Brain benders

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I reckon it was Alex for no logical reason really but because he was the only person to accuse people who did not accuse him.

 

OK, it's late, so I'm probably not thinking straight, but here I go.

It is Chris who is definitely lying.

If Chris was being truthful, then Brad must be lying.

If Brad is lying, then Alex is telling the truth.

If Alex is telling the truth, then Chris must be lying.

However, Chris started by telling the truth, so that is impossible. The only logical answer is that Chris is lying.

[Edit] Having thrown around my faulty logic in my head for a bit, I have decided that you know that they are all lying as they all contradict each other. [/Edit]

[This message has been edited by Wildfire (edited October 10, 2001).]

 

uhm Invoker. The one who is lying are we sure he is lying on both statements?

There are two approaches.

a) to say that if X lies for certain then the persons Y and Z he mentions must be telling the truth. If Y tells the truth the M and N lies. This approach shows no solutions.

b) To say that if X lies for certain then the persons who mentions X as a lier must tell the truth. This approach show no solutions.

Hence all are liers
Both are obviously very easy to prove so I don't think you intended it this way. hmmm


EDITED HERE: ahhh After reading Dargorn's post it has become clear what your aim is Invoker. Since we can't wager one statement over another it is imposible to determine anything except from the Alex case. But the only thing we can say about Alex is that it is LESS likely that he lies. or what?

[This message has been edited by Nobleman (edited October 10, 2001).]

 

Actually, Dargorn's simple solution is correct. The best way to solve this puzzle is by reductio ad absurdum -- i.e., assume the opposite and then follow it to its logical conclusion until it proves itself false. So assume Alex is telling the truth. Chris and Dave must be lying. Chris claims that Brad and Dave are lying; since he's right about Dave, Brad must actually be telling the truth. But Brad claims that Alex is lying. Therefore, Alex CANNOT be telling the truth. QED.

Brad can be telling the truth if Dave and only Dave is also telling the truth. Chris can be telling the truth if Eric and only Eric is also telling the truth.

 

Doh! Well done Capstone. Good puzzle Invoker. Sorry that I misunderstood.

Because I didn't flip flop (reductio ad absurdum ) it, I ran into the problem the first line in my post above says.. It was funny. At least I think it was.

[This message has been edited by Nobleman (edited October 10, 2001).]

 

A ranger, took his bow and fired an arrow at ten birds in a tree. (closely bunched together)
How many birds were left after he shot the arrow?

Err... this one's a bit easy .

HINT
It doesn't matter if the ranger hit a bird or not.

[This message has been edited by creudzfeldt-jakob (edited October 10, 2001).]

 

hmmmm. After he shot his arrow all ten of them would still be there. It is only when/if the arrow arrives that the morale of the birds matter.


EDITED HERE: Was it an arrow of detonation? Then it wouldn't matter if he hit them or not.

[This message has been edited by Nobleman (edited October 10, 2001).]

 

The crucial piece of information in this puzzle is that the firee is a ranger. From this insignificant clue we can deduce that all the birds will still be there -- nailed to the tree limb by the ranger's incredible hunting prowess.

 

wow you guys are great Well done!

here's another one if you feel like indulging, albeit a bit on the mathematical side:

there are 9 potions (each different)in a wizard's lab but their labels have fallen off. The wizard tells his apprentice to re-label them. The apprentice has no idea which is which but is too afraid to admit this to his master. So he decides to reaaply the labels randomly. What is the expected number of potions labeled correctly ? (Rounding the answer to an integer)

PS: what a greedy ranger

 

I hate math I really do.

SO I will take an "educated" haha guess, I think you can expect that the apprentice will only get one right and that's with rounding.

 

Speaking of experience in your own lab Invoker?

Anyway it is just slave work. I love it

The chance of all labelled right is: 1/9 x 1/8 x 1/7 x 1/6 x 1/5 x 1/4 x 1/3 x 1/2 x 1

The chance of 8 labelled right is:

8/9 (chance applied of missing one) x 1/8 x 1/7 x 1/6 x 1/5 x 1/4 x 1/3 x 1/2 x 1

The chance of 7 labelled right is:

8/9 x 7/8 (chance applied of missing two) x 1/7 x 1/6 x 1/5 x 1/4 x 1/3 x 1/2 x 1

.....................

Chace of Zero bottles labelled right is

8/9 x 7/8 x 6/7 x 5/6 x 4/5 x 3/4 x 2/3 x 1/2 x 1.

So the one with all the largest nominators next to the denominator is Zero bottles. But it is not at all very likely. Just the most likely hmmm I don't wanna calculate it.

 

lol my lab allright

well, my bad shouldn't have turned the thread into a stat mech class with a swift stroke of potion vials

Not a good choice of puzzle there sorry.

in case anyone wonders:

well nobleman you're righ but we want the probabilities to add up to 1. So you should also multiply those with combinations. (like 4 correct potions could be any combinations of potions)

So, for 8 potions, you should have P(8) = C(9,8)*(1/9)*(1/8)* etc..

P is probability
C is combination (how many eight potions you can pick out of 9)

then the expected number = 9*P(9) + 8*P(8) + ..

bah I wouldn't wanna calculate it either


[This message has been edited by Invoker (edited October 10, 2001).]

 

Probabilities and Combinations is one way to look at it. But *sigh why do I submit myself to such torture* I think there is another way. Something like there are 9 trials and each trial has 2 options-> a chance of failure and a chance of success. A bionomial distribution thing I learned, anywho that class is long gone and so is the formula in my head I still think expected value rounds to 1, without doing any calculations, just judging from past experience.

Lol it's binomial not bionomial, I always do that

[This message has been edited by Big B (edited October 10, 2001).]

 

Yeah I do it twice a day too Big B

Anyway Invoker I'll let you know if I have a week or two to spare to calculate it in

 

Well . . .
Nobleman's and Capstone's answers weren't exactly on the spot.
As I said, it doesn't matter if the ranger hit the birds or not, so even if he happened to kill all the birds (with skill or an arrow of detonation , it wouldn't matter)

By the way, the fact that he's a ranger doesn't matter. Also, when I mean "here", I mean alive and perching on a branch of the tree.

Well, I guess I'll give it a few more days.


[This message has been edited by creudzfeldt-jakob (edited October 10, 2001).]

 

Well I think Nobleman got it, but just to spell it out - they'd all fly away, being scared of the shot.

 

Actually, Nobleman, I don't believe that is the correct answer. For instance, the chance of getting exactly eight bottles labeled correctly is zero. If he's got eight right, the ninth one is going to be right too. The problem is these are not independent events, and therefore the math is a bit more complicated. Unfortunately, I've got an appointment in ten minutes, so I've got to leave, but I'll try to post further explanation tomorrow.

 

Yes you are right. I adressed the problem pretty superficial.

So there is this one instance to account for. Anyway Invoker has already told and explained most of it short and precise.

All the probabilities have to add up to 1.

So if we want to calculate the probality for, lets say 6 labelled correctly. We have to sum up the probability for all the combinations of bottles labelled wrong. that is an exhaustive work. It could be bottle 1 and two and 4 or bottle 3 and 4 and 6. Its too long time ago that I worked with probabilities so I trust Big B on this one. All I remember is that probability is some nasty stuff, so there might be some short cuts. If there are any I bet you find them Capstone