I think the riddle's already been solved. Answer's that number two knows. See, they all face left, let's say. 1 can't see any hats. 2 can see 1. 3 can see both 1 and 2. If 3 sees two whites, he knows that he has a black hat, but because he does not say anything, number 2 knows that he has a black hat, because if he did, 3 would know already.
Thanks - my pleasure. I can't believe I actually got one of those things. I made the tables like Kimble14 did (except not using a computer). At least that stuck with me from what I learned in the gifted program in elementary school. Most of these riddles have been crazy difficult... I'm impressed with the great intellects on this bbs.
You want to send a valuable object to a friend securely. You have a box which can be fitted with multiple locks, and you have several locks and their corresponding keys. However, your friend does not have any keys to your locks, and if you send a key in an unlocked box, the key could be copied en route. How can you send the object securely?
Technically, you send the object securely if you just lock it and send it. Of course, that does not mean the other guy can open it, but you have sent it securely... If you want to send it securely and want the other guy to be able to open it, the other guy should have his own locks. You lock the box, keep the key, send it to the guy. He attaches another lock of his own and sends it back to you. You take off your lock and send it back to him (his lock still attached). He can now open it with his own key to his own lock. And now I would like to know - what was your former user name, Martian Man?
that sounds like an answer to a similar riddle that you've heard before. Former user name? what makes you think I've had one?
Guy number 2. He waits for guy number 1 to answer who would answer Black if guy#2 wore white. When guy number 1 doesn't answer, guy number 2 knows guy number one sees a black hat on his head and can't answer.
It's kind of not fair, since I've worked out this problem before.. but here it goes. You line them up, take one bar from the first, 2 from the secodn guy, 3 from the third guy. At the end you should have 78 bars of gold, which multiplied with 12 is some big number (too lazy to grab a calculator). Then you subtract the acutal weight of the bars from that number, and how many kilo less it correspond to the position of the thief in the line.
Here's another one that my friend was actually asked in an interview. You have 3 switches connected to a room with 3 light bulbs. You can go into the room only once. But you may switch the switches anyway and any combination you want. How do you find out which switch is connected to the wich light bulb?
Are you sure there's not another part to this riddle?? Either way, it has to do with the heat of the light bulbs. Something like you leave one OFF, turn one on, wait a few minutes for ONE to be HOT AS HELL, then turn the other one on and run like HELL to the room... touch them and the one that is cooler is the last one you turned on?
The Monty Hall problem is a tricky question, but the answer is 50/50. The tricky part is people tend to regard the process as two related selections, but actually the process is divided into two individual selections. The first selection is to choose one door out of three doors. The chance for the right choice is 1/3, no problem there. However, after this selection Monty eliminates one empty door, then you get to CHOOSE AGAIN between two doors. It is starting all over as a new guess, because whether one chose the right door the first time makes no difference on the odds of the second selection. If one chose the right door the first time but doesn't know it, he faces the second selection with the same info he had. and vice versa. It's actually a different selection and therefore must be treated individually instead of cumulatively. Since there's only two choices left upon second try, the correct odds is 50/50. In a situation among which the choices is eliminated by one on each selection, the logical conclusion is that one's chance of choosing the right answer is increased after each selection, which is exactuly why the right answer of the second selection is 1/2, greater than the 1/3 of the first selection. The conclusion of switching the answer after each selection increases the odds of winning is not logical. For example, if your teacher gives you a multiple choice question, and eliminate one wrong answer after your first answer, can you say that changing your first answer increases the odds? When we don't have a clue about the correct answer, we eliminate the wrong answers we know for sure to increase our odds of choosing the right answer, we do not eliminate a choice we know that's wrong for sure in order to change our original guess. So next time when you eliminate a sure wrong choice, don't feel obligated to switch your choice.
Marylin's answer from the previous pages makes the most sense. If Monty tells you the door he eliminates is wrong, then the switch you make is your switched door's odds plus the eliminated door (1/3+1/3). Your example of the multiple choice question is different because it assumes you're coming in with extra knowledge which makes the odds of having a random answer uneven. If your teacher gave a shell game where you knew nothing from the beginning, then the circumstance would be the same as the Monty Hall problem and you'd have higher odds by switching.
Panda: If you guess the wrong door and always switch, you'll always win. Your chances of doing that are 2 in 3. (The longer version got Alt-F4ed due to an emergency boss visit.) Assume there are n choices. If you never switch, your probability of getting it right is 1/n. If you always switch after a wrong answer is revealed: * Case 1: You got it right the first time: You'll never win. * Case 2: You missed it the first time: You'll win 1/(n-2) of the time, since you've eliminated 2 choices (your first guess, by definition, and the one that was revealed.) The probability of case 1 occurring is 1/n. For case 2, it's (n-1)/n. Your total probability of winning if you always switch (the sum of the products of the probabilities of each case happening and the probability of winning in each case) is: (1/n)(0) + [(n-1)/n][1/(n-2)] = (n-1) / [n(n-2)] Since your probability of winning if you never switch is: 1/n = (n-2) / [n(n-2)] ...you're better off if you switch. For the Monty Hall problem, n = 3. If you never switch, you'll win 1/3 of the time. If you always switch, you'll win 2/3 of the time. If you switch 1/2 of the time -- that is, if you randomly choose after one of the doors is taken away -- you'll win 1/2 of the time.
