Your Favorite Brain Teaser
- Thread starter RedDeadFred
- Start date
Recommended Videos
Altheus_Necron said:
Well done, I'm all out of cookies but here's something to tide you over:skystryke said:
Here's one for you computer scientists and logicians; (This is an actual interview question from Google)
You have 1000 bottles of wine, and one of them is poisoned, but you don't know which one.
At your disposal are 10 prisoners who are slated to die today, which you can use to figure out which bottle is poisoned.
The poison is relatively fast-acting, meaning it will kill the prisoner after a little while, but one could take a drink from all 1000 bottles before the poison would take effect.
How do you figure out which bottle is poisoned?
There are two possible answers to this riddle, one is a lot better than the other.
You have 1000 bottles of wine, and one of them is poisoned, but you don't know which one.
At your disposal are 10 prisoners who are slated to die today, which you can use to figure out which bottle is poisoned.
The poison is relatively fast-acting, meaning it will kill the prisoner after a little while, but one could take a drink from all 1000 bottles before the poison would take effect.
How do you figure out which bottle is poisoned?
There are two possible answers to this riddle, one is a lot better than the other.
Well, 2^10 = 1024, so you could just do a binary search. I'm not sure what the better answer is, though.Finnboghi said:
Are people still stumped on mine?
the mathematical answer is simple. divide up the wine, give each prisoner a glass of wine from each of 100 separate bottles, then wait for one of them to die. when one dies, you take his 100 bottles and divide them again among the 9 remaining survivors, giving each 11 bottles (and one of them 12) and wait for one to die again. dividing the 11 (or 12) bottles between the 8 survivors gives 1 or 2 each; if the one that dies has 2 bottles, give one to one of 2 and the one that dies has the poisoned bottle. and you managed to save 6 or 7 prisoners' lives... at least until they are executed. (quicker, but more torturous, no?)Finnboghi said:
OR you can give each one a sample from each bottle, wait the requisite time and see if they die, 10 bottles at a time. It's significantly slower, but has a 1% chance (increasing with every round) of ending early. This is why regular brute force sucks...
the REAL scientific answer is even simpler. run a dropper-sized sample from each of the 1000 bottles (clearly marked of course) through a mass spectrometer, the poisoned wine will be chemically different in a significant way. the prisoners can help you or whatever, then you can let them share the one bottle, if you feel like killing them yourself. (you have access to prisoners, which means you have access to a prison, which grants you access to police officers, which in turn could grant you use of the local crime lab. this assumes the "experiment" takes place in modern times with modern communications and forensic equipment)
the best answer, of course, is to ignore the prisoners completely and sell all 1000 bottles. you don't get any of the wine, but you're 100% guaranteed not to drink the poisoned one... and you can use some of the cash you made selling the wine to settle the lawsuit against you from the family of whoever bought your poisoned bottle and the rest to buy non-poisoned wine directly from a vineyard. Caveat Emptor! (I think I spelled that right, my Latin is terrible)
Look at the hat as it's being put on your head. duh.SeaCalMaster said:
OR Answer correctly that you do not know the color of your hat. (although that will probably get you killed for being a smart-ass)
OR Look carefully at the line ahead of you and try to discern a pattern of distribution, while paying close attention to the answers behind you to confirm.
OR (Smart-ass answer #2): shake your head violently to remove the hat from your head, then answer correctly "I'm not wearing a hat". (you could modify this plan to get the hat into your hands, observe its color, then replace it)
a river
EDIT: oops wrong page
Concerning the hat thing, is there any fixed number of black and white hats? I guess I'd go to stand all the way at the front (so that I'd be last) and then listen to what other people say. Whoever is all the way at the back probably bets on him having a hat of whatever colour there are most of, or which there are least of if he counts 49/50.
My fav riddle:
The billiard balls. There's 12 of them and one is either slightly lighter than the others or slightly heavier. You have a set of scales (ones that strike one way or the other, not indicate weight) and you get to weigh three times. By the end you have to be able to determine which ball is the odd one out and whether it is heavier or lighter.
Obviously the difference is too small for you to tell by guessing manually.
EDIT: oops wrong page
Concerning the hat thing, is there any fixed number of black and white hats? I guess I'd go to stand all the way at the front (so that I'd be last) and then listen to what other people say. Whoever is all the way at the back probably bets on him having a hat of whatever colour there are most of, or which there are least of if he counts 49/50.
My fav riddle:
The billiard balls. There's 12 of them and one is either slightly lighter than the others or slightly heavier. You have a set of scales (ones that strike one way or the other, not indicate weight) and you get to weigh three times. By the end you have to be able to determine which ball is the odd one out and whether it is heavier or lighter.
Obviously the difference is too small for you to tell by guessing manually.
my mom's a surgeon xDYou can see the hats of people in front of you. And the people behind you CAN communicate with you (you can overhear their answer to the nutjob), it's just the only information they can give is White or Black. The guy at the back names the colour which he sees an even number of, say, black. He has a 50/50 chances of dying. The guy in front counts the hats in front of him. If he sees an odd number of blacks, he must be black. The next person counts, if he sees an even number of blacks, he must be black. And so on. 99 people guaranteed to live, 50/50 chance of all 100 living.2xDouble said:
Here's my brain teaser:
Three women share a hotel room. They order a bottle of chapagne from room service. The room service guy knocks at their door, gives them the bottle, and gets a ten-dollar bill each from them since the champagne costs $30. When the room-service guy gets down to the lobby, the guy in the lobby says "You idiot, it costs $25, not $30! Go up and return $5!" So our friend room-service guy does that, but pockets $2 as a tip, since they won't be able to split $5 evenly between three women anyway. He gives a dollar to each woman.
That means every woman paid $10. Every woman got $1 back, so in the end every woman paid $9. And $9 times 3 equals $27. Add to that the $2 the rooms-ervce guy pocketed as a "tip" and we've got $29. What happened to the last dollar?
Three women share a hotel room. They order a bottle of chapagne from room service. The room service guy knocks at their door, gives them the bottle, and gets a ten-dollar bill each from them since the champagne costs $30. When the room-service guy gets down to the lobby, the guy in the lobby says "You idiot, it costs $25, not $30! Go up and return $5!" So our friend room-service guy does that, but pockets $2 as a tip, since they won't be able to split $5 evenly between three women anyway. He gives a dollar to each woman.
That means every woman paid $10. Every woman got $1 back, so in the end every woman paid $9. And $9 times 3 equals $27. Add to that the $2 the rooms-ervce guy pocketed as a "tip" and we've got $29. What happened to the last dollar?
The math in this brainteaser is really stupid! if you devide the womens' money by 3, they each payed 9 and 1/3 of a dollar, it ads up to 28, plus the 2 dolars the kid took as a tip. It's pretty simple math that they payed 30 dollars and the room-service guy comes back with 5 dollars because it was only $25. it's when you do the deviding that you get confused, because you shouldn't devide like that! BTW I'm not mad at you Queen Michael, it's just that I really don't like this kind of calculating. It's the riddle I'm mad at, not you...Queen Michael said:
I LOL'd.madbird-valiant said:
OK, a turtle walks towards a door 2 metres away at a speed of 1cm a minute. After 5 minutes, a man tries to overtake him. He covers half the distance between him and the turtle, but in that time the turtle has moved forward a small amount.
The man breaks into a run, covering half the distance again, but the turtle has moved.
Now sprinting, he covers half the distance again, but it's still in front.
How can the man ever overtake the turtle?
1st weigh in: 6 balls on one scale - 6 balls on the other, whichever bunch are heaviest keep those 6 for the next weigh inTazTheTerrible said:
2nd weigh in: 3 balls on one scale - 3 balls on the other, whichever bunch are heaviest keep those 3 for the next weigh in
3rd weigh in: 1 ball on each scale, keep the third on the table, scales will tell you which is the heavier ball, if both balls are the same than the one left on the table must be the heaviest.