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
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
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
The Serial Killer Husband
A man kills his wife. Many people watch him doing so. Yet no one will ever be able to accuse him of murder. Why?
Hint:
Snowman Baby Crib Riddle
Hint:
I'm Not Used To Smile
I;m a unit that measures length
Although Im not a mile
Theres two of me on your body
But Im not used to smile
I am?
Although Im not a mile
Theres two of me on your body
But Im not used to smile
I am?
Hint:
Different Shapes And Sizes
I come in different shapes and sizes but I'm not a snowflake
I can be broken but Im not a window
I can be brittle but Im not peanut butter
I'm sometimes humerus but Im not funny
I am?
Hint:
An Electric Train Riddle
An electric train is moving north at 100mph and a wind is blowing to the west at 10mph. Which way does the smoke blow?
Hint:
A Man Was Outside Taking A Walk When It Began To Rain
A man was outside taking a walk when it began to rain. He did not have an umbrella and he wasn't wearing a hat. His clothes were soaked, yet not a single hair on his head got wet. How could this happen?
Hint:
Going To London Riddle
While I was going to London I met a men with 7 wives, 7 sacks, 7 kids and 7 cats. How many were going to London?
Hint:
Ticking In Its Weary Pace
Time is writ upon my face
My heart ticks in its weary pace.
Though wind and rain do leave their trace,
I watch them in their endless race.
What kind of clock is it, though?
My heart ticks in its weary pace.
Though wind and rain do leave their trace,
I watch them in their endless race.
What kind of clock is it, though?
Hint:
An Island That Has 3 Gods
There is an Island that has 3 gods. One god always tells a lie, and the other always tells the truth. The third god has a random behavior. To top it off, these three gods, being jerks, answer in their own languages such that you are unable to tell which word, between "ja" or "da", means "no" or "yes". You have 3 questions to work out the True god, the false god, and the Random god.
Hint:
Question 1: (To any of the three gods) If I were to ask you "Is that the random god," would your answer be "ja?" (This questions, no matter the answer, will enable you to tell which god is not random i.e. the god who is either False or True)
Question 2: (To either the True or False god) If I asked you "are you false," would your answer be "ja?"
Question 3: (To the same god you asked the second question) If I asked you "whether the first god I spoke to is random," would your answer be "ja?" Did you answer this riddle correctly?
YES NO
Question 2: (To either the True or False god) If I asked you "are you false," would your answer be "ja?"
Question 3: (To the same god you asked the second question) If I asked you "whether the first god I spoke to is random," would your answer be "ja?" Did you answer this riddle correctly?
YES NO
Going To St. Ives
As I was going to St. Ives,
I met a man with seven wives,
Each wife had seven sacks,
Each sack had seven cats,
Each cat had seven kits:
Kits, cats, sacks, and wives,
How many were there going to St. Ives?
I met a man with seven wives,
Each wife had seven sacks,
Each sack had seven cats,
Each cat had seven kits:
Kits, cats, sacks, and wives,
How many were there going to St. Ives?
Hint:
One. As John McClane learns, this is a classic trick question. If the narrator meets the group on the way to St. Ives, then they must be going in the opposite direction and the math calculations are simply a bit of trickery meant to misdirect. Did you answer this riddle correctly?
YES NO
YES NO
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