goat. No matter which door you choose, the one that you previously chose, or the other, you still have a 1 in 2 chance of winning the car. I don't understand the answers they came up with.
When you make your first selection, there is a 1/3 chance you will pick the winning door.

The host does you a favor by showing you which of the doors has nothing behind it. Now there is only one door left (the one you didn't choose)... but there is still a 2/3 chance that the item you want is behind that door.

The result confirms itself if you actually do enough tests -- you'll win 2/3 of the time by always choosing to switch.

7. Originally Posted by Jennifer
When you make your first selection, there is a 1/3 chance you will pick the winning door.

The host does you a favor by showing you which of the doors has nothing behind it. Now there is only one door left (the one you didn't choose)... but there is still a 2/3 chance that the item you want is behind that door.

The result confirms itself if you actually do enough tests -- you'll win 2/3 of the time by always choosing to switch.
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.

8. Originally Posted by oberon67
Okay, with the Wason card problem, you're trying to determine if the following statement is false: "If a card has a vowel on one side, it has an even number on the other side." So to determine if it is false, you have to find just one example of a card that does not follow the rule.

So you turn over the A to see if the number on the reverse is odd, and you turn over the 7 to see if the letter on the reverse is a vowel. If either or both is true, you've accomplished your purpose. The other two cards are irrelevant, because the B does not match the premise, and the presence of the 4 can only confirm, not void, the conclusion.

I confess, I fell victim to confirmation bias before I went back and thought it through.
This was my first thought as well...but after thinking about it, wouldn't you have to turn the B card over as well as the 7 and the A? It isn't stated that cards have a letter on one side and a number on the other, so the B could have a vowel on the opposite side, disproving the hypothesis. The 4 is the only card that has no chance of disproving the statement.

Edit: oh, it does say that. uh, nevermind . I coulda sworn the first time I saw this problem (somewhere else) it didn't specify that. Just the A and 7 then.

9. Originally Posted by athenian200
My understanding is that if you choose any particular door, and then one of the other non-car doors is eliminated, then you have two choices, one of which is the car, and one of which is the goat. No matter which door you choose, the one that you previously chose, or the other, you still have a 1 in 2 chance of winning the car. I don't understand the answers they came up with.
Yeah, I'm with you here. I don't see why it makes any difference at all whether you switch or not, both should have an equal chance of winning unless the contest is rigged.

This is the part that makes no sense to me:
Originally Posted by Jennifer
but there is still a 2/3 chance that the item you want is behind that door.
EDIT: I googled it. "The probability that you will pick an incorrect door on your first guess is 2/3. When this happens, Monty is forced to open the one remaining GOAT door. By switching, you will then get the winning door on the 2 out of 3 times that you were incorrect on the first guess. Thus by switching, you turn the odds in your favor." That makes sense, I guess.

10. 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.
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.

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