Brain Teasers Q&A Game - Answer the First Question Now!

[Your.Master]

I can quickly up their odds to 12.5%

Let's call Art History guy A, Biochemistry guy B, Law guy L, and Math guy M.

They decide to label the desk closest to the door as 1 and the furthest is 4. Alternatively, if row is aligned such that no desk is closest the door, then you choose the rightmost as 1 and the leftmost as 4. Or we could do something with cardinal directions. Anyway, we just need them labelled.

A chooses between 1 and 2. 50% he's wrong, and then we've failed. Everybody else operates under the assumption he's right in his choice.

B chooses between 1 and 2. Since we're assuming A was right (otherwise, we're boned anyway), he has a 25% chance of being right. 50% times 25% is 12.5%.

L and M each choose between 3 and 4. A and B have to already be right, or else it doesn't matter. If they are right, then 1 and 2 are already accounted for, L and M are both guaranteed to get the right exam.

This isn't good enough yet, clearly, just my start.

 

[dontdisturbmycircles]

We have 4 guys, art history, biochemistry, law, and mathematics.

Let me rename them please because I named them differently on my piece of paper and I don't want to convert the names

I named them Math, science, biology, and chemistry.

The main thing that I was going for is to ensure that they all sit at different tests while being flexible.

The plan is this. The Math guy goes in, and flips over paper #1.


When Math guy sits at paper #1

If it is math, he sits down and writes his test

If it is science, he goes to #2

If it is bioogy, he goes to # 3

If it is Chemisty, he goes to #4

When the Science guy sits down, he goes to paper #2

If it is math, he goes to #1

If it is science, he sits and writes his test

If it is chemistry, he goes to #4

If it is Biology, He goes to #3

When the Biology guy goes to write his test, he goes to table #3

If it is math, he goes to #1

If it is science, he goes to #2

If it is biology, he sits and writes his test

If it is Chemitry, he goes to #4

When the Chemistry guy goes, he looks at paper #4 first

If it is math, he goes to #1

If it is science, he goes to #2

If it is Bio, he goes to #3

If it is Chemistry, he sits down and writes his test.

I think that solving the probability for this strategy in itself is a brain teaser... lol, I know that they for sure have a 1/6 chance of a guaranteed win (math and science compose first 2 booklets) I can also see that if I can get the first one seated correctly, and that the second one sits at a different desk, then his chances of selecting the right desk is higher because when he first picks up the paper and finds out where he should sit, if the first guy is seated correctly, then there is only 2 seats to choose from. I am going to bed, goodnight, good luck to anyone who is trying to solve this :P.

Edit: Thought of a way to test probability... doing it now.

 

[dontdisturbmycircles]

I am correct, the probability of success using my system is 41.66666%. I will post proof in a second.

EDIT:

Proof(Not the most elegant/simple proof but it is a 'proof' none the less) You can either work through it or trust me! :-)):

Please note that the B with the erasing marks should be a C.

http://img207.imageshack.us/img207/2351/math01002sj1.jpg

http://img207.imageshack.us/img207/2171/math01003yk4.jpg

The underlined ones are what actually happens in the given situations. I tally up the successes in the right side and divide the successes by the total number in the sample space. Proving that my system will work 10 out of 24 of the possible times. Hurrah! That was one of the best puzzles I have ever seen DaveE, seriously.

Oh, and excuse the lack of a % sign after the 41 2/3 :P.

I don't know why it took me so long to think of writing out the entire sample space...

M=Math
S=Science
B=Biology
C=Chemistry

I started to use different names and then never changed. But obviously it is the same solution. :P

 

[davee123]

the probability of success using my system is 41.66666%.
...
Proving that my system will work 10 out of 24 of the possible times. Hurrah!

Correct!

Now, the thing that totally blew *my* mind about this problem was that even if the remaining students are allowed to watch *everything* that's going on in the test room (so they can see which tests are which as the current exam-taker reveals them), they STILL have a 41.666% chance of success! The only thing that it saves them is the number of wrong guesses! Explanation:

Possible situations (with other students watching):
A) 1st student picks wrong, then wrong = 3/4 * 2/3 = 1/2
nothing else matters, they fail

B) 1st student picks wrong, then correct = 3/4 * 1/3 = 1/4
now, the remaining students know the identity of one of the remaining tests, and that student can go straight for it without guessing. The two other students know that they can go for the other two options, and are guaranteed success.

C) 1st student picks correctly on his first try = 1/4
now, the remaining students have learned very little. New student goes in:
C1) 2nd student picks correctly on his first try = 1/3 (victory!)
C2) 2nd student picks correctly on his second try = 2/3 * 1/2 = 1/3 (victory!)
C3) 2nd student picks incorrectly on both = 2/3 * 1/2 = 1/3 (failure!)

So overall chance of success is the chance of B (1/4) plus the chance of success in C (1/4 * (1/3 + 1/3)). So 1/4 + 1/6 = 10/24 = 41.666%!

DaveE

 

[dontdisturbmycircles]

Thanks DaveE. I am not very good at probability so I just solved it from a logical standpoint. I knew that even if the test writers can't tell the others what they did, the future test writers can tell what they did by looking at the tests. For example, when the second guy goes in and looks at the second test. If it is not math then he knows that either they already failed or that the first desk was math. So he wants to go to a different desk obviously because if two of them write the same test, they fail.

Then for the third and fourth guys, they also know what happened. If third guy walks in and the 1st guy's test is on his desk, he knows that either they screwed it up, or the 3rd guy's test is sitting on the first desk. So yea, two different ways to solve it I guess. The only difficult part for me was trying to prove the probability... I need to improve my probability skills.

I'll think of another problem in a bit.

 

[dontdisturbmycircles]

Here is my main puzzle

Q: Place the same two letters in the exact centre of each word so that five longer words are formed. Which two letters must be used?

CARS EARS LAST PANT WARS

 

[Jimmy Snyder]

Q: Place the same two letters in the exact centre of each word so that five longer words are formed. Which two letters must be used?

CARS EARS LAST PANT WARS

Answer (hilite to see) ET works. eom.

 

[dontdisturbmycircles]

Nicely done JimmySnyder. Your turn to ask a question.

 

[vamfun]

Hi all, I wanted to post this comment re the second teasor. Is there a way to have it appear in a relavent place rather than here?

"I agree with this answer... But it must allow 0 hairs. I would reason a little differently.
To get colony started 1st man must have 0 hair to meet condition 3.
The second man must have 1 hair to meet conditions 1 and 3
The third man must have 2 hairs to meet conditions 1 and 3 etc
So there is always one more man than the max number of hairs on a head and the number of hairs increases by one with each new man added.

Since no man can have exactly 483207 hairs then the population stops at 483207 with the last man having exactly 483207-1 hairs."