Who is the liar in this logic puzzle?

[y3nis]

I got something nice too.
I am a Gabber with a bold head.
Am I lying or telling the truth? [?]


Kindest regard Dj You_nis

 

[marqq]

yes you are

 

[y3nis]

wrong, i aint a skinnhead I am the devil :P

 

[marqq]

Tell me y3nis where did i said that you were a skinhead ?

 

[y3nis]

you didn't. let's make this really philosophy. What exectly does skinhead mean? Someone that is bold i'd say. Else my previous post weren't correct

 

[marqq]

Did I really said smth?
Pls. quote me if did

 

[y3nis]

Originally posted by marqq
Did I really said smth?
Pls. quote me if did

Save My Tits Here=smth?

Well i'll tell you the truth. I aint a gabber and i aint bold.
Did i really tell the truth?

 

[marqq]

Originally posted by y3nis
Save My Tits Here=smth?

something

Btw...try to puzzle me by offering an corect answer to my 12 balls problem

 

[marqq]

Originally posted by marqq
so...we have 12 balls.Only one is different(lighter or heavyer)
We also have a balance...and only 3 trys to find the different ball
Enjoy

you didn't forget about that didn't you?

 

[marqq]

where r the signatures?

 

[tribdog]

next problem

You are the big winner of a television game show. The host says "Grand prize is behind door #1, 2 or 3. pick one." After you make your choice (lets say you pick #1) the host says "You may be right. It's either #1 or #2 but its not number 3. Now I'm going to give you one last chance do you want to stick with door #1 or would you like to pick again?" What should you do?

 

[leto]

I just took the time to answer that horribly long balance puzzle, and my post didn't work. I'm not certain I am right, I am tired and can't think anymore. There is more to it than this, but this is all I had in my clipboard and there is no way in hell I'm typing anymore:

I would take 8 of the 12 balls and weigh them with 4 on each side of the balance.

If they were even, then I would know the odd ball was one of the 4 I haven't weighed. If the balance tipped then I would label the balls on the tipped side as l1 l2 l3 and l4. The balls on the other h1 h2 h3 and h4.

I would then take l1, h4, h3 and weigh against h1, h2 and l2. If the balance was even, then I would know the odd ball was either l3 or l4. if the balance tipped to the first side, then i would know the odd ball was either h4, h3 or l2. if the scale tipped to the second side, then I would know the odd ball was either h1, h2 or l2.

If it was l3 or l4 then I would use my last balance to measure l3 again any ball other than l4. If the scale was even then l4 is the odd ball. If the scale dips to l3 then l3 is the odd ball.

It it was h4, h3, or l2 then I would use my last balance with h4 vs h3. If the balance dips to h4 then h4 is the odd ball. If the balance dips to h3 then h3 is the odd ball. if the scale is even then l2 is the odd ball.

If it was h1, h2, or l2, then I would do the same as directly above and find the odd ball.

If my initial weigh was balanced, then I would take 3 of the 4 remaining balls and weigh them against 3 other balls. If the balance stayed level, then I would know the odd ball was the one I didn't weigh. If it did not, then I would know whether the 3 balls were lighter or heavier. I would then take 2 of the 3 remaining balls and weigh them against each other. If the balance was level then the odd ball is the one I didn't weigh. Otherwise, it is the heavier/lighter one, depending on whether the 3 balls were lighter or heavier.

 

[marqq]

It it was h4, h3, or l2 then I would use my last balance with h4 vs h3. If the balance dips to h4 then h4 is the odd ball. If the balance dips to h3 then h3 is the odd ball. if the scale is even then l2 is the odd ball.

If it was h1, h2, or l2, then I would do the same as directly above and find the odd ball.

If my initial weigh was balanced, then I would take 3 of the 4 remaining balls and weigh them against 3 other balls. If the balance stayed level, then I would know the odd ball was the one I didn't weigh. If it did not, then I would know whether the 3 balls were lighter or heavier. I would then take 2 of the 3 remaining balls and weigh them against each other. If the balance was level then the odd ball is the one I didn't weigh. Otherwise, it is the heavier/lighter one, depending on whether the 3 balls were lighter or heavier.

isn't it cute? BUT IS WRONG! remember that u don't know if the ball is lighter or heavyer...

It it was h4, h3, or l2 then I would use my last balance with h4 vs h3. If the balance dips to h4 then h4 is the odd ball. If the balance dips to h3 then h3 is the odd ball. if the scale is even then l2 is the odd ball.

why?
but you had a good start!

 

[marqq]

You are the big winner of a television game show. The host says "Grand prize is behind door #1, 2 or 3. pick one." After you make your choice (lets say you pick #1) the host says "You may be right. It's either #1 or #2 but its not number 3. Now I'm going to give you one last chance do you want to stick with door #1 or would you like to pick again?" What should you do?

whatever you want...you have a 33% chance to find the prize...you pick #1...and you find that the door #3 is a fake...Your chances increase to 50%...so either you take #2 or you rest on #1 you heave the same chances to win the prise :)

 

[leto]

I had one typo, but the fact that I don't know whether it is heavier or lighter was taken into account. This puzzle would be simple if I knew if it was lighter or heavier.

"if the scale tipped to the second side, then I would know the odd ball was either h1, h2 or l2."

should of been:
"if the scale tipped to the second side, then I would know the odd ball was either h1, h2 or l1."


I can connects the dots if you still think I am wrong, but I'm too lazy to do that until then :P

 

[tribdog]

whatever you want...you have a 33% chance to find the prize...you pick #1...and you find that the door #3 is a fake...Your chances increase to 50%...so either you take #2 or you rest on #1 you heave the same chances to win the prise :)

no I don't think that is correct. its close, but not right. What if there are a million doors and after you choose Monty Hall/Bob Barker throws out 999998 of the bad doors and says it's either the one you chose or this one? do you think it is still a 50/50 chance then?

 

[marqq]

tribdog if there were 1milion doors you would have 1/1000000 chances to win the prise...if we-ll take off a fake one, either you change or you keep your door, you'll have a 1/999999 chances to win the prize...so...the answer is the same...you can do whatever you like...you have the same chances

 

[marqq]

leto ... let me give you a hint...try to use the 1st weigh...make 2 options: 1.what if it tippes ao a 1st side, and 2.what if it tippes on second side... You'll have to work longer, but you have a lot more chances to find the odd ball

 

[leto]

I disagree. By removing the 999998 doors, the chance of the door you didn't pick being the door increases. The door you did pick could not be removed, and is therefore less likely to be the prized door.

 

[leto]

There are three possibilities with the first weigh. Either it tips to the left, right or stays level. I accounted for all three of these.

Edit:

I think you are assuming I label them l1-l4 and h1 - h4 because I assume I know this. I only label them as such so that I know later that, if one of them is the odd ball, they have to follow that pattern. A ball that was on the heavy side will never be the lighter odd ball.

 

[tribdog]

I disagree. By removing the 999998 doors, the chance of the door you didn't pick being the door increases. The door you did pick could not be removed, and is therefore less likely to be the prized door.

correct. in the million door scenario the chances of the door you picked being the correct door is 1 in a million, therefore wouldn't the chance of the other door being the correct one be 999999 in a million? The point being it is always better odds to make the switch in a situation like this.
another way of looking at it is this: after you make your first choice you are given the option of keeping your door or taking all 999999 of the other doors.(which is the same thing as taking one door that may be right and throwing out 999998 wrong ones) Do you keep your 1 or take all 999999 of the others?

 

[marqq]

I disagree. By removing the 999998 doors, the chance of the door you didn't pick being the door increases. The door you did pick could not be removed, and is therefore less likely to be the prized door.

The problem allow us to eliminate only one door...From 3 doors or from 100000000 doors...we are able to remove only one door...
I'll reed about probabilyties and i'll give you an answer

 

[marqq]

There are three possibilities with the first weigh. Either it tips to the left, right or stays level. I accounted for all three of these.

Edit:

I think you are assuming I label them l1-l4 and h1 - h4 because I assume I know this. I only label them as such so that I know later that, if one of them is the odd ball, they have to follow that pattern. A ball that was on the heavy side will never be the lighter odd ball.

sorry ... i missunderstood your "l-f" definition...i was tired and i thought you ment left-right...and swiching the balls, an left becomes right and viceversa...therefore i missjudged your solution...i am deeply sorry...your answer is as corect as it can be... I'm really sorry about that

 

[marqq]

Another one..A funny one

The mother is 21 years older than the son.In 6 years, the son will be 5 times younger then his mother.Question: where is the father now?

 

[pig]

enjoying himself :)

 

[marqq]

enjoying himself :)

you are closer to the answer than you imagine

 

[pig]

you are closer to the answer than you imagine

i got the answer, that's why i wrote that ;)

 

[marqq]

would you please enlarge your answer for the others?

 

[verty]

Here's an easy one. See if you can solve this riddle in under 30 seconds...

Two Russians are walking down a street. One Russian is the father of the other Russian's son. How are they related?

 

[Chen]

Husband and wife? In second thought, I don't know how things work in Russia so my answer would be "they had sex".

 

[Chen]

no I don't think that is correct. its close, but not right. What if there are a million doors and after you choose Monty Hall/Bob Barker throws out 999998 of the bad doors and says it's either the one you chose or this one? do you think it is still a 50/50 chance then?

Ahh yes, why not? Because Bob says "it is either the one you chose or the other", which implies that one of the doors is the right one. He does not ask "would you like to change your pick now", which doesn't necessarily mean one of the doors is the right one. So yes, your chances would be 50/50.

 

[marqq]

Another funny one

We have a driver who has to make the distance between A and B in his own car...He drives the 1st 3rd of the distance with 40 km/h (he was reall calm) then he becomes angry and drives the 2nd 3rd with 120 km/h...then it calms down and asks himself what is the speed that he should use for haveing a decent average of 90 km/h

I'll ask you to try (before starting to calculate) an aproximation by intuition, and then make the difference to see it it was a small or a huge error

 

[Chen]

Is 120 km/sec a typo or did you really mean that? If it's a typo he should drive at 110 km/h, if it's not... I don't think he can reach an average of 90 km/h after driving at ~400,000 km/h.

 

[marqq]

If it's a typo he should drive at 110 km/h, if it's not...

sorry
I was thinkin at smth else
120 km/h
And about the 110km/h you are wrong !Hint : make an ecuation

 

[Chen]

I did...

$$90 = \frac{40 + 120 + x}{3}$$

$$x = 110$$