#1 - Probability Riddle Loaded Revolver

Henry has been caught stealing cattle, and is brought into town for justice. The judge is his ex-wife Gretchen, who wants to show him some sympathy, but the law clearly calls for two shots to be taken at Henry from close range. To make things a little better for Henry, Gretchen 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 Henry is still alive, she will then either take another shot, or spin the chamber again before shooting.

Henry 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 Gretchen loads the chambers, spins the revolver, and pulls the trigger. Whew! It was blank. Then Gretchen 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 Henry choose?

Probability Riddle Loaded Revolver

Henry should have Gretchen pull the trigger again without spinning.

We know that the first chamber Gretchen 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 Henry has Gretchen pull the trigger again, the probability that a bullet will be fired is 1/4.

If Gretchen spins the chamber again, the probability that she shoots Henry 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.

#2 - Alexander Puzzle

Alexander is stranded on an island covered in forest.

One day, when the wind is blowing from the west, lightning strikes the west end of the island and sets fire to the forest. The fire is very violent, burning everything in its path, and without intervention the fire will burn the whole island, killing the man in the process.

There are cliffs around the island, so he cannot jump off.

How can the Alexander survive the fire? (There are no buckets or any other means to put out the fire)

Alexander Puzzle

Alexander picks up a piece of wood and lights it from the fire on the west end of the island.

He then quickly carries it near the east end of he island and starts a new fire. The wind will cause that fire to burn out the eastern end and he can then shelter in the burnt area.

Alexander survives the fire, but dies of starvation, with all the food in the forest burnt....lolzzz

#3 - 2 Eggs 100 Floors Puzzle

-> You are given 2 eggs.
-> You have access to a 100-storey building.
-> Eggs can be very hard or very fragile means it may break if dropped from the first floor or may not even break if dropped from 100 th floor.Both eggs are identical.
-> You need to figure out the highest floor of a 100-storey building an egg can be dropped without breaking.
-> Now the question is how many drops you need to make. You are allowed to break 2 eggs in the process

2 Eggs 100 Floors Puzzle

answer is: 14

Let x be the answer we want, the number of drops required.

So if the first egg breaks maximum we can have x-1 drops and so we must always put the first egg from height x. So we have determined that for a given x we must drop the first ball from x height. And now if the first drop of the first egg doesn’t breaks we can have x-2 drops for the second egg if the first egg breaks in the second drop.

Taking an example, lets say 16 is my answer. That I need 16 drops to find out the answer. Lets see whether we can find out the height in 16 drops. First we drop from height 16,and if it breaks we try all floors from 1 to 15.If the egg don’t break then we have left 15 drops, so we will drop it from 16+15+1 =32nd floor. The reason being if it breaks at 32nd floor we can try all the floors from 17 to 31 in 14 drops (total of 16 drops). Now if it did not break then we have left 13 drops. and we can figure out whether we can find out whether we can figure out the floor in 16 drops.

Lets take the case with 16 as the answer

1 + 15 16 if breaks at 16 checks from 1 to 15 in 15 drops
1 + 14 31 if breaks at 31 checks from 17 to 30 in 14 drops
1 + 13 45 .....
1 + 12 58
1 + 11 70
1 + 10 81
1 + 9 91
1 + 8 100 We can easily do in the end as we have enough drops to accomplish the task


Now finding out the optimal one we can see that we could have done it in either 15 or 14 drops only but how can we find the optimal one. From the above table we can see that the optimal one will be needing 0 linear trials in the last step.

So we could write it as

(1+p) + (1+(p-1))+ (1+(p-2)) + .........+ (1+0) >= 100.

Let 1+p=q which is the answer we are looking for

q (q+1)/2 >=100

Solving for 100 you get q=14.
So the answer is: 14
Drop first orb from floors 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100... (i.e. move up 14 then 13, then 12 floors, etc) until it breaks (or doesn't at 100)

#4 - Popular Age Problem

Two old friends, Jack and Bill, meet after a long time.

Three kids
Jack: Hey, how are you, man?
Bill: Not bad, got married and I have three kids now.
Jack: That's awesome. How old are they?
Bill: The product of their ages is 72 and the sum of their ages is the same as your birth date.
Jack: Cool..But I still don't know.
Bill: My eldest kid just started taking piano lessons.
Jack: Oh, now I get it.

How old are Bill's kids?

Popular Age Problem

3,3,8

Lets break it down. The product of their ages is 72. So what are the possible choices?

2, 2, 18 sum(2, 2, 18) = 22
2, 4, 9 sum(2, 4, 9) = 15
2, 6, 6 sum(2, 6, 6) = 14
2, 3, 12 sum(2, 3, 12) = 17
3, 4, 6 sum(3, 4, 6) = 13
3, 3, 8 sum(3, 3, 8 ) = 14
1, 8, 9 sum(1,8,9) = 18
1, 3, 24 sum(1, 3, 24) = 28
1, 4, 18 sum(1, 4, 18) = 23
1, 2, 36 sum(1, 2, 36) = 39
1, 6, 12 sum(1, 6, 12) = 19

The sum of their ages is the same as your birth date. That could be anything from 1 to 31 but the fact that Jack was unable to find out the ages, it means there are two or more combinations with the same sum. From the choices above, only two of them are possible now.

2, 6, 6 sum(2, 6, 6) = 14
3, 3, 8 sum(3, 3, 8 ) = 14

Since the eldest kid is taking piano lessons, we can eliminate combination 1 since there are two eldest ones. The answer is 3, 3 and 8.

#5 - Brain Twister Puzzle

2+3=8,
3+7=27,
4+5=32,
5+8=60,
6+7=72,
7+8=??

Solve it?

98

2+3=2*[3+(2-1)]=8
3+7=3*[7+(3-1)]=27
4+5=4*[5+(4-1)]=32
5+8=5*[8+(5-1)]=60
6+7=6*[7+(6-1)]=72
therefore
7+8=7*[8+(7-1)]=98
x+y=x[y+(x-1)]=x^2+xy-x

#6 - Pyramid Number Puzzle

The below is a number puzzle. It should be read left to right, top to bottom.
1
1 1
2 1
1 2 1 1
1 1 1 2 2 1
? ? ? ? ? ?
? ? ? ? ? ? ? ?
Question 1: What is the next two rows of numbers?
Question 2: How was this reached?

Popular Number Sequence Puzzle
3 1 2 2 1 1
1 3 1 1 2 2 2 1

Line 1 is 'Two ones' (2 1)
Line 2 then becomes 'One two, and one one' (1 2 1 1)
Line 3 therefore is 'One one, one two and two ones' (1 1 1 2 2 1)
Line 4 is 'Three ones, two twos and one one' (3 1 2 2 1 1)
Line 5 is 'One three, one one, two twos and two ones' (1 3 1 1 2 2 1 1)

#7 - Challenging Puzzle

Outside a room there are three light switches. One of switch is connected to a light bulb inside the room.
Each of the three switches can be either 'ON' or 'OFF'.

You are allowed to set each switch the way you want it and then enter the room(note: you can enter the room only once)

Your task is to then determine which switch controls the bulb ??

Challenging Puzzle

Set the first switches on for abt 10min, and then switch on the second switch and then enter the room.
Three cases are possible
1.Bulb is on => second switch is the ans
2.Bulb is off and on touching bulb , you will find bulb to be warm
=>1st switch is the ans.
3.Bulb is off and on touching second bulb , you will find bulb to be normal(not warm)
=>3rd bulb is the ans.

#8 - Cipher Puzzle

What does this message say?

G T Y O R J O T E O U I A B G T


Hint
Count the letters and try splitting the letters up into groups.

'Great Job You Got It'

This type of code is known as a Caesar Box (Julius Caesar was the first to write codes this way.) To decipher the message, simply divide the code into four groups of four (you can also divide them into groups such as 5 groups of 5 or 6 groups of 6 depending on the number of letters in the phrase), and rearrange them vertically like this...
G T Y O
R J O T
E O U I
A B G T

Then you read vertically column by column.

#9 - Hard Math Riddle

Take 9 from 6, 10 from 9, 50 from 40 and leave 6.

How Come ??

SIX - 9 (IX) = S
9 (IX) - 10 (X) = I
40 (XL) - 50 (L) = X
===>SIX

#10 - Hardest Balance Logic Puzzle

You are given a set of scales and 12 marbles. The scales are of the old balance variety. That is, a small dish hangs from each end of a rod that is balanced in the middle. The device enables you to conclude either that the contents of the dishes weigh the same or that the dish that falls lower has heavier contents than the other.
The 12 marbles appear to be identical. In fact, 11 of them are identical, and one is of a different weight. Your task is to identify the unusual marble and discard it. You are allowed to use the scales three times if you wish, but no more.

Note that the unusual marble may be heavier or lighter than the others. You are asked to both identify it and determine whether it is heavy or light.

So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8.

There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second one possibility is to weigh 9,10,11 against 1,2,3

(1) They balance, in which case you know 12 is the different coin, and you just weigh it against any other to determine whether it is heavy or light.
(2) 9,10,11 is heavy. In this case, you know that the different coin is 9, 10, or 11, and that that coin is heavy. Simply weigh 9 against 10; if they balance, 11 is the heavy coin. If not, the heavier one is the heavy coin.
(3) 9,10,11 is light. Proceed as in the step above, but the coin you're looking for is the light one.

That was the easy part.

What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.

Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. The direction of the tilts can tell us whwther the offending coin is heavier or lighter.

Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin.

For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.