Best riddles

logictricky

A boat has a ladder that has six rungs, each rung is one foot apart. The bottom rung is one foot from the water. The tide rises at 12 inches every 15 minutes. High tide peaks in one hour. When the tide is at it's highest, how many rungs are under water?
None, the boat rises with the tide.
69.31 %
118 votes
clever

A very vicious dog is tied to a tree. If you came close enough, you'ed lose your head. If he sees you, he'll follow your every step. Your basket ball rolls past the dog. If you are alone, how do you get your ball back with out being mutilated?
Walk around the tree until there is no more rope left on the dogs leash. Then, grab your ball.
69.31 %
118 votes
simplecleanstory

Four cars come to a four way stop, all coming from a different direction. They can't decide who got there first, so they all entered the intersection at the same time. They do not crash into each other. How is this possible?
They all made right hand turns.
69.31 %
118 votes
logictricky

John Heysham Gibbon was most renowned surgeon of 1940-1970. More than 90% of his surgeries he performed are highly successful and still almost all of his patients die.
The surgery was performed way back, by now approx 90% of them have died by old age.
69.30 %
77 votes
cleanlogicmysterystory

A man owned a casino and invited some friends. It was a dark stormy night, and they all placed their money on the table right before the lights went out. When the lights came back on, the money was gone. The owner put a rooster in an old rusty tea kettle. He told everyone to get in line and touch the kettle after he turned the lights off, and the rooster will crow when the robber touched it. After everyone touched it, the rooster didn't crow, so the man told everyone to hold out their hands. After examining all the hands, he pointed out who the robber was. How did he know who stole the money?
Because the tea kettle was rusty, whoever touched it would have rust on their hands. The robber didn't touch the kettle, therefore he was the only one whose hands weren't rusty.
69.26 %
771 votes
logicmathclean

What is the least number of people that need to be in a room such that there is greater than a 50% chance that at least two of the people have the same birthday?
Only 23 people need to be in the room. Our first observation in solving this problem is the following: (the probability that at least 2 people have the same birthday + the probability that nobody has the same birthday) = 1.0 What this means is that there is a 100% chance that EITHER everybody in the room has a different birthday, OR at least two people in the room have the same birthday (and these probabilities don't add up to more than 1.0 because they cover mutually exclusive situations). With some simple re-arranging of the formula, we get: the probability that at least 2 people have the same birthday = (1.0 - the probability that nobody has the same birthday) So now if we can find the probability that nobody in the room has the same birthday, we just subtract this value from 1.0 and we'll have our answer. The probability that nobody in the room has the same birthday is fairly straightforward to calculate. We can think of this as a "selection without replacement" problem, where each person "selects" a birthday at random, and we then have to figure out the probability that no two people select the same birthday. The first selection has a 365/365 chance of being different than the other birthdays (since none have been selected yet). The next selection has a 364/365 chance of being different than the 1 birthday that has been selected so far. The next selection has a 363/365 chance of being different than the 2 birthdays that have been selected so far. These probabilities are multiplied together since each is conditional on the previous. So for example, the probability that nobody in a room of 3 people have the same birthday is (365/365 * 364/365 * 363/365) =~ 0.9918 More generally, if there are n people in a room, then the probability that nobody has the same birthday is (365/365 * 364/365 * ... * (365-n+2)/365 * (365-n+1)/365) We can plug in values for n. For n=22, we get that the probability that nobody has the same birthday is 0.524, and thus the probabilty that at least two people have the same birthday is (1.0 - 0.524) = 0.476 = 47.6%. Then for n=23, we get that the probability that nobody has the same birthday is 0.493, and thus the probabilty that at least two people have the same birthday is 1.0 - 0.493) = 0.507 = 50.7%. Thus, once we get to 23 people we have reached the 50% threshold.
69.25 %
190 votes