A Box of Defective Balls

Question: You have 10 boxes of balls (each ball weighing exactly10 gm) with one box with defective balls (each one of the defective balls weigh 9 gm). You are given an electronic weighing machine and only one chance at it. How will find out which box has the defective balls?

Answer: For convenience sake, let’s name the boxes from 1 to 10. In order to solve this problem, you have to leverage the fact that you know exactly what each good ball is supposed to weigh and what each defective ball is supposed to weigh. Many of us instinctively will take one ball out of each box and try to find a way to make it work but the trick to take different number of balls from each box.

The number of balls you pick from each bag is equal to the box number. For example, pick 1 ball from box 1, 2 balls from box 2 and so on. In total you will have 55 balls. If all of the boxes have good balls, then the total weight of these balls would be 550gm.

If box 1 has defective balls, then the total weight should be 1gm less than expected (only one ball weighing 9 gm). If box 2 has defective balls, then the total weight should be 2gm less than expected (two balls weighing 9 gm). So once you weigh the set of chosen balls, find out the difference between the total weight and the expected weight. That number represents the box number which contains the defective balls.

5 Pirates Fight for 100 Gold Coins

Question: Five pirates discover a chest containing 100 gold coins. They decide to sit down and devise a distribution strategy. The pirates are ranked based on their experience (Pirate 1 to Pirate 5, where Pirate 5 is the most experienced). The most experienced pirate gets to propose a plan and then all the pirates vote on it. If at least half of the pirates agree on the plan, the gold is split according to the proposal. If not, the most experienced pirate is thrown off the ship and this process continues with the remaining pirates until a proposal is accepted. The first priority of the pirates is to stay alive and second to maximize the gold they get. Pirate 5 devises a plan which he knows will be accepted for sure and will maximize his gold. What is his plan?

Answer: To understand the answer, we need to reduce this problem to only 2 pirates. So what happens if there are only 2 pirates. Pirate 2 can easily propose that he gets all the 100 gold coins. Since he constitutes 50% of the pirates, the proposal has to be accepted leaving Pirate 1 with nothing.

Now let’s look at 3 pirates situation, Pirate 3 knows that if his proposal does not get accepted, then pirate 2 will get all the gold and pirate 1 will get nothing. So he decides to bribe pirate 1 with one gold coin. Pirate 1 knows that one gold coin is better than nothing so he has to back pirate 3. Pirate 3 proposes {pirate 1, pirate 2, pirate 3} {1, 0, 99}. Since pirate 1 and 3 will vote for it, it will be accepted.

If there are 4 pirates, pirate 4 needs to get one more pirate to vote for his proposal. Pirate 4 realizes that if he dies, pirate 2 will get nothing (according to the proposal with 3 pirates) so he can easily bribe pirate 2 with one gold coin to get his vote. So the distribution will be {0, 1, 0, 99}.

Smart right? Now can you figure out the distribution with 5 pirates? Let’s see. Pirate 5 needs 2 votes and he knows that if he dies, pirate 1 and 3 will get nothing. He can easily bribe pirates 1 and 3 with one gold coin each to get their vote. In the end, he proposes {1, 0, 1, 0, 98}. This proposal will get accepted and provide the maximum amount of gold to pirate 5.

Boxes of Money

Question: You are given b boxes and n dollar bills. The money has to be sealed in the b boxes in a way such that without thereafter opening a box, you can give someone a requested whole amount of dollars from 0 to n. How should b be related to n for this to happen?  Answer: Stumped? Let’s think of an example to approach this problem.  Say we have $100. A good approach to distributing $100 would be the binary number system. So you’d have $1, $2, $4, $8, $16, $32 in the first six boxes. We can’t fill the next box with $64 dollars because we are only left with $37 dollars (from a total of $100). So we’d have to put $37 in the seventh box. To supply any requested amount, we’d have to use a combination of these boxes.  To find out the restrictions on the values of b and n, we have to think of different scenarios. For instance, with a million dollars and just one box, we would never be able to dispense any requested amount less than a million. However, if we are ever in a situation with more boxes than dollars, there is a never a problem.  Using this approach, we can create a table showing the best relationship between b and n  b = 1     n = up to $1 b = 2     n = up to $2 + $1 = $3 b = 3     n = up to $4 + $2 + $1 = $7 b = 4     n = up to $8 + $4 + $2 + $1 = $15

Is Your Husband a Cheat?

Question: A certain town comprises of 100 married couples. Everyone in the town lives by the following rule: If a husband cheats on his wife, the husband is executed as soon as his wife finds out about him. All the women in the town only gossip about the husbands of other women. No woman ever tells another woman if her husband is cheating on her.  So every woman in the town knows about all the cheating husbands in the town except her own. It can also be assumed that a husband remains silent about his infidelity. One day, the mayor of the town announces to the whole town that there is at least 1 cheating husband in the town. What do you think happens?

Answer: Stumped? Let’s solve this methodically. Say there was only 1 cheating husband in the town. There will be 99 women who know exactly who the cheater is. The 1 remaining woman, who is being cheated on, would have assumed there are no cheaters. But now that the mayor has confirmed that there is at least one cheater, she realizes that her own husband must be cheating on her. So her husband gets executed on the day of the announcement.

Now let’s assume there are 2 cheaters in the town. There will be 98 women in the town who know who the 2 cheaters are. The 2 wives, who are being cheated on, would think that there is only 1 cheater in the town.  Since neither of these 2 women know that their husbands are cheaters, they both do not report their husbands in on the day of the announcement. The next day, when the 2 women see that no husband was executed, they realize that there could only be one explanation – both their husbands are cheaters. Thus, on the second day, 2 husbands are executed.

Through induction, it can be proved that when this logic is applied to n cheating husbands, they all die on the n th day after the mayor’s announcement.

Chameleons

Question:On an island live 13 purple, 15 yellow and 17 maroon chameleons. When two chameleons of different colors meet, they both change into the third color. Is there a sequence of pairwise meetings after which all chameleons have the same color?

Solution: Let <p, y, m> denote a population of p purple, y yellow and m maroon chameleons. Can population <13, 15, 17> can be transformed into <45, 0, 0> or <0, 45, 0> or <0, 0, 45> through a series of pairwise meetings? Define function X(p, y, m) = (0p + 1y + 2m) mod 3. An interesting property of X is that its value does not change after any pairwise meeting because X(p, y, m) = X(p-1, y-1, m+2) = X(p-1, y+2, m-1) = X(p+2, y-1, m-1). Now X(13, 15, 17) equals 1. However, X(45, 0, 0) = X(0, 45, 0) = X(0, 0, 45) = 0. This means that there is no sequence of pairwise meetings after which all chameleons will have identical color.

blind bartender’s problem

Question: 

Four glasses are placed on the corners of a square table. Some of the glasses are upright (up) and some upside-down (down). You have to arrange the glasses so that they are all up or all down (while keeping your eyes closed all the time). The glasses may be re-arranged in turns subject to the following rules.

1. Any two glasses may be inspected in one turn and after feeling their orientation you may reverse the orientation of either, neither or both glasses.
2. After each turn table is rotated through a random angle.
3. At any point of time if all four glasses are of the same orientation a ring will bell

You have to come up with a solution to ensure that all glasses have the same orientation (either up or down) in a finite number of turns. The algorithm must be non-stochastic i.e. it must not depend on luck.

Answer:

1. On the first turn choose a diagonally opposite pair of glasses and turn both glasses up.
2. On the second turn choose two adjacent glasses. At least one will be up as a result of the previous step. If the other is down, turn it up as well. If the bell does not ring then there are now three glasses up and one down(3U and 1D).
3. On the third turn choose a diagonally opposite pair of glasses. If one is down, turn it up and the bell will ring. If both are up, turn one down. There are now two glasses down, and they must be adjacent.
4. On the fourth turn choose two adjacent glasses and reverse both. If both were in the same orientation then the bell will ring. Otherwise there are now two glasses down and they must be diagonally opposite.
5. On the fifth turn choose a diagonally opposite pair of glasses and reverse both. The bell will ring for sure.

Days of month using 2 dice

Question: How can you represent days of month using two 6 sided dice ?

You can write one number on each face of the dice from 0 to 9 and you have to represent days from 1 to 31, for example for 1, one dice should show 0 and another should show 1, similarly for 31 one dice should show 3 and another should show 1.

Answer is:
Dice 1: 0 1 2 3 5 7
Dice 2: 0 1 2 4 6 8 

How?
Basically you have to show 11, 22 so 1 and 2 should be present in both dices, similarly to show 01, 09 0 should be present in both dices, now the trick is for showing 9 you can use dice with 6 printed on one of the face.

How many people in party?

Question: At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?

Answer:

lets say there are n persons

first person shakes hand with everyone else: n-1 times(n-1 persons)
second person shakes hand with everyone else(not with 1st as its already done): n-2 times
3rd person shakes hands with remaining persons: n-3

So total handshakes will be = (n-1) + (n-2) + (n-3) +…… 0;
= (n-1)*(n-1+1)/2 = (n-1)*n/2 = 66
= n^2 -n = 132
=(n-12)(n+11) = 0;
= n = 12 OR n =-11
-11 is ruled out so the answer is 12 persons.

Divide a triangle into n triangles

Question: How to divide a triangle into n equal area triangles ?

Answer:

For any triangle its area can be calculated using the formula base * height/2. So we can divide a triangle into n triangles of equal area by simply dividing the base into n equal parts.

Explanations :
For example to divide the triangle in the image below into 4 equal triangles, we can simply divide its base of 8cms to 4 equal parts of 2cms each and form the 4 triangles of equal area.

Heavier ball

Question:You have 8 balls all look identical (in shape, color etc.). All of them have same weight except one defective ball which is heavier than others. You can use a two sided balance system (not the electronic one). Find the minimum no. of measures required to separate the defective ball and the way you separate it.

If you have solved it then try it for N no. of identical balls with one defective ball.

Solution:

The minimum no. of measures required are 2. Now can you solve it… ? Give it a try…

Here is the solution which tells you that how can you do it in 2 measurements.

Divide 8 balls into groups of 3, 3 and 2.

First weigh :
Weigh the two groups of 3 ball, now there are two possibilities
a) They are balanced (all 6 balls are of equal weight)
b) One side is heavier then the other.

case (a) :
The group of 2 has the defective ball. Weigh them with one on each side of balancing machine. The side which has more weigh has the heavier/defective ball.

case (b) :
We got the three balls of the side which is heavier. Now take any two balls and weigh them, then again there are two cases.
(i) They both are of equal weight.
(ii) One is heavier than the other.

for case (b)(i) :
The third ball is heavier/defective than all other balls

for case (b)(ii) :
The heavier ball is defective.
——————————————–

Solution for N no. of balls.

If there is 1 ball, No measurement is required.
If there are 2-3 balls, we need 1 measurement.[(3^0 +1) to 3^1]
If there are 4-9 balls, we need 2 measurement. [(3^1 +1) to 3^2]

Now if there are N balls
and lets say it lies between (3^(x-1) +1) to 3^x
then x = [logN/log3] = Greatest integer of logN to the base 3.
Then minimum no. of measures is equal to x = [logN/log3]

River crossing

Question: Three cannibals and three anthropologists have to cross a river. The boat they have is only big enough for two people. If at any point in time there are more cannibals on one side of the river than anthropologists, the cannibals will eat them. What plan can the anthropologists use for crossing the river so they don’t get eaten? (remember! the boat can’t cross the river by itself, someone has to be in it to row it across)
Note that if a boat with a cannibal and an anthropologist travels to a shore with one cannibal on it, then no. of cannibals > no. of anthropologists, even if you say the anthropologist immediately takes the boat back.

Answer:

Let

A = Anthropologist
C = Cannibal
B = the boat
W = the west shore (which they are all on)
and E = the east shore (where they want to go)Step 1 : A and C crosses
W [A, A, C, C]
E [A, C, B]Step 2 : A returns
W [A, A, A, C, C, B]
E [C]Step 3 : Two C crosses
W [A, A, A]
E [C, C, C, B]Step 4 : C returns
W [A, A, A, C, B]
E [C, C]Step 5 : Two A crosses
W [A, C]
E [A, A, C, C, B]Step 6 : A and C returns
W [A, A, C, C, B]
E [A, C]Step 7 : Two A crosses
W [C, C]
E [A, A, A, C, B]Step 8 : C returns
W [C, C, C, B]
E [A, A, A]Step 9 : Two C crosses
W [C]
E [A, A, A, C, C, B] 

Step 10 : C returns
W [C, C, B]
E [A, A, A, C]

Step 11 : Two C crosses
W [Empty]
E [A, A, A, C, C, C, B]

King and wine bottles

Question:

A bad king has a cellar of 1000 bottles of delightful and very expensive wine. A neighboring queen plots to kill the bad king and sends a servant to poison the wine. Fortunately (or say unfortunately) the bad king’s guards catch the servant after he has only poisoned one bottle. Alas, the guards don’t know which bottle but know that the poison is so strong that even if diluted 100,000 times it would still kill the king. Furthermore, it takes one month to have an effect. The bad king decides he will get some of the prisoners in his vast dungeons to drink the wine. Being a clever bad king he knows he needs to murder no more than 10 prisoners – believing he can fob off such a low death rate – and will still be able to drink the rest of the wine (999 bottles) at his anniversary party in 5 weeks time. Explain what is in mind of the king, how will he be able to do so ? (of course he has less then 1000 prisoners in his prisons)

Answer:

Think in terms of binary numbers. (now don’t read the solution without giving a try…)

Number the bottles 1 to 1000 and write the number in binary format.

bottle 1 = 0000000001 (10 digit binary)
bottle 2 = 0000000010
bottle 500 = 0111110100
bottle 1000 = 1111101000

Now take 10 prisoners and number them 1 to 10, now let prisoner 1 take a sip from every bottle that has a 1 in its least significant bit. Let prisoner 10 take a sip from every bottle with a 1 in its most significant bit. etc.

prisoner = 10 9 8 7 6 5 4 3 2 1
bottle 924 = 1 1 1 0 0 1 1 1 0 0

For instance, bottle no. 924 would be sipped by 10,9,8,5,4 and 3. That way if bottle no. 924 was the poisoned one, only those prisoners would die.
After four weeks, line the prisoners up in their bit order and read each living prisoner as a 0 bit and each dead prisoner as a 1 bit. The number that you get is the bottle of wine that was poisoned.
1000 is less than 1024 (2^10). If there were 1024 or more bottles of wine it would take more than 10 prisoners.

Jelly beans

Question:You have three jars that are all mislabeled. One contains peanut butter jelly beans, another grape jelly jelly beans and the third has a mix of both (not necessarily half-half mix). How many jelly beans would you have to pull out and out of which jars, to find out how to fix the labels on the jars?

Labels on jars are as follows
Jar 1 : Peanut butter
Jar 2 : Grape
Jar 3 : P.b. / Grape

Solution:

Only one jelly bean from the p.b./grape jar will do the trick.

The trick here is to realize that every jar is mislabeled. Therefore you know that the peanut butter jelly bean jar is not the peanut butter jelly bean jar and the same goes for the rest.
You also need to realize that it is the jar labeled p.b./grape, labelled as the mix jar, that is your best hope. If you choose a jelly bean out of there, then you will know whether that jar is peanut butter or grape jelly jelly beans. It can’t be the mix jar because i already said that every jar is mislabeled.
Once you know that jar 3 is either peanut butter, or grape jelly, then you know the other jars also. If it is peanut butter, then jar 2 must be mixed because it can’t be grape (as its labeled) and it can’t be peanut butter (that’s jar 3). Hence jar 1 is grape.
If jar 3 is grape, then you know jar 1 must be the mix because it can’t be p.b. (as its labeled) and it can’t be grape (that’s jar 3). Hence jar 2 is peanut butter.

Balls in a bag

Question:

You have 20 blue balls and 14 red balls in a bag. You put your hand in and remove 2 at a time. If they’re of the same color, you add a blue ball to the bag. If they’re of different colors, you add a red ball to the bag. (assume you have a big supply of blue & red balls for this purpose. note: when you take the two balls out, you don’t put them back in, so the number of balls in the bag keeps decreasing). What will be the color of the last ball left in the bag?

Once you tackle that, what if there are 20 blue balls and 13 red balls to start with?

Answer:

There are 3 possible cases of removing the two balls…

a) If we take off 1 RED and 1 BLUE, in fact we will take off 1 BLUE

b)If we take off 2 RED, in fact we will take off 2 RED (and add 1 BLUE)

c) If we take off 2 BLUE, in fact we will take off 1 BLUE

So In case of (a) or (c), we are only removing one blue ball, but we always take off red balls two by two.

1) 20 Blue, 14 Red balls

If there are 14 (even) number of red balls, we can not have one single red ball left in the bag, so the last ball will be blue.

2) 20 Blue, 13 Red balls

Now as the no. of red balls is odd, there will be one single red ball in the bag with other blue balls, and whenever we remove 1 red and 1 blue ball, we end up taking off only the blue ball. So the red ball will be the last ball in the bag.

Average salary

Question:Three coworkers would like to know their average salary. How can they do it, without disclosing their own salaries to other two?

 Solution:

Let the three persons are A, B and C. Now first A tells the sum of his salary and a random number to B , lets say (SA + A(random no. of A)). Now B adds the sum of his salary and a random number to the number given by A i. e. B makes the total = (SA +A) + (SB+B) and he passes this number to C. (at each step, they don’t show the no. to the third person). Now C does the same, adds his salary and a random number to the amount told by B. Now C passes this total equals to (SA+SB+SC+A+ B+C) to A. Now A subtracts his random no. (A), passes to B then B subtracts his random no. (B) and passes to C. Finally C subtracts his random no. (C) and tells everybody. They divide it by 3. Now they have the average salary.

Whiskey Problem

Question:I organized a small get together at my home.
In the party , i have a barrel with some whiskey in it.

Suddenly Guest 1 say 'I bet this barrel of whiskey is more than half full'.
'No, its less than half full' Guest 2 replied.

I don't have any measuring instrument and without removing whiskey from it , how can i determine which of the guest is right ?

Answer:

Tilt the barrel until the whiskey 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 whiskey, then it is more than half full.

25 Horses Puzzle

Question:

This is a classic puzzle which can look simple.
Given 25 horses, find the best 3 horses with minimum number of races. Each race can have only 5 horses. You don't have a timer.
Now the challange is how we can do it in 7 races.

Solution

We will have 5 races with all 25 horses
Let the results be
a1,a2,a3,a4,a5
b1,b2,b3,b4,b5
c1,c2,c3,c4,c5
d1,d2,d3,d4,d5
e1,e2,e3,e4,e5

Where a1 faster than a2 , a2 faster than a3 etc and

We need to consider only the following set of horses

a1,a2,a3,
b1,b2,b3,
c1,c2,c3,
d1,d2,d3,
e1,e2,e3,

Race 6
We race a1,b1,c1,d1 abd e1
Let speed(a1)>speed(b1)>speed(c1)>speed(d1)>speed(e1)

We get a1 as the fastest horse
We can ignore d1,d2,d3,e1,e2 and e3

a2,a3,
b1,b2,b3,
c1,c2,c3,

Race 7

Race a2,a3,b1,b2 and c1
The first and second will be second and third of the whole set

Three Boxes and a Ruby

Question: Alice places three identical boxes on a table. She has concealed a precious ruby in one of them. The other two boxes are empty. Bob is allowed to pick one of the boxes. Among the two boxes remaining on the table, at least one is empty. Alice then removes one empty box from the table. Bob is now allowed to open either the box he picked, or the box lying on the table. If he opens the box with the ruby, he gets a kiss from Alice (which he values more than the ruby, of course). What should Bob do?

Solution:

  This problem is known as the Monty Hall Problem.

 If Bob switches his choice, he wins with probability of 2/3. To get better intuition, it helps to consider a slightly different problem with 100 boxes with one box containing a ruby. Bob then picks one of the boxes at random. At least 98 of the remaining boxes are empty — these are removed by Alice. So now, we are left with two boxes: should Bob switch? Indeed, for he wins with probability 99/100 if he switches!

Hats and IIT students

Puzzle:

This is another hats puzzle same like as prisoners and hats puzzle. The puzzle is Nine IIT students were sitting in a classroom. Their professor wanted them to test. Next day the professor told all of his 9 students that he has 9 hats, The hats either red or black color. He also added that he has at least one hat with red color and the no. of black hats is greater than the no. of red hats. The professor keeps those hats on their heads and ask them tell me how many red and black hats the professor have? Obviously students can not talk to each other or no written communication, or looking into each other eyes; no such stupid options and no tricks.

Professor goes out and comes back after 20 minutes but nobody was able to answer the question. So he gave them 10 more minuets but the result was the same. So he decides to give them final 5 minutes. When he comes everybody was able to answer him correctly.

So what is the answer? and why?

Answer :

After 20 minutes :
Lets assume that their is 1 hat of red color and 8 hats of black color. The student with red hat on his head can see all 8 black hats, so he knows that he must be wearing a red hat.
Now we know that after first interval nobody was able to answer the prof that means our assumption is wrong. So there can not be 1 red and 8 black hats.After second interval of 10 minutes :
Assume that their are 2 hats of red color and 7 hats of black color. The students with red hat on their head can see all 7 black hats and 1 red hat, so they know that they must be wearing a red hat.
Now we know that after second interval nobody was able to answer the prof that means our assumption is again wrong. So there can not be 2 red and 7 black hats.

After third interval of final 5 minutes :
Now assume that their is 3 hats of red color and 6 hats of black color. The students with red hat on their head can see all 6 black hats and 2 red hats, so they know that they must be wearing a red hat.
Now we know that this time everybody was able to answer the prof that means our assumption is right.So there are 3 red hats and 6 black hats.

Now as everybody gave the answer so there can be a doubt that only those 3 students know about it how everybody came to know ?
Then here is what i think, the professor gave them FINAL 5 minutes to answer, so other guys will think that the professor expects the answer after 3rd interval (according to prof it must be solved after 3 intervals), so this is the clue for others.

Prisoners and three hats

Four prisoners are arrested for a crime, but the jail is full and the jailer has nowhere to put them. He eventually comes up with the solution of giving them a puzzle so if they succeed they can go free but if they fail they are executed.

The jailer puts three of the men sitting in a line. The fourth man is put behind a screen (or in a separate room). He gives all four men party hats (as in diagram). The jailer explains that there are two red and two blue hats. The prisoners can see the hats in front of them but not on themselves or behind. The fourth man behind the screen can't see or be seen by any other prisoner. No communication between the men is allowed.

If any prisoner can figure out and say (out loud) to the jailer what colour hat he has on his head all four prisoners go free. The puzzle is to find how the prisoners can escape.

The solution

For the sake of explanation let's label the prisoners in line order A B and C. Thus B can see A (and his hat colour) and C can see A and B.

The prisoners know that there are only two hats of each colour. So if C observes that A and B have hats of the same colour, C would deduce that his own hat is the opposite colour. However, If A and B have hats of different colours, then C can say nothing. The key is that prisoner B, after allowing an appropriate interval, and knowing what C would do, can deduce that if C says nothing the hats on A and B must be different. Being able to see A's hat he can deduce his own hat colour. (The fourth prisoner is irrelevant to the puzzle: his only purpose is to wear the fourth hat).

Puzzle : Prisoners and Four - hats.
Now the same puzzle with different condition. Instead of two red and two blue hats there are 3 hats of one colour and only 1 hat of another, and the 3 prisoners can see each other i.e. A sees B & C, B sees A & C and C sees A & B. As like shown in diagram (D again not to be seen and only there to wear the last hat)

The solution

There are two cases:

In first case, one of the three prisoners (in A, B, C) wears the single off-colour hat, thus the other two can easily deduce the colour of theirs. For example if A is single color then B can see that C and A have different color hats. From this he concludes that his color is not single so he shouts his color.

In second case, the three prisoners wear hats of the same colour, while D wears the off-colour hat. After a while, all four prisoners should be able to deduce that, since none of the others was able to state the colour of his own hat, D must wear the off-colour hat.

Puzzle : Prisoners and Five - hats.
Another different condition. only three prisoners and five hats (supposedly two black and three white) are involved. The three prisoners are ordered to stand in a straight line facing the front, with A in front and C at the back. They are told that there will be two black hats and three white hats. One hat is then put on each prisoner's head; each prisoner can only see the hats of the people in front of him and not on his own's. The first prisoner that is able to announce the colour of his hat correctly will be released. No communication between the prisoners is allowed. After some time, only A is able to announce (correctly) that his hat is white. Why is that so?

The solution

Assuming that A wears a black hat.

• If B wears a black hat as well, C can immediately tell that he is wearing a white hat after looking at the two black hats in front of him.

• If B does not wear a black hat, C will be unable to tell the colour of his hat (since there is a black and a white). Hence, B can deduce from A's black hat and C's response that he (B) is not wearing a black hat (otherwise the above situation will happen) and is therefore wearing a white hat.

This therefore proves that A must not be wearing a black hat.

Aeroplane

Puzzle :

The puzzle question is : On Bagshot Island, there is an airport. The airport is the homebase of an unlimited number of identical airplanes. Each airplane has a fuel capacity to allow it to fly exactly 1/2 way around the world, along a great circle. The planes have the ability to refuel in flight without loss of speed or spillage of fuel. Though the fuel is unlimited, the island is the only source of fuel.

What is the fewest number of aircraft necessary to get one plane all the way around the world assuming that all of the aircraft must return safely to the airport? How did you get to your answer?

Notes:

(a) Each airplane must depart and return to the same airport, and that is the only airport they can land and refuel on ground.

(b) Each airplane must have enough fuel to return to airport.

(c) The time and fuel consumption of refueling can be ignored. (so we can also assume that one airplane can refuel more than one airplanes in air at the same time.)

(d) The amount of fuel airplanes carrying can be zero as long as the other airplane is refueling these airplanes. What is the fewest number of airplanes and number of tanks of fuel needed to accomplish this work? (we only need airplane to go around the world)

solution for this puzzle is given below. Write your comments.. and any different answers.

Answer :

           As per the puzzle given ablove The fewest number of aircraft is 3! Imagine 3 aircraft (A, B and C). A is going to fly round the world. All three aircraft start at the same time in the same direction. After 1/6 of the circumference, B passes 1/3 of its fuel to C and returns home, where it is refuelled and starts immediately again to follow A and C.

C continues to fly alongside A until they are 1/4 of the distance around the world. At this point C completely fills the tank of A which is now able to fly to a point 3/4 of the way around the world. C has now only 1/3 of its full fuel capacity left, not enough to get back to the home base. But the first "auxiliary" aircraft reaches it in time in order to refuel it, and both "auxiliary" aircraft are the able to return safely to the home base.

Now in the same manner as before both B and C fully refuelled fly towards A. Again B refuels C and returns home to be refuelled. C reaches A at the point where it has flown 3/4 around the world. All 3 aircraft can safely return to the home base, if the refuelling process is applied analogously as for the first phase of the flight.

Hard Probability puzzle 1

Question:

Three people enter a room and have a green or blue hat placed on their head. They cannot see their own hat, but can see the other hats.
The color of each hat is purely random. They could all be green, or blue, or any combination of green and blue.
They need to guess their own hat color by writing it on a piece of paper, or they can write 'pass'.
They cannot communicate with each other in any way once the game starts. But they can have a strategy meeting before the game.
If at least one of them guesses correctly they win $50,000 each, but if anyone guess incorrectly they all get nothing.
What is the best strategy?

Solution
Simple strategy: Elect one person to be the guesser, the other two pass. The guesser chooses randomly 'green' or 'blue'. This gives them a 50% chance of winning.

Better strategy: If you see two blue or two green hats, then write down the opposite color, otherwise write down 'pass'.

It works like this ('-' means 'pass'):

Hats: GGG, Guess: BBB, Result: Lose
Hats: GGB, Guess: --B, Result: Win
Hats: GBG, Guess: -B-, Result: Win
Hats: GBB, Guess: G--, Result: Win
Hats: BGG, Guess: B--, Result: Win
Hats: BGB, Guess: -G-, Result: Win
Hats: BBG, Guess: --G, Result: Win
Hats: BBB, Guess: GGG, Result: Lose

Result: 75% chance of winning!

Probability Of Having Same Birthday

Question:

How many people must be gathered together in a room, before you can be certain that there is a greater than 50/50 chance that at least two of them have the same birthday?

Solution:

Only twenty-three people need be in the room, a surprisingly small number. The probability that there will not be two matching birthdays is then, ignoring leap years, 365x364x363x...x343/365 over 23 which is approximately 0.493. this is less than half, and therefore the probability that a pair occurs is greater than 50-50. With as few as fourteen people in the room the chances are better than 50-50 that a pair will have birthdays on the same day or on consecutive days.

Probability Riddle Loaded Revolver

Question: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?

Solution:

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

Famous Probability puzzle SHOOT

Question:

Mr. Black, Mr. Gray, and Mr. White are fighting in a truel. They each get a gun and take turns shooting at each other until only one person is left. Mr. Black, who hits his shot 1/3 of the time, gets to shoot first. Mr. Gray, who hits his shot 2/3 of the time, gets to shoot next, assuming he is still alive. Mr. White, who hits his shot all the time, shoots next, assuming he is also alive. The cycle repeats. If you are Mr. Black, where should you shoot first for the highest chance of survival?

Solution:   He should shoot at the ground.  If Mr. Black shoots the ground, it is Mr. Gray's turn. Mr. Gray would rather shoot at Mr. White than Mr. Black, because he is better. If Mr. Gray kills Mr. White, it is just Mr. Black and Mr. Gray left, giving Mr. Black a fair chance of winning. If Mr. Gray does not kill Mr. White, it is Mr. White's turn. He would rather shoot at Mr. Gray and will definitely kill him. Even though it is now Mr. Black against Mr. White, Mr. Black has a better chance of winning than before.

Life Or Death

Question:

You are a prisoner sentenced to death. The Emperor offers you a chance to live by playing a simple game. He gives you 50 black marbles, 50 white marbles and 2 empty bowls. He then says, 'Divide these 100 marbles into these 2 bowls. You can divide them any way you like as long as you use all the marbles. Then I will blindfold you and mix the bowls around. You then can choose one bowl and remove ONE marble. If the marble is WHITE you will live, but if the marble is BLACK... you will die.'

How do you divide the marbles up so that you have the greatest probability of choosing a WHITE marble?

Solution:

1 white marble in one bowl, and rest of the marbles in other bowl.

The chances of survival becomes:

(1/2)*1 + (1/2)*49/99

Four People on a Rickety Bridge

Question: Four people need to cross a rickety bridge at night. Unfortunately, they have only one torch and the bridge is too dangerous to cross without one. The bridge is only strong enough to support two people at a time. Not all people take the same time to cross the bridge. Times for each person:  1 min, 2 mins, 7 mins and 10 mins. What is the shortest time needed for all four of them to cross the bridge?

Answer: The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 mins. Is that it? No. That would make this question too simple even as a warm up question.

Let’s brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let’s put all this together.

1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)

Total time = 2 + 2 + 10 + 1 + 2 = 17 mins