Difficult riddles

logicmathprobability

The same birthday

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.
89.44 %
54 votes

cleanlogicmathshort

Number one

How do you make the number one disappear by adding to it?
Add the letter 'G' and it becomes Gone.
89.19 %
66 votes

logicmath

5 Pirates Fight for 100 Gold

There are 5 pirates in a ship. Pirates have hierarchy C1, C2, C3, C4 and C5.C1 designation is the highest and C5 is the lowest. These pirates have three characteristics : a. Every pirate is so greedy that he can even take lives to make more money. b. Every pirate desperately wants to stay alive. c. They are all very intelligent.There are total 100 gold coins on the ship. The person with the highest designation on the deck is expected to make the distribution. If the majority on the deck does not agree to the distribution proposed, the highest designation pirate will be thrown out of the ship (or simply killed). The first priority of the pirates is to stay alive and second to maximize the gold they get. Pirate 5 devises a plan which he knows will be accepted for sure and will maximize his gold. What is his plan?
To understand the answer,we need to reduce this problem to only 2 pirates. So what happens if there are only 2 pirates. Pirate 2 can easily propose that he gets all the 100 gold coins. Since he constitutes 50% of the pirates, the proposal has to be accepted leaving Pirate 1 with nothing. Now let’s look at 3 pirates situation, Pirate 3 knows that if his proposal does not get accepted, then pirate 2 will get all the gold and pirate 1 will get nothing. So he decides to bribe pirate 1 with one gold coin. Pirate 1 knows that one gold coin is better than nothing so he has to back pirate 3. Pirate 3 proposes {pirate 1, pirate 2, pirate 3} {1, 0, 99}. Since pirate 1 and 3 will vote for it, it will be accepted. If there are 4 pirates, pirate 4 needs to get one more pirate to vote for his proposal. Pirate 4 realizes that if he dies, pirate 2 will get nothing (according to the proposal with 3 pirates) so he can easily bribe pirate 2 with one gold coin to get his vote. So the distribution will be {0, 1, 0, 99}. Smart right? Now can you figure out the distribution with 5 pirates? Let’s see. Pirate 5 needs 2 votes and he knows that if he dies, pirate 1 and 3 will get nothing. He can easily bribe pirates 1 and 3 with one gold coin each to get their vote. In the end, he proposes {1, 0, 1, 0, 98}. This proposal will get accepted and provide the maximum amount of gold to pirate 5.
88.64 %
50 votes

clean

Extraordinary event

Something very extraordinary happened on the 6th of May, 1978 at thirty-four minutes past twelve a.m. What was it?
At that moment, the time and day could be written as 12:34, 5/6/78.
88.42 %
49 votes

logic

Hidden message

Find a short hidden message in the list of words below. carrot fiasco nephew spring rabbit sonata tailor bureau legacy corona travel bikini object happen soften picnic option waited effigy adverb report accuse animal shriek esteem oyster
Starting with the first two words, take the first and last letters, reading from left to right. Example: Carrot fiascO "from these pairs" the message is as follows: CONGRATULATIONS CODE BREAKER
88.42 %
49 votes

cleanlogicwhat am I

Silver-tongued, but never lie

I’m teary-eyed but never cry, silver-tongued, but never lie. double-winged, but never fly, air-cooled, but never dry. What am I?
Mercury. The element looks shiny, silver, and is wet. The god Mercury has two wings but only uses them to run.
88.34 %
61 votes

logicshort

The Missing Servant

A king has 100 identical servants, each with a different rank between 1 and 100. At the end of each day, each servant comes into the king's quarters, one-by-one, in a random order, and announces his rank to let the king know that he is done working for the day. For example, servant 14 comes in and says "Servant 14, reporting in." One day, the king's aide comes in and tells the king that one of the servants is missing, though he isn't sure which one. Before the other servants begin reporting in for the night, the king asks for a piece of paper to write on to help him figure out which servant is missing. Unfortunately, all that's available is a very small piece that can only hold one number at a time. The king is free to erase what he writes and write something new as many times as he likes, but he can only have one number written down at a time. The king's memory is bad and he won't be able to remember all the exact numbers as the servants report in, so he must use the paper to help him. How can he use the paper such that once the final servant has reported in, he'll know exactly which servant is missing?
When the first servant comes in, the king should write down his number. For each other servant that reports in, the king should add that servant's number to the current number written on the paper, and then write this new number on the paper. Once the final servant has reported in, the number on the paper should equal (1 + 2 + 3 + ... + 99 + 100) - MissingServantsNumber Since (1 + 2 + 3 + ... + 99 + 100) = 5050, we can rephrase this to say that the number on the paper should equal 5050 - MissingServantsNumber So to figure out the missing servant's number, the king simply needs to subtract the number written on his paper from 5050: MissingServantsNumber = 5050 - NumberWrittenOnThePaper
87.96 %
47 votes

logic

Four big houses

There are 4 big houses in my home town. They are made from these materials: red marbles, green marbles, white marbles and blue marbles. Mrs Jennifer's house is somewhere to the left of the green marbles one and the third one along is white marbles. Mrs Sharon owns a red marbles house and Mr Cruz does not live at either end, but lives somewhere to the right of the blue marbles house. Mr Danny lives in the fourth house, while the first house is not made from red marbles. Who lives where, and what is their house made from ?
From, left to right: #1 Mrs Jennifer - blue marbles #2 Mrs Sharon - red marbles #3 Mr Cruz - white marbles #4 Mr Danny - green marbles If we separate and label the clues, and label the houses #1, #2, #3, #4 from left to right we can see that: a. Mrs Jennifer's house is somewhere to the left of the green marbles one. b. The third one along is white marbles. c. Mrs Sharon owns a red marbles house d. Mr Cruz does not live at either end. e. Mr Cruz lives somewhere to the right of the blue marbles house. f. Mr Danny lives in the fourth house g. The first house is not made from red marbles. By (g) #1 isn't made from red marbles, and by (b) nor is #3. By (f) Mr Danny lives in #4 therefore by (c) #2 must be red marbles, and Mrs Sharon lives there. Therefore by (d) Mr Cruz must live in #3, which, by (b) is the white marbles house. By (a) #4 must be green marbles (otherwise Mrs Jennifer couldn't be to its left) and by (f) Mr Danny lives there. Which leaves Mrs Jennifer, living in #1, the blue marbles house.
87.96 %
47 votes

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