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okay here is post 30, 31 will be a new page :D

Rule Recap
If you already know the puzzle, don't blurt out the answer. Whoever guesses first will get a special picture reward =D I'll give hints if people don't figure it out.

Puzzle #2
This one is classic. Please excuse my first person narrative. (I'm not married, but it helps to tell the puzzle this way =D)

My wife and I were invited to a party recently, a party attented by four other couples. Some of the 10 knew some of the others and some did not, some were polite and some were not. As a result, a certain amount of handshaking took place in an unpredictable way, subject only to two obvious conditions: No one shook his or her own hand and no husband shook his wife's hand.

When it was all over, I was curious. I went around the party asking each person "How many hands did you shake?...And you?...And you?" What answers could I have received? Conceivably, some people could have said none, and others could have given me a number from 1 to 8. That's right, isn't it? Since self-handshakes and spouse-handshakes were rules out, eight is the maximum number of hands that any one of the party of 10 could have shaken.

I asked nine people (everybody, including my wife), and the answer could have been any one of the nine numbers 0 to 8 inclusive. I was interested to note that the nine different people gave me nine different answers: Someone said "none," someone said "one," and so on until someone, finally, said "eight."

So, to recap. there were 10 people at the party, I asked nine of them how many hands they shook. As it happened, one person shook 0 hands, one person shook 1, one person shook 2, and so on all the way up to eight.

Now I ask you: How many hands did my wife shake?

this is a classic? I honeslty have never heard it... it seems to have some math involved... i'll have to get out a pad of paper and work on it for a bit here...

are you missing any clues there.. that seems to be impossible to solve with out at least a little bit more info...

The wife shook 1 hand.

The second one was easier than I thought it would be.

The person who shaked 8 hands didn't shake his own hand or his wife's hand (who didn't shake anyone's hand, the mean...). So the person who shook 7 hands was with the person who shook 1 hand, 6 with 2, 5 with 3, and the person who shook 4 hands was the wife.

So the answer is 4.

At 4/21/05 02:52 AM, ZombieLincoln wrote: Now I ask you: How many hands did my wife shake?

Seven by chance?

At 4/21/05 03:03 AM, AvatarEgo wrote: this is a classic? I honeslty have never heard it... it seems to have some math involved... i'll have to get out a pad of paper and work on it for a bit here...

Classic as in, I love it, not as in, well-known. =) Glad it's new for you - and there's no math necessary, only logic.

At 4/21/05 03:09 AM, AvatarEgo wrote: are you missing any clues there.. that seems to be impossible to solve with out at least a little bit more info...

The beauty of this puzzle is how little information it gives. It IS possible!

At 4/21/05 03:11 AM, Bivouac_Tenure wrote: The wife shook 1 hand.

There are only nine possible answers here, and that means that process of elimination is a factor here. To combat that, you must PROVIDE YOUR REASONING with the answers. I won't confirm or deny yours until I see how you it =D

Sorry I didn't think of this in the first post.

At 4/21/05 03:12 AM, Qwoxyl wrote: The second one was easier than I thought it would be.

The person who shaked 8 hands didn't shake his own hand or his wife's hand (who didn't shake anyone's hand, the mean...). So the person who shook 7 hands was with the person who shook 1 hand, 6 with 2, 5 with 3, and the person who shook 4 hands was the wife.

So the answer is 4.

Whoa! Before I even got done replying, you bust out with the answer. Well done man! XD
The awards picture will be done in about 3 minutes, can't keep up with yer brain =D

okay I know its now been solved... I took to long on paper I guess..
but here is the long extended explanation

so 0,1,2,3,4,5,6,7,8
8 and 0 are a couple 8 shook hands with the wife +1
so take them out and reask the question
new answers would be 0,1,2,3,4,5,6
6 and 0 are a couple, 6 shook hands with the wife +1
take those two out, we have a remaining 6
new answers would be 0,1,2,3,4
4 and 0 are a couple 4 shook hands with wife +1
remaining 0,1,2
2 shook hands with his Wife +1
2 and 0 are a couple.. the 1 is his wife

add it up and She shook hands with 4 people

At 4/21/05 03:25 AM, AvatarEgo wrote: okay I know its now been solved... I took to long on paper I guess..
but here is the long extended explanation

Well solved, sir. *respect*

But your speed didn't match the glory and power of Qwoxyl, this time!

To Qwoxyl, the spoils!

if there are 5 white people and one gets stiff nipples... THIS BRAINTEASER IS OVER!

At 4/21/05 03:28 AM, ZombieLincoln wrote:

At 4/21/05 03:25 AM, AvatarEgo wrote: okay I know its now been solved... I took to long on paper I guess..
but here is the long extended explanation

Well solved, sir. *respect*

But your speed didn't match the glory and power of Qwoxyl, this time!

It sure didn't (No one mention that I posted my answer 20 minutes after the question was).

To Qwoxyl, the spoils!

LOL. Yay, I have a weird guy award. Keep the questions coming.

I think this one is somewhat well known. If you know it, be a sport and don't tell the answer. And if you know the puzzle and want to pretend you figured out the answer on your own, wait a minute for me to finish the prize =D

Include your reasoning with your answers, please.

Puzzle #3

Some bored guy decides to start a new thread on the NG BBS. He tries to imagine a thread that would be something lots of people would contribute to! After much consideration, he decides to make a birthday thread.

He opens with "hey every1! I thought it would be cool to make a thread for peeps who are having there bibrthdays. its ok if your birthday isnt today, just tell everyone what day it is and we can see who has teh same birthday as you LOL!"

How many people must post their birthdays on the thread for there to be a 50% chance that there is at least one duplicate birthday?

Good luck =)
*gets to work on prize animation*

Whatever.

At 4/21/05 03:46 AM, _MessiaH_ wrote: Whatever.

Good call.

19

With 2 people posting, they have a 2/365 chance of having the same b-day. Every person thereafter adds to the Numerator a number equal to the number of other people already posted, because he or she has a 1/365 chance of having the same b-day as any other person. Once you get to the 19th person you have reached a possibility of 191/365 which is greater than 50% (182.5 is exactly 50%).

19

With 2 people posting, they have a 2/365 chance of having the same b-day. Every person thereafter adds to the Numerator a number equal to the number of other people already posted, because he or she has a 1/365 chance of sharing a b-day with any other person. Once you get to the 19th person you have reached a possibility of 191/365 which is greater than 50% (182.5 is exactly 50%).

Edited for clarity... stoopid lack of an edit button >.<

At 4/21/05 04:18 AM, Ravariel wrote: 19

With 2 people posting, they have a 2/365 chance of having the same b-day. Every person thereafter adds to the Numerator a number equal to the number of other people already posted, because he or she has a 1/365 chance of having the same b-day as any other person. Once you get to the 19th person you have reached a possibility of 191/365 which is greater than 50% (182.5 is exactly 50%).

You're certainly on the right track, sir, but this "adding to the numerator" method isn't accurate. The math is actually a bit uglier, I'm afraid =\

BTW, this award is looking pretty snazzy =D

Oh, son of McCrap...

I know what it is... but Im going ot need to find my calculator methinks...

Hate figuring out factorials in fractions...

The award is finished and ready to be customized. This means two things:
1: I bump this thread
2: I go to sleep for some hours.

Good luck to anyone participating!

2 people. They have a 50% chance, their birthdays either match or they don't. 50-50 you see. I think that's true actually :/

At 4/21/05 05:41 AM, FeeL_Os_oFf wrote: 2 people. They have a 50% chance, their birthdays either match or they don't. 50-50 you see. I think that's true actually :/

With two people, there's a 1 in 365 chance that they have the same birthday. However, with 3 people, the probabiliy is trickier...

At 4/21/05 05:41 AM, FeeL_Os_oFf wrote: 2 people. They have a 50% chance, their birthdays either match or they don't. 50-50 you see. I think that's true actually :/

As an excuse to bump this, I'll answer you in some more detail.

There are only two options when you have 2 people - the birthdays either match or they don't. However, the probability of each option isn't equal.

If the first person to post his birthday is Hitler, then you have April 20th as yer birthday to match. The second person to post has one way to match (having a birthday on 4/20) and 364 ways to not match (having any other birthday)

So, out of the 365 possible birthdays, only 1 will make a match, making the probability 1/365. Keep guessing!

PS. Ignore leap years, they're silly.
PPS. If you'd like, I could start making the award pictures BBS Sig-sized, if you want the chance to gloat over your victory with every single post you make. =D

Those puzzles rape your mind.

Don't do puzzles children >: ( stay in school.

At 4/21/05 12:16 PM, -0706- wrote: Those puzzles rape your mind.

Don't do puzzles children >: ( stay in school.

Lies! Drop out of school and live on the streets, doing puzzles like mad! Live in a box covered plastered with old crossword puzzles and word searches! JOIN ME!

-0706-, you may like the award picture for this puzzle...it would look awesome with your name on it....*tempt tempt*

Will no one take up this cup? (the question is halway up this page)

To be fair, you will probably need a calculator or a powerful brain to get the answer...
But, so much glory to be had! =\ =/

who the fuck cares?

At 4/21/05 05:54 PM, TheChaosDemon wrote: who the fuck cares?

Fair enough...

Well, i added up all the probabilities (1/365 + 2/364 + 3/363 + 4/362 etc.), and got an answer of 19, but I believe this has been suggested before. The 18th person has a probability of just 48%, the 19th is 54%.