The Backstabber
Hint:
Put It In A Glove
This is something you might hold
Of a person that you love
When it's really cold outside
You might put it in a glove
Of a person that you love
When it's really cold outside
You might put it in a glove
Hint:
Born In Mourning
I have a name, but it isn't my name. My face shows signs of age. I always mean the same thing, no matter what I say. I'm born in mourning, and I last 'til the end of days. Men plant me, but I never grow. They run from me, but I never move. They look at me and see their future, rotting in the fields where I bloom. What am I?
Hint:
Mortal Privation Riddle
Marking mortal privation, when firmly in place.
An enduring summation, inscribed in my face.
What am I?
An enduring summation, inscribed in my face.
What am I?
Hint:
Bruce Wayne Is This Type Of Man
Its black, furry and flies about
Try not to be scared if you can
It likes to hang out upside down
And Bruce Wayne is this type of man
Who am I?
Try not to be scared if you can
It likes to hang out upside down
And Bruce Wayne is this type of man
Who am I?
Hint:
The Holy Trinity
Hint:
Full Of Holes
Hint:
Four Legs And A Back Riddle
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
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
The Secret Santa Exchange
A group of ten friends decide to exchange gifts as secret Santas. Each person writes his or her name on a piece of paper and puts it in a hat. Then each person randomly draws a name from the hat to determine who has him as his or her secret Santa. The secret Santa then makes a gift for the person whose name he drew.
When it's time to exchange presents, each person walks over to the person he made the gift for and holds his or her left hand in his right hand.
What is the probability that the 10 friends holding hands form a single continuous circle?
When it's time to exchange presents, each person walks over to the person he made the gift for and holds his or her left hand in his right hand.
What is the probability that the 10 friends holding hands form a single continuous circle?
Hint: It's not as difficult as it seems.
It's the number of ways the friends can form a circle divided by the number of ways the names can be drawn out of the hat.
1/10
For a group of n friends, there are n! (n factorial) ways to draw the names out of the hat. Since a circle does not have a beginning and end, choose one person as the beginning and end of the circle. There are now (n-1)! ways to distribute the remaining people around the circle. Thus the probability of forming a single circle is
(n-1)! / n!
Since n! = (n-1)! * n (for n > 1), this can be rewritten as
(n-1)! / (n*(n-1)!)
Factoring out the (n-1)! from the numerator and denominator leaves
1/n
as the probability. Did you answer this riddle correctly?
YES NO
For a group of n friends, there are n! (n factorial) ways to draw the names out of the hat. Since a circle does not have a beginning and end, choose one person as the beginning and end of the circle. There are now (n-1)! ways to distribute the remaining people around the circle. Thus the probability of forming a single circle is
(n-1)! / n!
Since n! = (n-1)! * n (for n > 1), this can be rewritten as
(n-1)! / (n*(n-1)!)
Factoring out the (n-1)! from the numerator and denominator leaves
1/n
as the probability. 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
A Thought In Your Mind
I can bring a smile to your face, a tear to your eye or even a thought to your mind, but I can't be seen. What Am I?
Hint:
I Have A Tail Riddle
Hint:
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