logicprobabilityYour enemy challenges you to play Russian Roulette with a 6-cylinder pistol (meaning it has room for 6 bullets). He puts 2 bullets into the gun in consecutive slots, and leaves the next four slots blank. He spins the barrel and hands you the gun. You point the gun at yourself and pull the trigger. It doesn't go off. Your enemy tells you that you need to pull the trigger one more time, and that you can choose to either spin the barrel at random, or not, before pulling the trigger again. Spinning the barrel will position the barrel in a random position.
Assuming you'd like to live, should you spin the barrel or not before pulling the trigger again?

You are better off shooting again without spinning the barrel.
Given that the gun didn't fire the first time, it was pointing to one of the four empty slots. Because your enemy spun the cylinder randomly, it would have been pointing to any of these empty slots with equal probability. Three of these slots would not fire again after an additional trigger-pull, and one of them would. Thus, by not spinning the barrel, there is a 1/4 chance that pulling the trigger again would fire the gun.
Alternatively, if you spin the barrel, it will point to each of the 6 slots with equal probability. Because 2 of these 6 slots have bullets in them, there would be a 2/6 = 1/3 chance that the gun would fire after spinning the barrel.
Thus, you are better off not spinning the barrel.

logicprobabilityYour friend shows you two jars, one with 100 red marbles in it, the other with 100 blue marbles in it.
He proposes a game. He'll put the two jars behind his back and tell you to pick one of them at random. You'll then close your eyes, he'll hand you the jar you picked, and you'll pick a random marble from that jar.
You win if the marble you pick is blue, and you lose otherwise.
To give you the best shot at winning, your friend gives you the two jars before the game starts and says you can move the marbles around however you'd like, as long as all 200 marbles are in the 2 jars (that is, you can't throw any marbles away).
How should you move the marbles around to give yourself the best chance of picking a blue marble?

Put one blue marble in one jar, and put the rest of the marbles in the other jar. This will give you just about a 75% chance of picking a blue marble.

cleanlogicThree people check into a hotel. They pay $30 to the manager and go to their room. The manager finds out that the room rate is $25 and gives $5 to the bellboy to return. On the way to the room, the bellboy reasons that $5 would be difficult to share among three people, so he pockets $2 and gives $1 to each person. Now, each person paid $10 and got back $1. So they paid $9 each, totalling $27. The bellboy has $2, totalling $29. )
Where is the remaining dollar?

Each person paid $9, totalling $27. The manager has $25 and the bellboy has $2. The bellboy's $2 should be added to the manager's $25 or substracted from the tenant's $27, not added to the tenants' $27.

logicHandel has been killed and Beethoven is on the case. He has interviewed the four suspects and their statements are shown below. Each suspect has said two sentences. One sentence of each suspect is a lie and one sentence is the truth. Help Beethoven figure out who the killer is.
Joplin: I did not kill Handel. Either Grieg is the killer or none of us is.
Grieg: I did not kill Handel. Gershwin is the killer.
Strauss: I did not kill Handel. Grieg is lying when he says Gershwin is the killer.
Gershwin: I did not kill Handel. If Joplin did not kill him, then Grieg did.
Who is the killer?

Strauss is the one who killed Handel. You need to take turns assuming someone is the killer; that means everyone's second sentence is a lie. If Joplin was the killer, Grieg's lie mixed with Strauss' counteracts the other. If Grieg was the killer, Gershwin would need to be a killer too. If Gershwin was the killer, Gershwin would need to be a killer too. If Gershwin was the killer, Grieg and Strauss counter each other again, but with Strauss, everything would fit in.

logicYou have just purchased a small company called Company X. Company X has N employees, and everyone is either an engineer or a manager. You know for sure that there are more engineers than managers at the company.
Everyone at Company X knows everyone else's position, and you are able to ask any employee about the position of any other employee. For example, you could approach employee A and ask "Is employee B an engineer or a manager?" You can only direct your question to one employee at a time, and can only ask about one other employee at a time. You're allowed to ask the same employee multiple questions if you want.
Your goal is to find at least one engineer to solve a huge problem that has just hit the company's factory. The problem is so urgent that you only have time to ask N-1 total questions.
The major problem with questioning the employees, however, is that while the engineers will always tell you the truth about other employees' roles, the managers may lie to you if they like. You can assume that the managers will do their best to confuse you.
How can you find at least one engineer by asking at most N-1 questions?

You can find at least one engineer using the following process:
Put all of the employees in a conference room. If there happen to be an even number of employees, pick one at random and send him home for the day so that we start with an odd number of employees. Note that there will still be more engineers than managers after we send this employee home.
Then call them out one at a time in any order. You will be forming them into a line as follows:
If there is nobody currently in the line, put the employee you just called out in the line.
Otherwise, if there is anybody in the line, then we do the following. Let's call the employee currently at the front of the line Employee_Front, and call the employee who we just called out of the conference room Employee_Next.
So ask Employee_Front if Employee_Next is a manager or an engineer.
If Employee_Front says "manager", then send both Employee_Front and Employee_Next home for the day.
However, if Employee_Front says "engineer", then put Employee_Next at the front of the line.
Keep doing this until you've called everyone out of the conference room. Notice that at this point, you'll have asked N-1 or less questions (you asked at most one question each time you called an employee out except for the first employee, when you didn't ask a question, so that's at most N-1 questions).
When you're done calling everyone out of the conference room, the person at the front of the line is an engineer. So you've found your engineer!
But the real question: how does this work?
We can prove this works by showing a few things.
First, let's show that if there are any engineers in the line, then they must be in front of any managers.
We'll show this with a proof by contradiction. Assume that there is a manager in front of an engineer somewhere in the line. Then it must have been the case that at some point, that engineer was Employee_Front and that manager was Employee_Next. But then Employee_Front would have said "manager" (since he is an engineer and always tells the truth), and we would have sent them both home. This contradicts their being in the line at all, and thus we know that there can never be a manager in front of an engineer in the line.
So now we know that after the process is done, if there are any engineers in the line, then they will be at the front of the line. That means that all we have to prove now is that there will be at least one engineer in the line at the end of the process, and we'll know that there will be an engineer at the front.
So let's show that there will be at least one engineer in the line. To see why, consider what happens when we ask Employee_Front about Employee_Next, and Employee_Front says "manager". We know for sure that in this case, Employee_Front and Employee_Next are not both engineers, because if this were the case, then Employee_Front would have definitely says "engineer". Put another way, at least one of Employee_Front and Employee_Next is a manager. So by sending them both home, we know we are sending home at least one manager, and thus, we are keeping the balance in the remaining employees that there are more engineers than managers.
Thus, once the process is over, there will be more engineers than managers in the line (this is also sufficient to show that there will be at least one person in the line once the process is over). And so, there must be at least one engineer in the line.
Put altogether, we proved that at the end of the process, there will be at least one engineer in the line and that any engineers in the line must be in front of any managers, and so we know that the person at the front of the line will be an engineer.

logicIt was a Pink Island. There were 201 individuals lived in the island. Among them 100 people were blue eyed people, 100 were green eyed people and the leader was a black eyed one.
Except the leader, nobody knew how many individuals lived in the island. Neither have they known about the color of the eyes. The leader was a very strict person. Those people can never communicate with others. They even cannot make gestures to communicate. They can only talk and communicate with the leader. It was a prison for those 200 individuals.
However, the leader provided an opportunity to leave the island forever but on one condition. Every morning he questions the individuals about the color of the eyes! If any of the individuals say the right color, he would be released. Since they were unaware about the color of the eyes, all 200 individuals remained silent. When they say wrong color, they were eaten alive to death. Afraid of punishment, they remained silent.
One day, the leader announced that "at least 1 of you has green eyes! If you say you are the one, come and say, I will let you go if you are correct! But only one of you can come and tell me!"
How many green eyed individuals leave the island and in how many days?

All 100 green eyed individuals will leave in 100 days.
Consider, there is only one green eyed individual lived in the island. He will look at all the remaining individuals who have blue eyes. So, he can get assured that he has green eyes! If there were more than one green eyed people, when the first man looks at the second one with green eyes, the person didn’t leave on day the first day. It means he also has green eyes and the same rule applies to each green eyed man.

logicshortHow far can you run into the woods?

Half way. If you go in any further, you’d be running OUT of the woods.

logicmathA women walks into a bank to cash out her check.
By mistake the bank teller gives her rupee amount in change, and her paise amount in rupees.
On the way home she spends 5 paise, and then suddenly she notices that she has twice the amount of her check.
How much was her check amount ?

The check was for Rupees 31.63.
The bank teller gave her Rupees 63.31
She spent .05, and then she had Rupees 63.26, which is twice the check.
Let x be the rupees of the check, and y be the paise.
The check was for 100x + y paise
He was given 100y + x paise
Also
100y + x - 5 = 2(100x + y)
Expanding this out and rearranging, we find:
98y = 199x + 5
or 199x â‰¡ -5 (mod 98)
or 98*2*x + 3x â‰¡ -5 (mod 98)
3x â‰¡ -5 â‰¡ 93 (mod 98)
this quickly leads to x = 31

logicFrank and some of the boys were exchanging old war stories. James offered one about how his grandfather (Captain Smith) led a battalion against a German division during World War I. Through brilliant maneuvers he defeated them and captured valuable territory. Within a few months after the battle he was presented with a sword bearing the inscription: "To Captain Smith for Bravery, Daring and Leadership, World War One, from the Men of Battalion 8." Frank looked at James and said, "You really don't expect anyone to believe that yarn, do you?" 7
What is wrong with the story?

It wasn't valled World War One until much later. It was called the Great War at first, because they did not know during that war and immediately afterward that there would be a second World War (WW II).

logicJack and Joe were on vacation and driving along a deserted country road from the town of Kaysville to the town of Lynnsville. They came to a multiple fork in the road. The sign post had been knocked down and they were faced with choosing one of five different directions. Since they had left their map at the last gas station and there was no one around to ask, how could Jack and Joe find their way to Lynnsville?

They need to stand the signpost up so that the arm reading Kaysville points in the direction of Kaysville, the town they had just come from. With one arm pointing the correct way, the other arms will also point in the right directions.