The Last Cookie Riddle
Mike and James are arguing over who gets the last cookie in the jar, so their dad decides to create a game to settle their dispute. First, Mike flips a coin twice, and each time James calls heads or tails in the air. If James gets both calls right, he gets the last cookie. If not, Mike picks a number between one and six and then rolls a die. If he gets the number right, he gets the last cookie. If not, James picks two numbers between one and five, then spins a spinner with numbers one through five on it. If the spinner lands on one of James' two numbers, he gets the last cookie. If not, Mike does.
Who is more likely to win the last cookie, Mike or James? And what is the probability that person wins it?
Who is more likely to win the last cookie, Mike or James? And what is the probability that person wins it?
Hint: Their dad is a very smart person.
Believe it or not, both Mike and James have a 1/2 chance of winning.
James wins if:
-he calls both coin flips right = 1/2 x 1/2 = 1/4
OR
-he does not call both coin flips right, Mike does not call the die roll correctly, and he guesses the number on the spinner right = 3/4 x 5/6 x 2/5 = 30/120 = 1/4
1/4 + 1/4 = 1/2
Mike wins if:
-James does not call both coin flips right and he calls the die roll correctly = 3/4 x 1/6 = 3/24 = 1/8
OR
-James does not call both coin flips right, he does not call the die roll correctly, and Mike does not guess the number on the spinner right = 3/4 x 5/6 x 3/5 = 45/120 = 3/8
1/8 + 3/8 = 1/2
Of course, dad could have just flipped a coin Did you answer this riddle correctly?
YES NO
James wins if:
-he calls both coin flips right = 1/2 x 1/2 = 1/4
OR
-he does not call both coin flips right, Mike does not call the die roll correctly, and he guesses the number on the spinner right = 3/4 x 5/6 x 2/5 = 30/120 = 1/4
1/4 + 1/4 = 1/2
Mike wins if:
-James does not call both coin flips right and he calls the die roll correctly = 3/4 x 1/6 = 3/24 = 1/8
OR
-James does not call both coin flips right, he does not call the die roll correctly, and Mike does not guess the number on the spinner right = 3/4 x 5/6 x 3/5 = 45/120 = 3/8
1/8 + 3/8 = 1/2
Of course, dad could have just flipped a coin 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
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
Yahtzee Riddle
The game of Yahtzee is played with five dice. On the first turn, a player rolls all five dice, and then may decide to keep any, all, or none of the dice aside before rolling again. Each player has a maximum of three rolls to try to get a favorable combination of dice "kept" on the side.
If a player rolls two 2s and two 4s on his/her first roll, and keeps all four of these dice aside, what is the probability of getting a full house (three of one value and two of another) in one of his/her next two rolls? (ie what is the probability of getting either a 2 or a 4 in one of the next two rolls?)
If a player rolls two 2s and two 4s on his/her first roll, and keeps all four of these dice aside, what is the probability of getting a full house (three of one value and two of another) in one of his/her next two rolls? (ie what is the probability of getting either a 2 or a 4 in one of the next two rolls?)
Hint: Think of the probability of NOT getting a full house.
5/9
The answer is NOT 2/3 because you cannot add probabilities. On each roll, the probability of getting a 2 or a 4 is 1/3, so therefore, the probability of not getting a 2 or a 4 is 2/3. Since the die is being rolled twice, square 2/3 to get a 4/9 probability of NOT getting a full house in two rolls. The probability of getting a full house is therefore 1 - 4/9, or 5/9. Did you answer this riddle correctly?
YES NO
The answer is NOT 2/3 because you cannot add probabilities. On each roll, the probability of getting a 2 or a 4 is 1/3, so therefore, the probability of not getting a 2 or a 4 is 2/3. Since the die is being rolled twice, square 2/3 to get a 4/9 probability of NOT getting a full house in two rolls. The probability of getting a full house is therefore 1 - 4/9, or 5/9. 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
Russian Roulette Riddle
You are in a game of Russian Roulette with a revolver that has 3 bullets placed in three consecutive chambers. The cylinder of the gun will be spun once at the beginning of the game. Then, the gun will be passed between two players until it fires. Would you prefer to go first or second?
Hint:
Russian Roulette
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
Puzzle ID: #17681
Fun: *** (2.59)
Difficulty: ** (2.07)
Category: Probability
Submitted By: JMCLEOD****
Corrected By: cnmne
You are in a game of Russian Roulette with a revolver that has 3 bullets placed in three consecutive chambers. The cylinder of the gun will be spun once at the beginning of the game. Then, the gun will be passed between two players until it fires. Would you prefer to go first or second?
Answer
Label the chambers 1 through 6. Chambers 1 through 3 have bullets and chambers 4 through 6 are empty. After you spin the cylinder there are six possible outcomes:
1. Chamber 1 is fired first: Player 1 loses
2. Chamber 2 is fired first: Player 1 loses
3. Chamber 3 is fired first: Player 1 loses
4. Chamber 4 is fired first: Player 2 loses (First shot, player 1, chamber 4 empty. Second shot player 2, chamber 5, empty. Third shot player 1, chamber 6 empty. Fourth shot player 2, chamber 1 not empty.)
5. Chamber 5 is fired first: Player 1 loses (First shot, player 1, chamber 5 empty. Second shot player 2, chamber 6, empty. Third shot player 1, chamber 1 not empty.)
6. Chamber 6 is fired first: Player 2 loses (First shot, player 1, chamber 6 empty. Second shot, player 2, chamber 1, not empty)
Therefore player 2 has an 4/6 or 2/3 chance of winning. Did you answer this riddle correctly?
YES NO
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
Puzzle ID: #17681
Fun: *** (2.59)
Difficulty: ** (2.07)
Category: Probability
Submitted By: JMCLEOD****
Corrected By: cnmne
You are in a game of Russian Roulette with a revolver that has 3 bullets placed in three consecutive chambers. The cylinder of the gun will be spun once at the beginning of the game. Then, the gun will be passed between two players until it fires. Would you prefer to go first or second?
Answer
Label the chambers 1 through 6. Chambers 1 through 3 have bullets and chambers 4 through 6 are empty. After you spin the cylinder there are six possible outcomes:
1. Chamber 1 is fired first: Player 1 loses
2. Chamber 2 is fired first: Player 1 loses
3. Chamber 3 is fired first: Player 1 loses
4. Chamber 4 is fired first: Player 2 loses (First shot, player 1, chamber 4 empty. Second shot player 2, chamber 5, empty. Third shot player 1, chamber 6 empty. Fourth shot player 2, chamber 1 not empty.)
5. Chamber 5 is fired first: Player 1 loses (First shot, player 1, chamber 5 empty. Second shot player 2, chamber 6, empty. Third shot player 1, chamber 1 not empty.)
6. Chamber 6 is fired first: Player 2 loses (First shot, player 1, chamber 6 empty. Second shot, player 2, chamber 1, not empty)
Therefore player 2 has an 4/6 or 2/3 chance of winning. Did you answer this riddle correctly?
YES NO
Old Man On London Bridge
I met an old man on London bridge,
As the sun set on the ridge,
He tipped his hat and drew his name,
And cheated at the guessing game.
What was the mans name?
As the sun set on the ridge,
He tipped his hat and drew his name,
And cheated at the guessing game.
What was the mans name?
Hint:
Andrew. In the third line, and drew his name. It works better when you say it. Did you answer this riddle correctly?
YES NO
YES NO
Cinderella Playing Soccer
Hint:
Failed Salute Riddle
Hint:
Saluting indoors is forbidden except when formally reporting to a superior officer. Did you answer this riddle correctly?
YES NO
YES NO
A Terrorist Took Over A Plane Riddle
A terrorist hijacks a plane with 10 passengers and there is lots of gold in the plane.
After talking the gold, he asked the government officials for 11 parachutes.
He killed all the passenger so that no one can identify him, take one parachute and jumps off.
Was he stupid to ask for 11 parachutes if he need only one?
After talking the gold, he asked the government officials for 11 parachutes.
He killed all the passenger so that no one can identify him, take one parachute and jumps off.
Was he stupid to ask for 11 parachutes if he need only one?
Hint:
He was genius.
Officials must have thought he was jumping with a hostage, they would never risk giving him a faulty parachute. Did you answer this riddle correctly?
YES NO
Officials must have thought he was jumping with a hostage, they would never risk giving him a faulty parachute. Did you answer this riddle correctly?
YES NO
7B91011 Riddle
Mrs. Barbara Whitcombe, Ray's wife
Mr. Jason McCubbins, Ray's business partner
Mr. Harold Nichols, Ray's best friend
All three visited Mr. Whitcombe the day of his murder, but all three provided the police with stories of explanation as to the reason for their visit.
Police found Mr. Whitcombe with his wrist watch still on his right arm, a torn up picture of his wife laying on the floor beside the trash can, and an ink pen in his right hand. On the desk, the police found a name plate, a telephone that was off the hook, and a personal calendar turned to the July 5th page with 7B91011 written on it. After examining this evidence, the police knew their suspect.
Who was it?
Hint:
Jason McCubbins, Ray's business partner.
The calendar was the clue to solving this murder. The police realized that since Mr. Whitcombe was wearing his watch on his right arm, he must have been left-handed. Realizing that the number on the calendar was written in a hurry and with his opposite hand, police matched the written number with the months of the year. So the B was an 8, thereby giving us 7-8-9-10-11: July, August, September, October, November. Use the first letter of each month and it spells J-A-S-O-N. Did you answer this riddle correctly?
YES NO
The calendar was the clue to solving this murder. The police realized that since Mr. Whitcombe was wearing his watch on his right arm, he must have been left-handed. Realizing that the number on the calendar was written in a hurry and with his opposite hand, police matched the written number with the months of the year. So the B was an 8, thereby giving us 7-8-9-10-11: July, August, September, October, November. Use the first letter of each month and it spells J-A-S-O-N. Did you answer this riddle correctly?
YES NO
Three People At A Bus Stop Riddle
1. An old woman who looks as if she is about to die.
2. An old friend who once saved your life.
3.The perfect partner you have been dreaming about (your soulmate).
Knowing that you only have room for one passenger in your car (its a really small car), which one would you choose to offer a ride to? And why?
Hint:
I would give the car keys to my old friend, and let him take the old woman to the hospital. Then I would stay behind and wait for the bus with the partner of my dreams. Did you answer this riddle correctly?
YES NO
YES NO
Polar Bear Dice Riddle
Polar bears around an ice hole, like petals around a rose. The game is in the name, and the name is in the game. How many polar bears are there?" Johnny asked as he rolled the five dice. The first roll produced 4, 6, 1, 3, 2. "Six," said Billy. "No, two," Johnny replied. The next roll was 5, 1, 5, 2, 4. "Four?" said Billy. "No, eight," Johnny said. The next rolls were 3, 5, 3, 1, 2. There were 8 polar bears. The next rolls were 6, 2, 1, 2, 4. There were no polar bears. How does Johnny figure out the number of polar bears?
Hint: A rose by any other name...could be a die?
Dice all look the same. On a die, the 1, 3, and 5 all have a dot in the center. The 3 has 2 dots on either side of the center dot, and the 5 has 4 dots around the center dot. Johnny simply counted the number of dots around the outside. A "3" has 2 "petals around the rose, or polar bears around an ice hole." The "5" has 4 "petals" or "polar bears." Roll some dice and it will become clear!! Did you answer this riddle correctly?
YES NO
YES NO
With Eyes That See I Cant Be Found Riddle
Hint:
Prisoner Hat Riddle
There is a wall between prisoners C and D (which cannot be seen through or around). Prisoner A can see prisoners B and C (by moving his head to the side). Prisoner B can see prisoner C. Prisoners C and D see only the wall.
The prisoners are immobilized in the ground and can't twist their body to see the person behind them. The warden shows them two black hats and two white hats and then puts the hats in a bag to conceal them. He then stands behind each prisoner, chooses a hat from the bag, and puts it on their head. The color of each prisoner's hat is shown in the image above.
The rules are simple. If any prisoner can figure out the color of the hat on his head, all four prisoners will be set free. But they must be sure, if one of them simply guesses and is wrong, they will all be shot dead! The prisoners are not allowed to talk to each other and they have 10 seconds.
The warden counts down "ten, nine, eight, seven". All four prisoners are silent. The warden smiles, knowing that he put the hats on in such a way that no prisoner could possibly know the color of the hat they had on. He continues "six, five, four, thr.."
"I know the color of my hat!" one of the prisoners finally blurts out.
Which prisoner called out and why is he 100% certain of the color of his hat?
Hint:
Prisoner B.
If prisoners B and C had the same color hat on, prisoner A would have know immediately that his hat was the other color (there are only two hats of each color). Since prisoner A was silent, prisoners B and C must have different colored hats. Prisoner B realized this and knew that his hat was not the same color as prisoner C, therefore his hat must be black! Did you answer this riddle correctly?
YES NO
If prisoners B and C had the same color hat on, prisoner A would have know immediately that his hat was the other color (there are only two hats of each color). Since prisoner A was silent, prisoners B and C must have different colored hats. Prisoner B realized this and knew that his hat was not the same color as prisoner C, therefore his hat must be black! Did you answer this riddle correctly?
YES NO
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