Dimples And A Nose
What has dimples and a nose
Ten fingers and ten toes
Sweet fat little cheeks
And in *** months you will meet
Ten fingers and ten toes
Sweet fat little cheeks
And in *** months you will meet
Hint:
Under The Cup Riddle
You decide to play a game with your friend where your friend places a coin under one of three cups. Your friend would then switch the positions of two of the cups several times so that the coin under one of the cups moves with the cup it is under. You would then select the cup that you think the coin is under. If you won, you would receive the coin, but if you lost, you would have to pay.
As the game starts, you realise that you are really tired, and you don't focus very well on the moving of the cups. When your friend stops moving the cups and asks you where the coin is, you only remember a few things:
He put the coin in the rightmost cup at the start.
He switched two of the cups 3 times.
The first time he switched two of the cups, the rightmost one was switched with another.
The second time he switched two of the cups, the rightmost one was not touched.
The third and last time he switched two of the cups, the rightmost one was switched with another.
You don't want to end up paying your friend, so, using your head, you try to work out which cup is most likely to hold the coin, using the information you remember.
Which cup is most likely to hold the coin?
As the game starts, you realise that you are really tired, and you don't focus very well on the moving of the cups. When your friend stops moving the cups and asks you where the coin is, you only remember a few things:
He put the coin in the rightmost cup at the start.
He switched two of the cups 3 times.
The first time he switched two of the cups, the rightmost one was switched with another.
The second time he switched two of the cups, the rightmost one was not touched.
The third and last time he switched two of the cups, the rightmost one was switched with another.
You don't want to end up paying your friend, so, using your head, you try to work out which cup is most likely to hold the coin, using the information you remember.
Which cup is most likely to hold the coin?
Hint: Write down the possibilities. Remember that there are only three cups, so if the rightmost cup wasn't touched...
The rightmost cup.
The rightmost cup has a half chance of holding the coin, and the other cups have a quarter chance.
Pretend that Os represent cups, and Q represents the cup with the coin.
The game starts like this:
OOQ
Then your friend switches the rightmost cup with another, giving two possibilities, with equal chance:
OQO
QOO
Your friend then moves the cups again, but doesn't touch the rightmost cup. The only switch possible is with the leftmost cup and the middle cup. This gives two possibilities with equal chance:
QOO
OQO
Lastly, your friend switches the rightmost cup with another cup. If the first possibility shown above was true, there would be two possibilities, with equal chance:
OOQ
QOO
If the second possibility shown above (In the second switch) was true, there would be two possibilities with equal chance:
OOQ
OQO
This means there are four possibilities altogether, with equal chance:
OOQ
QOO
OOQ
OQO
This means each possibility equals to a quarter chance, and because there are two possibilities with the rightmost cup having the coin, there is a half chance that the coin is there. Did you answer this riddle correctly?
YES NO
The rightmost cup has a half chance of holding the coin, and the other cups have a quarter chance.
Pretend that Os represent cups, and Q represents the cup with the coin.
The game starts like this:
OOQ
Then your friend switches the rightmost cup with another, giving two possibilities, with equal chance:
OQO
QOO
Your friend then moves the cups again, but doesn't touch the rightmost cup. The only switch possible is with the leftmost cup and the middle cup. This gives two possibilities with equal chance:
QOO
OQO
Lastly, your friend switches the rightmost cup with another cup. If the first possibility shown above was true, there would be two possibilities, with equal chance:
OOQ
QOO
If the second possibility shown above (In the second switch) was true, there would be two possibilities with equal chance:
OOQ
OQO
This means there are four possibilities altogether, with equal chance:
OOQ
QOO
OOQ
OQO
This means each possibility equals to a quarter chance, and because there are two possibilities with the rightmost cup having the coin, there is a half chance that the coin is there. Did you answer this riddle correctly?
YES NO
Roll The Dice
A gambler goes to bet. The dealer has 3 dice, which are fair, meaning that the chance that each face shows up is exactly 1/6.
The dealer says: "You can choose your bet on a number, any number from 1 to 6. Then I'll roll the 3 dice. If none show the number you bet, you'll lose $1. If one shows the number you bet, you'll win $1. If two or three dice show the number you bet, you'll win $3 or $5, respectively."
Is it a fair game?
The dealer says: "You can choose your bet on a number, any number from 1 to 6. Then I'll roll the 3 dice. If none show the number you bet, you'll lose $1. If one shows the number you bet, you'll win $1. If two or three dice show the number you bet, you'll win $3 or $5, respectively."
Is it a fair game?
Hint: What will happen if there are 6 gamblers, each of whom bet on a different number?
It's a fair game. If there are 6 gamblers, each of whom bet on a different number, the dealer will neither win nor lose on each deal.
If he rolls 3 different numbers, e.g. 1, 2, 3, the three gamblers who bet 1, 2, 3 each wins $1 while the three gamblers who bet 4, 5, 6 each loses $1.
If two of the dice he rolls show the same number, e.g. 1, 1, 2, the gambler who bet 1 wins $3, the gambler who bet 2 wins $1, and the other 4 gamblers each loses $1.
If all 3 dice show the same number, e.g. 1, 1, 1, the gambler who bet 1 wins $5, and the other 5 gamblers each loses $1.
In each case, the dealer neither wins nor loses. Hence it's a fair game. Did you answer this riddle correctly?
YES NO
If he rolls 3 different numbers, e.g. 1, 2, 3, the three gamblers who bet 1, 2, 3 each wins $1 while the three gamblers who bet 4, 5, 6 each loses $1.
If two of the dice he rolls show the same number, e.g. 1, 1, 2, the gambler who bet 1 wins $3, the gambler who bet 2 wins $1, and the other 4 gamblers each loses $1.
If all 3 dice show the same number, e.g. 1, 1, 1, the gambler who bet 1 wins $5, and the other 5 gamblers each loses $1.
In each case, the dealer neither wins nor loses. Hence it's a fair game. Did you answer this riddle correctly?
YES NO
Fighting In A Truel
Mr. Black, Mr. Gray, and Mr. White are fighting in a truel. They each get a gun and take turns shooting at each other until only one person is left. Mr. Black, who hits his shot 1/3 of the time, gets to shoot first. Mr. Gray, who hits his shot 2/3 of the time, gets to shoot next, assuming he is still alive. Mr. White, who hits his shot all the time, shoots next, assuming he is also alive. The cycle repeats. All three competitors know one another's shooting odds. If you are Mr. Black, where should you shoot first for the highest chance of survival?
Hint: Think from the points of view of Mr. Gray and Mr. White, not just Mr. Black.
He should shoot at the ground.
If Mr. Black shoots the ground, it is Mr. Gray's turn. Mr. Gray would rather shoot at Mr. White than Mr. Black, because he is better. If Mr. Gray kills Mr. White, it is just Mr. Black and Mr. Gray left, giving Mr. Black a fair chance of winning. If Mr. Gray does not kill Mr. White, it is Mr. White's turn. He would rather shoot at Mr. Gray and will definitely kill him. Even though it is now Mr. Black against Mr. White, Mr. Black has a better chance of winning than before. Did you answer this riddle correctly?
YES NO
If Mr. Black shoots the ground, it is Mr. Gray's turn. Mr. Gray would rather shoot at Mr. White than Mr. Black, because he is better. If Mr. Gray kills Mr. White, it is just Mr. Black and Mr. Gray left, giving Mr. Black a fair chance of winning. If Mr. Gray does not kill Mr. White, it is Mr. White's turn. He would rather shoot at Mr. Gray and will definitely kill him. Even though it is now Mr. Black against Mr. White, Mr. Black has a better chance of winning than before. Did you answer this riddle correctly?
YES NO
Two In A Row Riddle
A certain mathematician, his wife, and their teenage son all play a fair game of chess. One day when the son asked his father for 10 dollars for a Saturday night date, his father puffed his pipe for a moment and replied, "Let's do it this way. Today is Wednesday. You will play a game of chess tonight, tomorrow, and a third on Friday. If you win two games in a row, you get the money."
"Whom do I play first, you or mom?"
"You may have your choice," said the mathematician, his eyes twinkling.
The son knew that his father played a stronger game than his mother. To maximize his chance of winning two games in succession, should he play father-mother-father or mother-father-mother?
"Whom do I play first, you or mom?"
"You may have your choice," said the mathematician, his eyes twinkling.
The son knew that his father played a stronger game than his mother. To maximize his chance of winning two games in succession, should he play father-mother-father or mother-father-mother?
Hint: Who does he need to beat to win?
Father-mother-father
To beat two games in a row, it is necessary to win the second game. This means that it would be to his advantage to play the second game against the weaker player. Though he plays his father twice, he has a higher chance of winning by playing his mother second. Did you answer this riddle correctly?
YES NO
To beat two games in a row, it is necessary to win the second game. This means that it would be to his advantage to play the second game against the weaker player. Though he plays his father twice, he has a higher chance of winning by playing his mother second. Did you answer this riddle correctly?
YES NO
Pearl Problems Riddle
"I'm a very rich man, so I've decided to give you some of my fortune. Do you see this bag? I have 5001 pearls inside it. 2501 of them are white, and 2500 of them are black. No, I am not racist. I'll let you take out any number of pearls from the bag without looking. If you take out the same number of black and white pearls, I will reward you with a number of gold bars equivalent to the number of pearls you took."
How many pearls should you take out to give yourself a good number of gold bars while still retaining a good chance of actually getting them?
How many pearls should you take out to give yourself a good number of gold bars while still retaining a good chance of actually getting them?
Hint: If you took out 2 pearls, you would have about a 50% chance of getting 2 gold bars. However, you can take even more pearls and still retain the 50% chance.
Take out 5000 pearls. If the remaining pearl is white, then you've won 5000 gold bars! Did you answer this riddle correctly?
YES NO
YES NO
The Same Birthday Riddle
How many people must be gathered together in a room, before you can be certain that there is a greater than 50/50 chance that at least two of them have the same birthday?
Hint:
Only twenty-three people need be in the room, a surprisingly small number. The probability that there will not be two matching birthdays is then, ignoring leap years, 365x364x363x...x343/365 over 23 which is approximately 0.493. this is less than half, and therefore the probability that a pair occurs is greater than 50-50. With as few as fourteen people in the room the chances are better than 50-50 that a pair will have birthdays on the same day or on consecutive days. Did you answer this riddle correctly?
YES NO
YES NO
Matching Socks Riddle
Mismatched Joe is in a pitch dark room selecting socks from his drawer. He has only six socks in his drawer, a mixture of black and white. If he chooses two socks, the chances that he draws out a white pair is 2/3. What are the chances that he draws out a black pair?
Hint: Three pairs of matching socks... maybe not!!!
He has a ZERO chance of drawing out a black pair.
Since there is a 2/3 chance of drawing a white pair, then there MUST be 5 white socks and only 1 black sock. The chances of drawing two whites would thus be: 5/6 x 4/5 = 2/3 . With only 1 black sock, there is no chance of drawing a black pair. Did you answer this riddle correctly?
YES NO
Since there is a 2/3 chance of drawing a white pair, then there MUST be 5 white socks and only 1 black sock. The chances of drawing two whites would thus be: 5/6 x 4/5 = 2/3 . With only 1 black sock, there is no chance of drawing a black pair. Did you answer this riddle correctly?
YES NO
Four Balls In A Bowl
This is a famous paradox probability riddle which has caused a great deal of argument and disbelief from many who cannot accept the correct answer.
Four balls are placed in a bowl. One is Green, one is Black and the other two are Yellow. The bowl is shaken and someone draws two balls from the bowl. He looks at the two balls and announces that at least one of them is Yellow. What are the chances that the other ball he has drawn out is also Yellow?
Four balls are placed in a bowl. One is Green, one is Black and the other two are Yellow. The bowl is shaken and someone draws two balls from the bowl. He looks at the two balls and announces that at least one of them is Yellow. What are the chances that the other ball he has drawn out is also Yellow?
Hint:
1/5
There are six possible pairings of the two balls withdrawn,
Yellow+Yellow
Yellow+Green
Green+Yellow
Yellow+Black
Black+Yellow
Green+Black.
We know the Green + Black combination has not been drawn.
This leaves five possible combinations remaining. Therefore the chances tbowl the Yellow + Yellow pairing has been drawn are 1 in 5.
Many people cannot accept tbowl the solution is not 1 in 3, and of course it would be, if the balls had been drawn out separately and the color of the first ball announced as Yellow before the second had been drawn out. However, as both balls had been drawn together, and then the color of one of the balls announced, then the above solution, 1 in 5, must be the correct one. Did you answer this riddle correctly?
YES NO
There are six possible pairings of the two balls withdrawn,
Yellow+Yellow
Yellow+Green
Green+Yellow
Yellow+Black
Black+Yellow
Green+Black.
We know the Green + Black combination has not been drawn.
This leaves five possible combinations remaining. Therefore the chances tbowl the Yellow + Yellow pairing has been drawn are 1 in 5.
Many people cannot accept tbowl the solution is not 1 in 3, and of course it would be, if the balls had been drawn out separately and the color of the first ball announced as Yellow before the second had been drawn out. However, as both balls had been drawn together, and then the color of one of the balls announced, then the above solution, 1 in 5, must be the correct one. Did you answer this riddle correctly?
YES NO
The Gardners Riddle
Gretchen and Henry were discussing their new neighbors, the Gardners. Gretchen mentioned that she met two of the daughters, and they each had blond hair.
"I have met all of the sisters," replied Henry, "and the probability that both of the girls you met would have had blond hair, assuming you were equally likely to meet any of the sisters, is exactly 50%. Do you know how many children there are?"
After thinking for a minute, Gretchen asks if the family is abnormally large. When Henry replies that it is not, Gretchen tells him how many girls are in the family. What number did she say?
"I have met all of the sisters," replied Henry, "and the probability that both of the girls you met would have had blond hair, assuming you were equally likely to meet any of the sisters, is exactly 50%. Do you know how many children there are?"
After thinking for a minute, Gretchen asks if the family is abnormally large. When Henry replies that it is not, Gretchen tells him how many girls are in the family. What number did she say?
Hint:
Gretchen said that there were 4 girls in the family, three of whom were blond.
This would make the probability that she saw two blonds (3/4) * (2/3), which equals 1/2.
Other numbers would work, but the next pair would lead to a rather large family. Did you answer this riddle correctly?
YES NO
This would make the probability that she saw two blonds (3/4) * (2/3), which equals 1/2.
Other numbers would work, but the next pair would lead to a rather large family. Did you answer this riddle correctly?
YES NO
The Blue And Red Dice Riddle
Timothy and Urban play a game with two dice. But they do not use the numbers. Some of the faces are painted red and the others blue. Each player throws the dice in turn. Timothy wins when the two top faces are the same color. Urban wins when the colors are different. Their chances are even.
The first die has 5 red faces and 1 blue face. How many red and how many blue are there on the second die?
The first die has 5 red faces and 1 blue face. How many red and how many blue are there on the second die?
Hint:
Each die has 6 faces. When two dice are thrown, there are 36 equally possible results. For chances to be even, there must be 18 ways of getting the same color on top. Let X be the number of red faces on the second die. We have: 18 = 5X + 1(6 - X)
X = 3
The second die must have 3 red faces and 3 blue faces. Did you answer this riddle correctly?
YES NO
X = 3
The second die must have 3 red faces and 3 blue faces. Did you answer this riddle correctly?
YES NO
Little Billy's Calculator
Little Billy has a calculator with 15 buttons. He has 10 keys for 0-9, a key for addition, multiplication, division, and subtraction. Finally, he has an = sign. However, Mark the Meanie messed up the programming on Billy's calculator. Now, whenever Billy presses any of the number keys, it comes up with a random single-digit number. The same goes for the four operations keys (+,-,x, /). So whenever Billy tries to press the + button, the calculator chooses randomly between addition, multiplication, subtraction, and division. The only key left untouched was the = sign.
Now, if Billy were to press one number key, one operation key, then another number key, then the = button, what are the chances the answer comes out to 6?
Now, if Billy were to press one number key, one operation key, then another number key, then the = button, what are the chances the answer comes out to 6?
Hint: Think about how many ways he could possibly get 6.
There is a 4% chance.
There are 16 possible ways to get 6.
0+6
1+5
2+4
3+3
6+0
5+1
4+2
9-3
8-2
7-1
6-0
1x6
2x3
6x1
3x2
6/1
There are 400 possible button combinations.
When Billy presses any number key, there are 10 possibilities; when he presses any operation key, there are 4 possibilities.
10(1st#)x4(Operation)x10(2nd#)=400
16 working combinations/400 possible combinations= .04 or 4% Did you answer this riddle correctly?
YES NO
There are 16 possible ways to get 6.
0+6
1+5
2+4
3+3
6+0
5+1
4+2
9-3
8-2
7-1
6-0
1x6
2x3
6x1
3x2
6/1
There are 400 possible button combinations.
When Billy presses any number key, there are 10 possibilities; when he presses any operation key, there are 4 possibilities.
10(1st#)x4(Operation)x10(2nd#)=400
16 working combinations/400 possible combinations= .04 or 4% Did you answer this riddle correctly?
YES NO
100 Blank Cards Riddle
Someone offers you the following deal:
There is a deck of 100 initially blank cards. The dealer is allowed to write ANY positive integer, one per card, leaving none blank. You are then asked to turn over as many cards as you wish. If the last card you turn over is the highest in the deck, you win; otherwise, you lose.
Winning grants you $50, and losing costs you only the $10 you paid to play.
Would you accept this challenge?
There is a deck of 100 initially blank cards. The dealer is allowed to write ANY positive integer, one per card, leaving none blank. You are then asked to turn over as many cards as you wish. If the last card you turn over is the highest in the deck, you win; otherwise, you lose.
Winning grants you $50, and losing costs you only the $10 you paid to play.
Would you accept this challenge?
Hint: Perhaps thinking in terms of one deck is the wrong approach.
Yes!
A sample strategy:
Divide the deck in half and turn over all lower 50 cards, setting aside the highest number you find. Then turn over the other 50 cards, one by one, until you reach a number that is higher than the card you set aside: this is your chosen "high card."
Now, there is a 50% chance that the highest card is contained in the top 50 cards (it is or it isn't), and a 50% chance that the second-highest card is contained in the lower 50. Combining the probabilities, you have a 25% chance of constructing the above situation (in which you win every time).
This means that you'll lose three out of four games, but for every four games played, you pay $40 while you win one game and $50. Your net profit every four games is $10.
Obviously, you have to have at least $40 to start in order to apply this strategy effectively. Did you answer this riddle correctly?
YES NO
A sample strategy:
Divide the deck in half and turn over all lower 50 cards, setting aside the highest number you find. Then turn over the other 50 cards, one by one, until you reach a number that is higher than the card you set aside: this is your chosen "high card."
Now, there is a 50% chance that the highest card is contained in the top 50 cards (it is or it isn't), and a 50% chance that the second-highest card is contained in the lower 50. Combining the probabilities, you have a 25% chance of constructing the above situation (in which you win every time).
This means that you'll lose three out of four games, but for every four games played, you pay $40 while you win one game and $50. Your net profit every four games is $10.
Obviously, you have to have at least $40 to start in order to apply this strategy effectively. Did you answer this riddle correctly?
YES NO
The Coin Toss Riddle
You are in a bar having a drink with an old friend when he proposes a wager.
"Want to play a game?" he asks.
"Sure, why not?" you reply.
"Ok, here's how it works. You choose three possible outcomes of a coin toss, either HHH, TTT, HHT or whatever. I will do likewise. I will then start flipping the coin continuously until either one of our combinations comes up. The person whose combination comes up first is the winner. And to prove I'm not the cheating little weasel you're always making me out to be, I'll even let you go first so you have more combinations to choose from. So how about it? Is $10.00 a fair bet?"
You know that your friend is a skilled trickster and usually has a trick or two up his sleeve but maybe he's being honest this time. Maybe this is a fair bet. While you try and think of which combination is most likely to come up first, you suddenly hit upon a strategy which will be immensely beneficial to you. What is it?
"Want to play a game?" he asks.
"Sure, why not?" you reply.
"Ok, here's how it works. You choose three possible outcomes of a coin toss, either HHH, TTT, HHT or whatever. I will do likewise. I will then start flipping the coin continuously until either one of our combinations comes up. The person whose combination comes up first is the winner. And to prove I'm not the cheating little weasel you're always making me out to be, I'll even let you go first so you have more combinations to choose from. So how about it? Is $10.00 a fair bet?"
You know that your friend is a skilled trickster and usually has a trick or two up his sleeve but maybe he's being honest this time. Maybe this is a fair bet. While you try and think of which combination is most likely to come up first, you suddenly hit upon a strategy which will be immensely beneficial to you. What is it?
Hint: Think what would be most likely to happen if you chose HHH, would this be a good decision?
The answer is to let your friend go first. This puzzle is based on an old game/scam called Penny Ante. No matter what you picked, your friend would be able to come up with a combination which would be more likely to beat yours. For example, if you were to choose HHH, then unless HHH was the first combination to come up you would eventually lose since as soon as a Tails came up, the combination THH would inevitably come up before HHH. The basic formula you can use for working out which combination you should choose is as follows. Simply take his combination (eg. HHT) take the last term in his combination, put it at the front (in this case making THH) and your combination will be more likely to come up first. Try it on your friends! Did you answer this riddle correctly?
YES NO
YES NO
Blue Eyes Riddle
Both of my parents have brown eyes, as do I. My brother and my wife have blue eyes. Using the simple brown-blue model (two genes; a brown gene dominates blue gene), what are the chances of my first child having blue eyes?
Hint: Given my brother's blue eyes, what are the odds on my pair of eye-color genes?
1 in 3.
Since my brother has blue eyes (bb), both of my parents carry one brown and one blue gene (Bb). The three possibilities for my genotype, equally likely, are BB, Bb, and bB. Thus, there is a 2/3 chance that I carry a blue gene.
If I carry a blue gene, there is a 50% chance I will pass it on to my first child (and, obviously, 0% if I carry two brown genes).
Since my child will certainly get a blue gene from my wife, my gene will determine the eye color.
Multiplying the probabilities of those two independent events, there is a chance of 1/2 x 2/3 = 1/3 of my passing on a blue gene. Did you answer this riddle correctly?
YES NO
Since my brother has blue eyes (bb), both of my parents carry one brown and one blue gene (Bb). The three possibilities for my genotype, equally likely, are BB, Bb, and bB. Thus, there is a 2/3 chance that I carry a blue gene.
If I carry a blue gene, there is a 50% chance I will pass it on to my first child (and, obviously, 0% if I carry two brown genes).
Since my child will certainly get a blue gene from my wife, my gene will determine the eye color.
Multiplying the probabilities of those two independent events, there is a chance of 1/2 x 2/3 = 1/3 of my passing on a blue gene. Did you answer this riddle correctly?
YES NO
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