Riddle #687

logicmathclever

You and a friend are standing in front of two houses. In each house lives a family with two children. "The family on the left has a boy who loves history, but their other child prefers math," your friend tells you. "The family on the right has a 7-year old boy, and they just had a new baby," he explains. "Does either family have a girl?" you ask. "I'm not sure," your friend says. "But pick the family that you think is more likely to have a girl. If they do have a girl, I'll give you $100." Which family should you pick, or does it not matter?
You should pick the house on the left. Specifically, there is a 2/3 chance that the family on the left has a girl, whereas there's only a 1/2 chance that the house on the right has a girl. This is a very counterintuitive riddle. It seems like there should always be a 1/2 chance that a given child is a girl. And in fact there is. The key word there is "given". Because we are not asking about a "given" child for the house on the left. We are asking about what could be either child. Whereas for the house on the right, we are asking about a "given" child...specifically, we're asking about the younger child. There are 3 possibilities for the children in the first house: Younger Older Girl Boy Boy Girl Boy Boy There is no "Girl, Girl" option because we know the house on the left has at least one boy. Since each of these 3 options is equally likely, and 2 of them have one girl, there is a 2/3 chance of there being a girl in the house on the left. For the house on the right, because we already know the older child is a boy, there are only two possibilities: Younger Older Girl Boy Boy Boy And as we can see, there is a 1/2 chance for the house on the right having a girl. Search for: Boy or Girl paradox
62.98 %
66 votes

Similar riddles

See also best riddles or new riddles.

logicclever

Justin Case and Auntie Bellum are fellow con artists who deliver coded messages to each other to communicate. Recently Auntie Bellum was put in jail for stealing a rare and expensive diamond. Only a few days after this, Justin Case sent her a friendly letter asking her how she was. On the inside of the envelope of the letter, he hid a code. Yesterday, Auntie Bellum escaped and left the envelope and the letter inside the jail cell. The police did some research and found the code on the inside of the envelope, but they haven't been able to crack it. Could you help the police find out what the message is? This is the code: llwatchawtfeclocklnisksundialcirbetimersool
The message was "loose bricks in left wall." The message was put backward with words related to time in between. This is how the message looks when separated: ll watch awtfe clock Inisk sundial cirbe timer sool If you take out watch, clock, sundial, and timer, this is what is left: llawtfelniskcirbesool Look at this backwards and this is what you have: loose bricks in left wall Auntie Bellum took out the bricks and escaped in the night. Then, she put the bricks back where they were.
83.62 %
59 votes
cleanlogicsimpleclever

Two fathers and two sons went fishing one day. They were there the whole day and only caught 3 fish. One father said, that is enough for all of us, we will have one each. How can this be possible?
There was the father, his son, and his son's son. This equals 2 fathers and 2 sons for a total of 3!
82.79 %
71 votes
logicmath

There are 1 million closed school lockers in a row, labeled 1 through 1,000,000. You first go through and flip every locker open. Then you go through and flip every other locker (locker 2, 4, 6, etc...). When you're done, all the even-numbered lockers are closed. You then go through and flip every third locker (3, 6, 9, etc...). "Flipping" mean you open it if it's closed, and close it if it's open. For example, as you go through this time, you close locker 3 (because it was still open after the previous run through), but you open locker 6, since you had closed it in the previous run through. Then you go through and flip every fourth locker (4, 8, 12, etc...), then every fifth locker (5, 10, 15, etc...), then every sixth locker (6, 12, 18, etc...) and so on. At the end, you're going through and flipping every 999,998th locker (which is just locker 999,998), then every 999,999th locker (which is just locker 999,999), and finally, every 1,000,000th locker (which is just locker 1,000,000). At the end of this, is locker 1,000,000 open or closed?
Locker 1,000,000 will be open. If you think about it, the number of times that each locker is flipped is equal to the number of factors it has. For example, locker 12 has factors 1, 2, 3, 4, 6, and 12, and will thus be flipped 6 times (it will end be flipped when you flip every one, every 2nd, every 3rd, every 4th, every 6th, and every 12th locker). It will end up closed, since flipping an even number of times will return it to its starting position. You can see that if a locker number has an even number of factors, it will end up closed. If it has an odd number of factors, it will end up open. As it turns out, the only types of numbers that have an odd number of factors are squares. This is because factors come in pairs, and for squares, one of those pairs is the square root, which is duplicated and thus doesn't count twice as a factor. For example, 12's factors are 1 x 12, 2 x 6, and 3 x 4 (6 total factors). On the other hand, 16's factors are 1 x 16, 2 x 8, and 4 x 4 (5 total factors). So lockers 1, 4, 9, 16, 25, etc... will all be open. Since 1,000,000 is a square number (1000 x 1000), it will be open as well.
82.65 %
63 votes
logicmathclever

You can easily "tile" an 8x8 chessboard with 32 2x1 tiles, meaning that you can place these 32 tiles on the board and cover every square. But if you take away two opposite corners from the chessboard, it becomes impossible to tile this new 62-square board. Can you explain why tiling this board isn't possible?
Color in the chessboard, alternating with red and blue tiles. Then color all of your tiles half red and half blue. Whenever you place a tile down, you can always make it so that the red part of the tile is on a red square and the blue part of the tile is on the blue square. Since you'll need to place 31 tiles on the board (to cover the 62 squares), you would have to be able to cover 31 red squares and 31 blue squares. But when you took away the two corners, you can see that you are taking away two red spaces, leaving 30 red squares and 32 blue squares. There is no way to cover 30 red squares and 32 blue squares with the 31 tiles, since these tiles can only cover 31 red squares and 31 blue squares, and thus, tiling this board is not possible.
82.20 %
54 votes
logicmath

You have been given the task of transporting 3,000 apples 1,000 miles from Appleland to Bananaville. Your truck can carry 1,000 apples at a time. Every time you travel a mile towards Bananaville you must pay a tax of 1 apple but you pay nothing when going in the other direction (towards Appleland). What is highest number of apples you can get to Bananaville?
833 apples. Step one: First you want to make 3 trips of 1,000 apples 333 miles. You will be left with 2,001 apples and 667 miles to go. Step two: Next you want to take 2 trips of 1,000 apples 500 miles. You will be left with 1,000 apples and 167 miles to go (you have to leave an apple behind). Step three: Finally, you travel the last 167 miles with one load of 1,000 apples and are left with 833 apples in Bananaville.
82.12 %
61 votes
interviewlogicmath

Three ants are sitting at the three corners of an equilateral triangle. Each ant starts randomly picks a direction and starts to move along the edge of the triangle. What is the probability that none of the ants collide?
So let’s think this through. The ants can only avoid a collision if they all decide to move in the same direction (either clockwise or anti-clockwise). If the ants do not pick the same direction, there will definitely be a collision. Each ant has the option to either move clockwise or anti-clockwise. There is a one in two chance that an ant decides to pick a particular direction. Using simple probability calculations, we can determine the probability of no collision. P(No collision) = P(All ants go in a clockwise direction) + P( All ants go in an anti-clockwise direction) = 0.5 * 0.5 * 0.5 + 0.5 * 0.5 * 0.5 = 0.25
82.02 %
46 votes
logicmathsimpleclever

We all know that square root of number 121 is 11. But do you know what si the square root of the number "12345678987654321" ?
111111111 Explanation: It's a maths magical square root series as : Square root of number 121 is 11 Square root of number 12321 is 111 Square root of number 1234321 is 1111 Square root of number 123454321 is 11111 Square root of number 12345654321 is 111111 Square root of number 1234567654321 is 1111111 Square root of number 123456787654321 is 11111111 Square root of number 12345678987654321 is 111111111 (answer)
81.84 %
60 votes
logicmath

Two words are anagrams if and only if they contain the exact same letters with the exact same frequency (for example, "name" and "mean" are anagrams, but "red" and "deer" are not). Given two strings S1 and S2, which each only contain the lowercase letters a through z, write a program to determine if S1 and S2 are anagrams. The program must have a running time of O(n + m), where n and m are the lengths of S1 and S2, respectively, and it must have O(1) (constant) space usage.
First create an array A of length 26, representing the counts of each letter of the alphabet, with each value initialized to 0. Iterate through each character in S1 and add 1 to the corresponding entry in A. Once this iteration is complete, A will contain the counts for the letters in S1. Then, iterate through each character in S2, and subtract 1 from each corresponding entry in A. Now, if the each entry in A is 0, then S1 and S2 are anagrams; otherwise, S1 and S2 aren't anagrams. Here is pseudocode for the procedure that was described: def areAnagrams(S1, S2) A = new Array(26) A.initializeValues(0) for each character in S1 arrayIndex = mapCharacterToNumber(character) //maps "a" to 0, "b" to 1, "c" to 2, etc... A[arrayIndex] += 1 end for each character in S2 arrayIndex = mapCharacterToNumber(character) A[arrayIndex] -= 1 end for (i = 0; i < 26; i++) if A[i] != 0 return false end end return true end
81.65 %
45 votes
logicmathtricky

A train leaves from Halifax, Nova Scotia heading towards Vancouver, British Columbia at 120 km/h. Three hours later, a train leaves Vancouver heading towards Halifax at 180 km/h. Assume there's exactly 6000 kilometers between Vancouver and Halifax. When they meet, which train is closer to Halifax?
Both trains would be at the same spot when they meet therefore they are both equally close to Halifax.
81.65 %
45 votes