Hussey has been caught stealing goats, and is brought into court for justice. The judge is his ex-wife Amy Hussey, who wants to show him some sympathy, but the law clearly calls for two shots to be taken at Hussey from close range.
To make things a little better for Hussey, Amy Hussey tells him she will place two bullets into a six-chambered revolver in successive order. She will spin the chamber, close it, and take one shot.
If Hussey is still alive, she will then either take another shot, or spin the chamber again before shooting. Hussey is a bit incredulous that his own ex-wife would carry out the punishment, and a bit sad that she was always such a rule follower.
He steels himself as Amy Hussey loads the chambers, spins the revolver, and pulls the trigger. Whew! It was blank. Then Amy Hussey asks, 'Do you want me to pull the trigger again, or should I spin the chamber a second time before pulling the trigger?'
What should Hussey choose?
Hussey should have Amy Hussey pull the trigger again without spinning.
We know that the first chamber Amy Hussey fired was one of the four empty chambers. Since the bullets were placed in consecutive order, one of the empty chambers is followed by a bullet, and the other three empty chambers are followed by another empty chamber. So if Hussey has Amy Hussey pull the trigger again, the probability that a bullet will be fired is 1/4.
If Amy Hussey spins the chamber again, the probability that she shoots Hussey would be 2/6, or 1/3, since there are two possible bullets that would be in firing position out of the six possible chambers that would be in position.
There are five vowels in English language (a, e, i, o, u, and sometimes y).
Easy version: Can you tell us a word contains of these 5 vowels?
Hard: Can you tell us a word contains of all these 6 vowels with y?
Very difficult: Can you tell us a word contains of all these vowels with y in their alphabetical order?
Unquestionably
In their alphabetical order:
Facetiously
Abstemiously
Adventitiously
There is a barrel with no lid and some wine in it.
"This barrel of wine is more than half full," said Curly.
Moe says, "No it's not. It's less than half full."
Without any measuring implements and without removing any wine from the barrel, how can they easily determine who is correct?
Tilt the barrel until the wine barely touches the lip of the barrel. If the bottom of the barrel is visible then it is less than half full. If the barrel bottom is still completely covered by the wine, then it is more than half full.
A wise man lived on a hill above a small town. The townspeople often approached him to solve their difficult problems and riddles. One day, two lads decided to fool him. They took a dove and set off up the hill. Standing before him, one of the lads said "Tell me, wise man, is the dove I hold behind my back dead or alive?" The man smiled and said "I cannot answer your question correctly". Even though the wise man knew the condition of the dove, why wouldn't he state whether it was dead or alive?
The man told the two lads, "If I say the dove is alive, you will the bird and show me that it is dead. If I say that it is dead, you will release the dove and it will fly away. So you see I cannot answer your question.
Search: Schrödinger's cat
Create a number using only the digits 4,4,3,3,2,2,1 and 1.
So I can only be eight digits.
You have to make sure the ones are separated by one digit, the twos are separated by two digits the threes are separated with three digits and the fours are separated by four digits.
Six jugs are in a row. The first three are filled with coke, and the last three are empty. By moving only one glass, can you arrange them so that the full and the empty glasses alternate?
Move and then pour all coke from second glass to fifth glass.
A boy goes and buys a fishing pole that is 6' 3" long. As he goes to get on the bus, the driver stops him. The driver tells him that he can't take anything longer than 6' onto the bus. The boy goes back into town, purchases one more thing, and the driver allows the boy on the bus. What did the boy buy, and what did he do with it?
The boy bought 6' long box. He put the fishing pole in diagonally and the entire package was only 6'!
One day, Emperor Akbar posed a question to Birbal. He asked him what Birbal would choose if he offered either justice or a gold coin.
"The gold coin," said Birbal without hesitation.
On hearing this, Akbar was taken aback. "You would prefer a gold coin to justice?" he asked, not believing his own ears.
"Yes," said Birbal.
The other courtiers were amazed by Birbal's display of idiocy. They were full of glee that Birbal had finally managed himself to do what these courtiers had not been able to do for a long time - discredit Birbal in the emperor's eyes!
"I would have been disappointed if this was the choice made even by my lowliest of servants," continued the emperor. "But coming from you it's not only disappointing, but shocking and sad. I did not know you were so debased!"
How did Birbal justify his answer to the enraged and hurt Emperor?
"One asks for what one does not have, Your Majesty." said Birbal, smiling gently and in quiet tones.
"Under Your Majesty´s rule, justice is available to everybody. But I am a spendthrift and always short of money and therefore I said I would choose the gold coin."
The answer immensely pleased the emperor and respect for Birbal was once again restored in the emperor's eyes.
You have twelve balls, identical in every way except that one of them weighs slightly less or more than the balls.
You have a balance scale, and are allowed to do 3 weighings to determine which ball has the different weight, and whether the ball weighs more or less than the other balls.
What process would you use to weigh the balls in order to figure out which ball weighs a different amount, and whether it weighs more or less than the other balls?
Take eight balls, and put four on one side of the scale, and four on the other.
If the scale is balanced, that means the odd ball out is in the other 4 balls.
Let's call these 4 balls O1, O2, O3, and O4.
Take O1, O2, and O3 and put them on one side of the scale, and take 3 balls from the 8 "normal" balls that you originally weighed, and put them on the other side of the scale.
If the O1, O2, and O3 balls are heavier, that means the odd ball out is among these, and is heavier. Weigh O1 and O2 against each other. If one of them is heavier than the other, this is the odd ball out, and it is heavier. Otherwise, O3 is the odd ball out, and it is heavier.
If the O1, O2, and O3 balls are lighter, that means the odd ball out is among these, and is lighter. Weigh O1 and O2 against each other. If one of them is lighter than the other, this is the odd ball out, and it is lighter. Otherwise, O3 is the odd ball out, and it is lighter.
If these two sets of 3 balls weigh the same amount, then O4 is the odd ball out. Weight it against one of the "normal" balls from the first weighing. If O4 is heavier, then it is heavier, if it's lighter, then it's lighter.
If the scale isn't balanced, then the odd ball out is among these 8 balls.
Let's call the four balls on the side of the scale that was heavier H1, H2, H3, and H4 ("H" for "maybe heavier").
Let's call the four balls on the side of the scale that was lighter L1, L2, L3, and L4 ("L" for "maybe lighter").
Let's also call each ball from the 4 in the original weighing that we know aren't the odd balls out "Normal" balls.
So now weigh [H1, H2, L1] against [H3, L2, Normal].
-If the [H1, H2, L1] side is heavier (and thus the [H3, L2, Normal] side is lighter), then this means that either H1 or H2 is the odd ball out and is heavier, or L2 is the odd ball out and is lighter.
-So measure [H1, L2] against 2 of the "Normal" balls.
-If [H1, L2] are heavier, then H1 is the odd ball out, and is heavier.
-If [H1, L2] are lighter, then L2 is the odd ball out, and is lighter.
-If the scale is balanced, then H2 is the odd ball out, and is heavier.
-If the [H1, H2, L1] side is lighter (and thus the [H3, L2, Normal] side is heavier), then this means that either L1 is the odd ball out, and is lighter, or H3 is the odd ball out, and is heavier.
-So measure L1 and H3 against two "normal" balls.
-If the [L1, H3] side is lighter, then L1 is the odd ball out, and is lighter.
-Otherwise, if the [L1, H3] side is heavier, then H3 is the odd ball out, and is heavier.
If the [H1, H2, L1] side and the [H3, L2, Normal] side weigh the same, then we know that either H4 is the odd ball out, and is heavier, or one of L3 or L4 is the odd ball out, and is lighter.
So weight [H4, L3] against two of the "Normal" balls.
If the [H4, L3] side is heavier, then H4 is the odd ball out, and is heavier.
If the [H4, L3] side is lighter, then L3 is the odd ball out, and is lighter.
If the [H4, L3] side weighs the same as the [Normal, Normal] side, then L4 is the odd ball out, and is lighter.
