unreality

sorry Martini. ignore my posts. I was in a bad mood. lol.

there is a lot of confusion between the two sides of this problem, and i've read all ur arguements, and it isnt about conditionals or conjunctionals or anything its about wording. this is the problem. let me requote the important part (middle line) "They have two kids, one of them is a girl, what is the probability that the other kid is also a girl." "they have two kids." They already have them. They're not expecting kids... they already have both of them born. The kids could be 20 and 18 years old for all we care. "One of them is a girl" okay... so if we have two kids, already born, and we can rule out B/B: G/B B/G G/G "what is the probability that the other kid is also a girl." not: "what is the probability that the other kid will be a girl." In that case it would be 1/2, since the other child does not affect the probability. But it doesnt say that. It said "is also a girl". The child is already born. it all comes down to two sides: Side A (answer is 1/2): We're looking at the second child's birth as an independent event. Side B (answer is 1/3): We're looking at both together. and I am sorry to say Side B is right (as I was on Side A for quite some time) Why? Because the second child has already been born. We dont care what was the probability that it was boy/girl while it was in its mother's womb. We know thats 1/2. The fact is, the gender has already been determined. We're just trying to figure out the probability they are both girls. Since the options are: G/B B/G G/G we can safely say, G/G has a 1/3 chance. There is no fallacy. There is no bla-bla-bla. It all comes down to wording: the "other kid" is already born. The solution doesnt call for the probability that the other kid will be a girl. It calls for the probability that they are (as in right now) BOTH girls. We're not looking at one girl's chances. In that case there would be two options: B G but we're not. BECAUSE OF THE WORDING, we are looking at both children at once: G/B B/G G/G It is 1/3. It pains me to say this (I defended 1/2 for a while until i realized i was wrong) but I'm not being bitter and I'm trying to prove what I now see is right. its a fine line, based on skale's wording. but the answer is 1/3. I rest my case. I think this topic should be done. The discussion cant go much elsewhere.

when there's lots of arches next to each other..? lol.

Someone remind me to never lend money to Aaron Burr with the caveat that he is to pay me in x number of days. there are more solutions than just one. Aaron Burr is just taking advantage of the wording... what's wrong with that? lol btw aaron, i love the mary had a little lamb thing

You shouldnt be going around saying that like its fact. Maybe thats what you believe, but dont try to say that like its a true fact or something. Your decisions are based on your brain nature and past experiences, etc, we still don't know how the brain works, and if you go around thinking you dont have free will, life has no meaning. It's the LazyBones paradox. Life is based on free will. On unpredictability. So get over it and go have some fun! i dont believe in destiny. or fate. But thats what i believe. and even if everything is chemically fixed inside our brains, nobody can determine that, so you wouldnt know you wer going to die in 2 days anyway. Your brain doesnt have knowledge like that. That would have to be a combination of millions of ppl's brains and their actions to get you into that spot. yeah. all i'm saying is that its just a riddle, nothing more

lol k

look at the last four steps: 4 - 13 - 3 17 - 0 - 0 17 - 3 - 0 10 -3- 7 am I the only person who caught this? the OP said clearly there was 20 liters total. No pouring out into a sink and no getting water from anywhere. I think you meant to do this: 4 - 13 - 3 17 - 0 - 3 17-3-0 10-3-7 that works. good job tho!

exactly. But notice my OP- it said "THE 7 MUST BE EXACT". It can't be guesswork. my Solution #2 wasn't guesswork. It didn't have to be even double pouring to work. It could've been all outta one cup and none outta the other and it would have still worked. Remember, it must be exactly 7. guys, dont fret yourselves to hell though, I'm not even sure there's a solution better than my #3 one.

k thanks! Jar A: 0/19 Jar B: 13/13 Jar C: 7/7 B into C Jar A: 13/19 Jar B: 0/13 Jar C: 7/7 C into B Jar A: 13/19 Jar B: 7/13 Jar C: 0/7 A into B until B is full Jar A: 7/19 Jar B: 13/13 Jar C: 0/7 B into C 7/19 6/13 7/7 C into A 14/19 6/13 0/7 B into C 14/19 0/13 6/7 A into B 1/19 13/13 6/7 B into C 1/19 12/13 7/7 C into A 8/19 12/13 0/7 lol you've probably realized by now i'm confused as hell...

wow these are great!

can the jars hold as much as they want?

ask them if the first man is the liar: if the first man says Yes, you know he's the Random man ®, because neither the Liar (L) or Truth-teller (T) could say "yes" to that. now if the first man said no, he is either L or T. If he is L, T will say "yes he's L" and R will say either. If he is T, L will say "yes he's L". R will say either. Hmm. Is this the right start to solving it?

in ur drawing, the first two are 3 oz cups, the last is the 8 oz cup, so how did you come up with this: 2-3-8 Refill the 8 oz 7-3-3 Overspill 1 oz from the 8 to the 3 the 3oz cup cant hold 7 oz...?

so we got: 0/19 13/13 7/7 ?? if thats the case i'll get right on it. i like these puzzles!

is it real?

this is a water-measuring puzzle i made up, however i dont have the answer past my three solutions... You have 2 cups that can hold 3 oz, and one cup that holds 8 oz. You have a sink where you can pour out as much water as you need, and a place to dump water. The goal is to have 7 oz of water in the 8oz cup, with the minimum number of ounces dumped and wasted. THE 7 OZ MUST BE EXACT. Solution 1: fill up a 3oz cup and pour it in the 8oz cup. Do this twice, so the 8oz cup has 6 oz in it. Fill up the 3oz cup a third time and pour it in the 8oz cup until the 8 oz cup is full- and you still have 1 ounce in the cup. Now dump out all 8 oz in the 8oz cup and use the other 3oz cup twice to get the 8oz cup to 6 oz... then pour in the 1oz from the other cup to get 7 oz in the 8oz cup, and the two smaller cups are empty, and there was only a 8oz loss! Solution 2: Fill up both 3 oz cups and pour them into 8 oz cup so there are 6 oz in it, leaving 2 oz still free in the 8 oz cup. Fill both cups and slowly pour them into the 8 oz cup, pouring slowly and evenly. When the 8 oz cup is full, 2 oz is left in both smaller cups. Well no matter if it wasn't even, there will still be 4oz total over both cups, because together they held 6, and 2 was put. Even if you poured just 1 cup in there, one cup would have 3 and the other 1. So together they have 4 oz no matter what, but it will be closer to 2 each. Ideally they both have 2. Empty the 8 oz and pour both cups into the 8oz container- there are now 4 oz in the 8 oz cup. Now fill up one 3 oz cup and pour it in to make 7 oz in the 8 oz cup. This solution has the same end result as Solution 1: both smaller cups empty, the larger has 7 oz, and only 8 oz were wasted total. Can you find a solution where less than 8 oz have been dumped? WAIT A MINUTE... I just thought of something: Solution 3: Fill the 8 oz cup up with one cup so its finally full and 1 oz is left over in the 3oz cup. Now, with the 8 oz cup, pour 3 oz into the other 3 oz cup, filling it up and reducing the amount in the 8oz cup to 5 oz. So now we have 5/8, 3/3 and 1/3 in our cups. Dump out the 5 oz and pour the full 3 oz cup inside, then use that same cup to pour another 3 oz into the large cup to bring it 6 oz, then add the 1 oz! You now have 7 oz in the 8 oz with the other two empty, and only 5 ounces spilled this time! Can anyone find something less than 5? I think thats the absolute minimum with two 3oz cups... with a third 3 oz cup I could get it to 2 oz being dumped and two of the three 3oz cups full to pour in, with the last one pouring the 1 in. So you could get 2 oz wasted if you had three 3 oz cups... but you only have two of course. So is it possible to use yet a different method to waste less than 5 oz?

3^6, whatever that is

yes we've established that already, but the only options are female, so its Daughter

good one hunee9! that would be two ppl... guest #2 was the baby, so guess #3 (the pregnant mother) went into room 2 (and obviously so did her baby). or, the bellboy was bringing in their luggage from the car, and it was late and his shift was almost over, and he was thinking about his own cozy room at home that he shared with his brother, but he wanted his own room. since the bellboy was considered "guest 13" he put guest 2 in the last room and then went back to his own home ;D lol i know the riddle's supposed to be an impossible "paradox with a catch" in that he counts the thirteenth person as the second person, but i was having some fun

umm quoting myself...

two

for a guy, yes. But there is no male options, meaning the person is Barbera's daughter

exactly what i said on the first page! its 1/2! I'm just saying, they could reword the problem to be looking in retrospect, saying "one of two children is a girl, what is the prob they are both girls?" would be a good rewording of it, where the answer is 1/3 the way they worded it, the answer is clearly 1/2 if they said this instead: "one of two children is a girl, what is the prob they are both girls?" the answer for that is 1/3

i see what you're saying... but.... And he died of thirst. It was B's doing. But you're saying B didn't kill C because C already didnt have good water. That's like saying shooting an old man in the street is okay because he was going to die soon anyway. Maybe he was going to die soon. Maybe he wanted to die. But the fact is, its not the shooter's call. It's not the shooter's judgement that matters. Nobody decides for the old man if he should be killed because he's old. It would still be murder to shoot him in the head. Sorry that was a confusing anology. Forget that. This is what the arguement boils down to: Side 1: It does not matter what was in the water. It couldve been poison or the Water of Eternal Youth. It doesnt matter. B took it away from A and A died because he had nothing to drink. B's intent was to kill C by dehydration and it worked. He murdered C. Whatever A did to the water was insignificant because the water never came into play, because of B. Side 2: It does matter what the water was, because the water is what B took away from C. If the water was undrinkable, B was not responsible because C would have died anyway because the water was bad. My response to side 2: yes, C would have died anyway... but then A would be responsible, not B. And also, the only ppl that knew of the poison was A. B didnt know and C didnt know. So B wasn't "saving C from poison" and C didn't "die of thirst because he didn't want to drink poisoned water". Nobody knew about the poisoned water in their motives, except A. B's motives were different. And B's motives suceeded. Though its a nice point, barbera sans