# Thread: Simple puzzles to stump people

1. Originally Posted by FMWarner
Same here. In the Wason card problem, I don't see any reasoning for a card other than A, and the Monty Hall problem seems to be a simple 50/50 chance. The only complication I can see is a tell from the host, which isn't addressed. Is it possible that this is a case of lies, damned lies, and statistics?
For Wason:Think through the scenarios yourself, if you flip an odd number, is there a potential implication for the truth or falsehood of the general statement.

For Monty Hall:You can run the experiment yourself if you want. Take three playing cards, say 1 Queen and 2 Aces, and a friend.

Take turns being Monty Hall(you get real insight being him), or the contestant. Record the results.

I had my little brother do this with his pokemon cards and some friends of his and it was amazing how close to 2/3 of the time "Switching" was the correct strategy was (The little kids seemed to love playing the game).

2. Originally Posted by Economica
(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?
Okay, this made me understand. I accept now that it works this way, but I still think it's a funhouse mirror. If the universe is dress, we just saw the seam.

3. Originally Posted by ygolo
I don't think that's quite correct.
...
It doesn't add up.
Fudge, you're right -- I misread my grid. I'll have to rework.

(i hate details)

4. My solution.. spoiler!
There is pt and everyone else (E), 21 people, totalling 22 persons (A).
everyone in E has a different number of handshakes, from 0 to 20 handshakes.
lets call everyone in E by the number of their handshakes, from e0 to e20.

e20 must have shaken hands with everyone else in A but his/her partner.
Hence e20's partner must be e0, because that person is only person possible with no handshakes.

e19 has shaken hands with e20 but not e0, and with everyone from {pt, e1, .., e18} except his/her partner.

e19's partner hence has 1 handshake from e20, whereas others from {pt, e1, .., e18} have 2, one from e20 and one from 19.
Hence e19's partner must be e1, because that person is the only person possible with 1 handshake.

We can see that in every step, we decide one with the most handshakes and one with the least handshakes at any one point. We continue the same reasoning until we have solved e11..e20 and e0..e9.

At that point, e10 has shaken hands with everyone from {e11, .. ,e20}, but with no-one from {e0, .. ,e9}. These two groups include each other's partners, so e10's partner must be pt.

We thus identify e10 as Non, and see that Non has shaken hands with 10 people.

I would have formulated this rigorously if I had had the time. The puzzle was fun tho! Thanks!
It was nice exercise, but I would benefit more from the exercise of tidying my writings. I'll do that now
edit: on a second thought, I'll leave the explanation in it's untidy state.

5. Originally Posted by ptgatsby
I don't think I can come up with a practical analogy to the card problem...

Let's say the drinking age is 21 where you live.

If you are drinking (if a card has a vowel on one side), you must be 21 or older (then it has an even number on the other side).

There are four people you need to check for legality....

Person 1) "A" You know she is drinking alcohol, but don't know her age.
Person 2) "B" You know he isn't drinking alcohol, but don't know his age.
Person 3) "4" You know she is twenty-three but don't know what she's drinking.
Person 4) "7" You know he is 19 but don't know what he's drinking.

Flipping a card is akin to checking age, or for alcoholic beverage.

Now try it again, I bet it'll be easier.

6. Originally Posted by Santtu
My solution.. spoiler!
You got it.

7. Originally Posted by ygolo
Now try it again, I bet it'll be easier.
Ahhh... that made a lot more sense all of a sudden. Thanks!

8. Originally Posted by ygolo
You got it.
Whee!

9. Disclaimer: I haven't read this whole thread yet, so I apologize if someone has made these same comments.

On The Wason Card Problem I didn't see the solution readily displayed in the link, but this is how I reasoned through it. "If a card has a vowel on one side, then it has an even number on the other side."
From this statement it should be obvious that "A" is one of the solutions, since we know for sure that it has a vowel on one side. The contrapostivie of this statement is logically equivalent "If a card does not have an even number on one side, then it does not have a vowel on the other side." From this we can conclude that 7 is the other card that needs to be flipped.

On the Monty Hall problem, the solution depends on what assumptions are made about the problem. Logic problems are often created by INTP's and therefore there is an underlying assumption that the situation happens in a logical vacuum where "Let's Make a Deal" only airs once and/or this contestant has no prior knowledge about the "Let's Make a Deal" Show. According to these assumptions you should switch since there is a probabilty of 2/3 that the car is behind the other door.

However anyone who has studied game theory, and specifically Prisoner's Dilemma, knows that if you make assumptions that the game is repeated then the optimal strategy often can change. In this case if we assume that "Let's Make a Deal" is broadcast 5 days a week and that a great majority of the contestants are regular watchers of the show, then they would be able to observe that roughly 2/3 of the time it is advantageous to switch. Economically we should assume that the producers want to maximize profits and therefore give away as few cars as possible. It is therefore in the producers' best interest to make sure that the probability of getting the car by switching is as close to 1/2 as possible. (Because if the probability is far from 1/2 then daily observation of the show will let contestants know which strategy is optimal.)

One way to make the probability of guessing 1/2 is to open all of the doors 1/2 of the time whenever the contestant guesses incorrectly. This means that 1/3 of the time the contestant's initial guess is correct, 1/3 of the time it's incorrect with a chance to switch and 1/3 of the time the doors are opened immediately (instant failure). So if we assume that the contestant is given the option of switching the probability that the car is in the selected door is 1/2 and the probability that the car is in the other door is 1/2.

EDIT: Fixed a couple of spelling/grammar errors since this post was made in a hurry.

10. Originally Posted by The_Liquid_Laser
Disclaimer: I haven't read this whole thread yet, so I apologize if someone has made these same comments.

On The Wason Card Problem I didn't see the solution readily displayed in the link, but this is how I reasoned through it. "If a card has a vowel on one side, then it has an even number on the other side."
From this statement it should be obvious that "A" is one of the solutions, since we know for sure that it has a vowel on one side. The contapostivie of this statement is logically equivalent "If a card does not have an even number on one side, then it does not have a vowel on the other side." From this we can conclude that 7 is the other card that needs to be flipped.

On the Monty Hall problem, the solution depends on what assumptions are made about the problem. Logic problems are often created by INTP's are therefore there is an underlying assumption that the situation happens in a logical vacuum where "Let's Make a Deal" only airs once and/or this contestant has no prior knowledge about the "Let's Make a Deal" Show. According to these assumptions you should switch since there is a probabilty of 2/3 that the car is behind the other door.

However anyone who has studied game theory, and specifically Prisoner's Dilemma, knows that if you make assumptions that the game is repeated then the optimal strategy often can change. In this case if we assume that "Let's Make a Deal" is broadcast 5 days a week and that a great majority of the contestants are regular watchers of the show, then they would be able to observe that roughly 2/3 of the time it is advantageous to switch. Economically we should assume that the producers want to maximize profits and therefore give away as few car's as possible. It is therefore in the producers' best interest to make sure that the probability of getting the car by switching is as close to 1/2 as possible. (Because if the probability is far from 1/2 then daily observation of the show will let contestants know which strategy is optimal.)

One way to make the probability of guessing 1/2 is to open all of the doors 1/2 of the time whenever the contestant guesses incorrectly. This means that 1/3 of the time the contestant's initial guess is correct, 1/3 of the time it's incorrect with a chance to switch and 1/3 of the time the doors are opened immediately (instant failure). So if we assume that the contestant is given the option of switching the probability that the car is in the selected door is 1/2 and the probability that the car is in the other door is 1/2.
Another way to get a probablity of 1/2 is to simply say the producers will once again randomly place the desired prize behid the remaining doors after one door is opened.

Also....

if we were to modify the problem so that you are allowed three options, stay with your door or switch to either of the two other doors. Let's say now, due to theatrics by the host, you were able to accidentally glimpse behind a door you didn't pick to see a booby prize, instead of what you wanted. Obviously, you wouldn't pick the door with the booby prize, but would you keep your door, or switch? Does it make a difference?

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•