Clean riddles

cleanfunnylogictricky

If a green man lives in a green house, a purple man lives in a purple house, a blue man lives in a blue house, a yellow man lives in a yellow house, a black man lives in a black house. Who lives in a White house?
The President.
69.38 %
81 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
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.
69.27 %
768 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