Puzzle Quiz

SA - salary of A
SB - salary of B
Sc - salary of C
X, Y, Z - random numbers

Person A writes a (SA + X) and hands it to B, without showing C. B write (SB + Y + result from A) and passes to C. C adds SC + Z and passes to A. Now A subtracts X, passes to B. B and C each subtract their random number and pass. After C is done, he shows the result and they divide by 3.

Next. :D
 
Originally posted by Gainy bo
They could each just tell some random other guy, and get him to figure it out. He would then tell 'em the average. Correct?
Click to expand...
:) Why didn't I realize that? But I was thinking of something not involving a third party. You're technically correct, though.
Originally posted by betazed
SA - salary of A
SB - salary of B
Sc - salary of C
X, Y, Z - random numbers

Person A writes a (SA + X) and hands it to B, without showing C. B write (SB + Y + result from A) and passes to C. C adds SC + Z and passes to A. Now A subtracts X, passes to B. B and C each subtract their random number and pass. After C is done, he shows the result and they divide by 3.

Next. :D
Click to expand...
:thumbsup: Hmm, that works, but that's not what I had in mind. What I was thinking was person A collects the ones digit from everyone, person B the tens digit, etc. Then someone adds the digits together (making sure to multiply the tens digit by 10, etc.), and divides this sum by 3.

How about Gainy bo gets to ask the next question, since I assume you (betazed) don't want to ask a question.
 
I don't know any decent questions, i'll leave it to someone else if that's ok :)

(maybe you again Will ;))

Edit: S'ok by me :)
 
There are 5 pirates who have just captured a treasure chest of 100 coins. Now they must decide on a way to split it.

Each pirate is infinitely intelligent and greedy and knows the others are too.

Pirate E gets to propose a deal, then D, C, B, A, etc.

If the majority of the pirates approve of the deal, then the money is divided up accordingly. If not [or if there's a tie] the pirate is killed and the next person gets to propose a deal.

Obviously, dead pirates get no money! The pirates are infinitely bloodthirsty and have no qualms about offing a fellow buccaneer. For example, suppose B and A are the only pirates left. No matter what deal B proposes [even 99 for A and 1 for B], A will vote against it - because then B will die and A will get ALL the gold! From this, we can deduce that it is not in B's interest to be left alone with A.

Pirates value their own lives above any amount of money and any amount of money above any other pirate's life.

So: what deal did the pirates eventually agree on?
 
The question is too vague. Are these pirates happy with less gold than another? Are we to find the ideal deal for Pirate E?(Perfection's answer) If neither, then there are multiple answers. Any different three numbers which add up to 100 and none of which are zero could be a potential answer.
 
Since Pontiuth has logged off, I'll ask the next question since I believe the present one is faulty and/or ridiculously easy. This one I will post is not too difficult, but I will require explanation.

Edit:Sorry, messed up in the typing. Modified.

There are three men, named John, Joe, and Jim. They have a dispute and obviously they decide to have a shootout. They draw straws to find the order of the shooting. They will take turns and the last man standing...wins, I guess. When one of them has a choice of who to shoot, he always chooses in his best interest. He can only shoot one person at a time.

You see, John and Joe both are perfect shots. All the time. 100%. Jim, unfortunately, only hits the mark half the time.

So, who has the best chance to survive the shootout?
 
Originally posted by Syterion
Since Pontiuth has logged off, I'll ask the next question since I believe the present one is faulty and/or ridiculously easy. This one I will post is not too difficult, but I will require explanation.

There are three men, named John, Joe, and Jim. They have a dispute and obviously they decide to have a shootout. They draw straws to find the order of the shooting. They will take turns and the last man standing...wins, I guess. When one of them has a choice of who to shoot, he always chooses randomly.

You see, John and Joe both are perfect shots. All the time. 100%. Jim, unfortunately, only hits the mark half the time.

So, who has the best chance to survive the shootout?
Click to expand...
Joe and John have equal chances. I worked it out below (took a while). Below is each possibility of the straw-drawings listed, and below each of these possibilities is the list of sub-possibilities, and below each of these is the chances for that particular straw-drawing.:


Edit: DAMN YOU, all that work for nothing!


John, Joe, Jim
50%: John shoots Joe, 50%: John shoots Jim. If John shoots Joe, then Jim tries to shoot John and 50% succeeds (and survives), 50% fails (and dies by being shot by John). If John shoots Jim, Joe shoots John and Joe survives.
CHANCES: John: 25%, Joe: 50%, Jim: 25%

John, Jim, Joe
50%: John shoots Jim, 50%: John shoots Joe. If John shoots Jim, Joe then shoots John and Joe survives. If John shoots Joe, Jim then shoots John and 50% succeeds (and survives) and 50% fails (and dies from being shot by John).
CHANCES: John: 25%, Joe: 50%, Jim: 25%

Joe, John, Jim
Same as first one, but alternate John/Joe.

Joe, Jim, John
Same as second one, but alternate John/Joe.

Jim, John, Joe
50% Jim tries to shoot John, 50% Jim tries to shoot Joe. If Jim tries to shoot John, 50% succeeds and 50% fails. If Jim tries to shoot Joe, 50% succeeds and 50% fails. If Jim shoots John, Joe shoots Jim and survives. If Jim shoots Joe, John shoots Jim and survives. If Jim fails to shoot anyone, 50% John shoots Jim and 50% John shoots Joe. If John shoots Jim, Joe shoots John and survives. If John shoots Joe, Jim shoots John and 50% succeeds (and survives) and 50% fails (and dies from being shot by John).
CHANCES: John: 37.5%, Joe: 50%, Jim: 12.5%

Jim, Joe, John
Same as above, but alternate John/Joe.
 
Originally posted by Syterion
Look at it now, I've edited it.
Click to expand...
Well then it's Jim becasue the target him less often (he wins 5/12 of the time)
 
Hate to telly you, WillJ, that that is wrong. If all your calculations are right, John and Joe still have the same chance. Unfortunately, the edit erases all that.
 
Originally posted by Syterion
Hate to telly you, WillJ, that that is wrong. If all your calculations are right, John and Joe still have the same chance. Unfortunately, the edit erases all that.
Click to expand...
I think you missed my edit. ;)
 
Sorry for the confusion, people, here's a problem where I can't accidentally word it wrong.

Two people meet on a plane. We'll call them Tim and Tom.
"You have three sons, correct? What are their ages today?" said Tom, to Tim.
"The product of their ages is 36, and the sum of their ages is exactly today's date."said Tim.
"I'm sorry Tim, but that doesn't tell me the age of your sons." said Tom.
"Oh I forgot to tell you, my youngest son has red hair." said Tim.
"Ah now it's clear," says Tim, "I know exactly how old your sons are."

How old were the sons?
 
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