Toggler Riddle
Played 2,718 times
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July 1st, 2010
VN:RO [1.9.1_1087] You are in a room with five people. One of them will always tell you the truth. The other four will toggle between telling you the truth, and lying to you. However they can either tell you the truth OR lie on the first question you ask them. After that they must always toggle between that and the opposite. For example. I see one has a mustache, and he is a toggler.
I ask - "Do you have a mustache?"
"Yes."
I ask again.
"No."
I ask again.
"Yes."
OR
I ask - "Do you have a mustache?"
"No."
I ask again.
"Yes."
I ask again.
"No."
The challenge of the riddle is to ask exactly TWO questions to anyone. You can ask both questions to the same person or ask questions to two different people, and at the end of your two questions you must find out who the truth-teller is.
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hint
You must ask both questions to the same person.
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answer
Question 1: Are you the truth teller?
Question 2: (a) If the person responds "Yes". They are either the truth teller or they are a toggler that has just lied to you that will now tell you the truth. Either way the next answer will be truthful. Therefore ask who is the truth teller.
Question 2 (b) if the answer to question 1 is "no". They have to be a toggler that has just told you the truth. Therefore the next question will be a lie. Therefore the question " who is a toggler" will reveal the truth teller.
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Missing Dollar
Played 1,979 times
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June 24th, 2010
VN:RO [1.9.1_1087] Three men stroll into a restaurant and share a $30 meal. After paying for the meal, the owner of the restaurant realizes the waiter overcharged the men, and orders the waiter to return $5 back to the men. Not knowing how to divide the money the waiter decides to keep $2 and give each of the men $1 back. So each man ended up paying $9 ($10-$1), the waiter kept $2. So thats $9+$9+$9+$2 = $29. Where did the last dollar end up?
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hint
How much money did the restaurant take?
How much did the waiter take?
How much did the men take?
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answer
The math at the end of the question makes you add up numbers wrong
The Men started with $30
They were returned $1 each so they payed $27.
The restaurant took $25
The waiter took the last $2
$30-$3 = $27 ( $25 to restaurant, $2 to the waiter)
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June 4th, 2010
VN:RO [1.9.1_1087] I am looking at a picture of somebody. It has a caption. The caption reads: "Brothers and sisters have I none, but that man's father is my father's son. Who is that man?"
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hint
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answer
My father's son is yourself. Therefore. That man's father is yourself. Therefore it is your son.
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Card Proposition
Played 617 times
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June 4th, 2010
VN:RO [1.9.1_1087] You are given a betting proposition by an anonymous dealer. The dealer tells you he has a deck of 1000 cards, each card has a random number on it. The number could be any number, negative or positive, the numbers have no limit. The dealer says he'll flip over each card one by and one and you'll have to tell to him when to stop. If you stop and it turns out the card he just flipped is the highest card in the deck he'll give you 5 times your bet. How can you come up with a technique that will lead to a positive outcome?
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hint
Calculate what the chances are that the second highest card has already been turned over.
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answer
Technique: Once you've reached halfway, if you see a card that's higher than all the other cards, tell the dealer to stop.
If you take note of the second highest card in the first half of the deck. and then pick any card higher than the this card in the second half of the deck you will have a 1/4 chance of selecting the highest card.
Explanation:
There is a 1/2 chance that the second highest card is in the first half of the deck. There is a 1/2 chance that the highest card is in the second half of the deck. Therefore there is a (1/2)*(1/2) = 1/4 chance that both these conditions hold true. If both these conditions hold true you will always pick out the highest card.
Amazingly the outcome has nothing to do with the size of the deck!
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Weighing Coins
Played 568 times
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June 4th, 2010
VN:RO [1.9.1_1087] You are given 27 coins and a balance scale. All the coins weigh the same amount except one coin is counterfeit and weights slightly less. What is the least amount of times that you have to use the scale to ensure you find the counterfeit coin. And how is this possible?
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hint
You will only have to use the scale three times.
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answer
Divide the coins into three groups of 9 and put 2 of the groups on either end the scale. If they weight the same amount, it means the counterfeit coin is in the pile not on the scale. Of course if one side goes up the counterfeit coin is in that stack of 9. Now take that stack of 9 and divide evenly again into stacks 3 groups of three and put two of groups of the scale. Use the same technique. You will find a group of 3 be left with a group of 3 coins. Now weigh two of those coins and voila. The counterfeit coin!
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Pirate’s Dilemma
Played 588 times
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June 4th, 2010
VN:RO [1.9.1_1087] After pillaging another ship, five blood thirsty pirate's arrive on their boat with a chest of 100 gold coins. The pirates have a very unique way of dividing the gold. The pirates all have ranks '1' through '5'. The pirate's allow 'pirate 1' to divide up the gold however he wants. But then the pirates vote whether they are in favor or against the outcome. If they are against the outcome. 'Pirate 1' is thrown overboard. Then 'Pirate 2' is given a chance to divide up the gold. And so forth.
-All the pirate have perfect logic and will make the decision to grant themselves the most gold
-during the voting if it is a tie the pirate is thrown overboard.
-these are bloodthirsty pirate, if it's turns out they'll end up with the same amount of gold they'll chose to kill.
You are pirate 1 and must divide up the gold. How much gold must you give each other pirate to stay alive?
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hint
Try working backwards. Picture what would happen if only 2 pirates were left. Then picture what would happen if 3 pirates were left.
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answer
Let P represent the pirates.
If 1 pirate were left 'P5' receives all the gold.
If 2 pirates were left 'P5' would vote against 'P4', thus killing 'P4'. 'P5' knows that if he votes against 'P4', next round he will receive all the gold. P4 = 0gold P5 = 100gold
If 3 Pirates were left. 'P3' knows that if it gets down to 2 pirates, 'P4' is a dead man. Therefore he knows 'P4' will vote for him not matter what therefore he can keep all the gold himself. And stay alive with a vote from 'P4'
P3= 100 gold P4=0 gold P5= 0 gold
With 4 pirates left. 'P4' needs 2 votes to stay alive. By giving 'P2' and 'P1' 1 gold each, they are guaranteed more than if it goes another round.
P2=98 P3=0 P4=1 P5=1
With 5 pirates left. 'P5' (you) needs 2 more votes. Giving 'P4' 2 gold and 'P3' 1 gold will guarantee your survival.
P1=97 P2=0 P3=1 P4=2 P5=0
Or
P1=97 p2=0 p3=1 p4=0 p5=2
Therefore you can safely keep 97 gold.
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Rope Burning Riddle
Played 3,195 times
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June 3rd, 2010
VN:RO [1.9.1_1087] Description You are given two ropes and a lighter.
Both ropes take exactly one hour to burn. Some parts of the ropes are thicker
than other parts of the rope, therefore not all parts of the rope will burn
at the same speed. Consequently burning half the rope will not necessarily
take 30 minutes.
Goal: Find a way to Measure 45 minutes of time.
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hint
All the rope will be burnt in 45 minutes
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answer
Light both ends of rope 'A'.
At the same time light one end of rope 'B'. Once rope A has burnt out light the remaining end of rope 'B'.
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Three Women
Played 533 times
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June 3rd, 2010
VN:RO [1.9.1_1087] Jane is looking at Martha and Martha is looking at Elizabeth. Jane is married and Elizabeth is not married.
Is a married person looking a non-married person?
- Yes
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No
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Cannot tell with the information given
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hint
Go over the possibilities of Martha being married.
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answer
Yes a married person is looking at a non-married person.
Martha is either married or not married.
If Martha is not married then Jane(married) is looking at Martha(not married. If Martha is married then Martha(married) is looking at Jane(not Married).
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