• Why is a manhole cover round?

  • There are quite a few possible answers to this question. Some are:
    1. The Man hole cover being roundAnswer
    The Man hole cover is round so it will not fall into the hole. If it were square it would be possible for the cover to fall into the hole if it was tilted at the proper angle. Being round there is no way without breaking the cover for it to fall into the hole ( sent by Tony Bowman)
    2. Because manholes are round. (Simple and seemingly obvious, yet absolutely true)
    3. Round covers can be transported by one person, because they can be rolled on their edge. (Ease of use).
    4. A round cover doesn't need to be rotated to fit over a hole

• Given a rectangular cake with a rectangular piece removed (any size or orientation), how would you cut the remainder of the cake into two equal halves with one straight cut of a knife ?

  • Find the centers of both the original cake and the removed piece. Cut the reminder along the line connecting these two centers. This is true because this line cut both the original cake and the removed piece in half,thus the remainder into two halves.

   

• If you had an infinite supply of water and a 5 quart and 3 quart pail, how would you measure exactly 4 quarts?
  • Measure 5 quart using the 5 quart pail, Then fill the 3 quart pail from 5 quat pail. You have 2 quart left in the 5 quart pail. So, measure 2 quarts. Repeat theprocess and you will have measured four quarts.

   

• One train leaves Los Angeles at 15 MPH heading for New York. Another train leaves from New York at 20mph heading for Los Angeles on the same track. If a bird, flying at 25mph, leaves from Los Angeles at the same time as the train and flies back and forth between the two trains until they collide, how far will the bird have traveled?

  • 1. Let's say the distance between LA and NY is d miles.
  • 2. The time before which the trians will collide : d / (15+20) hours.
  • 3. Distance travelled by the bird in that time : (d / 35) * 25 = 5*d/7 miles.

• There are four people who need to cross a bridge at night. The bridge is only wide enough for two people to cross at once. There is only one flashlight for the entire group. When two people cross, they must cross at the slower member's speed. All four people must cross the bridge in 17 minutes, since the bridge will collapse in exactly that amount of time. Here are the times each member takes to cross the bridge:

  Person A: 1 minute

  Person B: 2 minutes

  Person C: 5 minutes

  Person D: 10 minutes

  
  So, if Person A and C crossed the bridge initally, 5 minutes would elapse, because Person C takes 5 minutes to cross. Then, Person A would have to come back to the other side of the bridge, taking another minue, or six minutes total. Now, if Person A and D crossed the bridge next, it would take them 10 minutes, totalling to 16 minutes. If A came back, it would take yet another minute, totally 17... the bridge would collapse with A and B on the wrong side. How can all four people get across the bridge within 17 minutes? Note: there is no trick-answer to this problem. You can't do tricky stuff like throwing the flashlight back from one end of the bridge to the other. This problem can be solved!

  • Yeah, this problem is little bit more tricky...but that's wht's makes it interesting.. :)
  • 1. A and B cross together. Total Time: 2 Minutes
    2. A comes back. Total Time: 3 Minutes.
    3. C and D cross together. Total Time: 13 Minutes
    4. B comes back. Total Time: 15 Minutes
    5. A and B cross together. Total Time: 17 Minutes

   

• You have someone working for you for seven days and a gold bar to pay them. The gold bar is segmented into seven connected pieces. You must give them a piece of gold at the end of every day. If you are only allowed to make two breaks in the gold bar, how do you pay your worker?

  • Day 1 : Break of a single segment and give it to the worker.
    Day 2 : Break two segments from the bar, Ask the worker to return the segment you gave him on day 1 and give him the segment with Two connected pieces.
    Day 3 : GIve the worker the 1 piece segment you have.
    Day 4 : Ask the worker to return all three pieces he has and give him the segment with 4 pieces on it.
    Day 5 : Give him the the segment with 1 piece.
    Day 6 : Ask him to return the 1 piece segment and give him the segment with two pieces.
    Day 7 : Give him the 1 piece segment you have.
  • Important point is to notice this pattern here :
    1
    10
    11
    100
    101
    110
    111
  • Now, you can answer this question for any bar with "n" segments.

   

• You have 5 jars of pills. Each pill weighs 10 gram, except for contaminated pills contained in one jar, where each pill weighs 9 gm. Given a scale, how could you tell which jar had the contaminated pills in just one measurement?

  • 1. Mark the jars with numbers 1, 2, 3, 4, and 5.
    2. Take 1 pill from jar 1, take 2 pills from jar 2, take 3 pills from jar 3, take 4 pills from jar 4 and take 5 pills from jar 5.
    3. Put all of them on the scale at once and take the measurement.
    4. Now, subtract the measurment from 150 ( 1*10 + 2*10 + 3*10 + 4*10 + 5*10)
    5. The result will give you the jar number which has contaminated pill.

• You have 12 balls. All of them are identical except one, which is either heavier or lighter than the rest - it is either hollow while the rest are solid, or solid while the rest are hollow. You have a simple two-armed scale, and are permitted three weighings. Can you identify the odd ball, and determine whether it is hollow or solid? ( This question is same as above but with an additional twist).

  • I had already worked this problem out for someone else using pills. The setting of the problem, as I remember it, goes something like
    this.
  • A wise man has committed a capital crime and is to be put to death.The king decides to see how wise the man is. He gives to the man 12
    pills which are identical in size, shape, color, smell, etc. However,they are not all exactly the same. One of them has a different weight. The wise man is presented with a balance and informed that all the pills are deadly poison except for the one which has a different weight. The wise man can make exactly three weighings and then must take the pill of his choice.
  • Remember that the only safe pill has a different weight, but the wise man doesn't know whether it is heavier or lighter.
  • Most people seem to think that the thing to do is weigh six pills against six pills, but if you think about it, this would yield you no information concerning the whereabouts of the only safe pill.
    
    So that the following plan can be followed, let us number the pills from 1 to 12. For the first weighing let us put on the left pan pills
    1,2,3,4 and on the right pan pills 5,6,7,8.
  • There are two possibilities. Either they balance, or they don't. If
    they balance, then the good pill is in the group 9,10,11,12. So for
    our second weighing we would put 1,2 in the left pan and 9,10 on the
    right. If these balance then the good pill is either 11 or 12.
  • Weigh pill 1 against 11. If they balance, the good pill is number 12.
    If they do not balance, then 11 is the good pill.
  • If 1,2 vs 9,10 do not balance, then the good pill is either 9 or 10.
    Again, weigh 1 against 9. If they balance, the good pill is number
    10, otherwise it is number 9.
  • 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 pills could be the safe pill. 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 good pill is either 3 or 4.
    Weigh 4 against 9, a known bad pill. If they balance then the good
    pill is 3, otherwise it is 4.
  • 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 good, heavy pill, or 1 is a good, light pill.
  • For the third weighing, weigh 7 against 8. Whichever side is heavy is
    the good pill. If they balance, then 1 is the good pill. 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 good heavy pill or 2 is a light good pill. Weigh 5
    against 6. The heavier one is the good pill. If they balance, then 2
    is a good light pill.
  • I think that one could write all of this out in a nice flow chart,
    but I'm not sure that the flow chart would show up correctly over
    e-mail.
  • After trying to solve the problem without cheating, I found a solution. It covers the case Mr. Manning does not cover and is hopefully easier to follow than Dr. Robert's solution.
  • If you have balls numbered 1-12 and if 1-4 and 5-8 don't balance (otherwise see above) then remove 1-3 from the left side. Replace with 5-7 (now containing 4-7). Add 9-11 to the right side (now containing 8-11).
  • If the scales balance then 1-3 contains the odd ball (heavier or lighter depending on whether the left side was heavier or lighter).
  • If the scales reverse then 5-7 contain the odd ball, which is deduced to be heavier or lighter as above.
  • In either case you have 3 balls and since you know if you are looking for a heavy or a light ball you can determine which ball it is in one weighing.
  • If the scales are unchanged then one of the unmoved balls (4 or 8) are odd. Weigh the heavier (say 4) against a known ball (say 12). If they balance, 8 is the odd (in this case lighter) ball. If they don't balance then the heavier ball is the odd ball.

   

• If you are on a boat and you throw out a suitcase, will the level of water increase?

  • The answer is: it depends on the density of the briefcase.
  • While the briefcase is in the boat, it will displace it's weight in water.
  • If the briefcase floats after you toss it out, then it still displaces its weight in water, and there is no change.
  • However, if the briefcase is heavier than water it will displace its volume, not its weight. The amount of displaced water will be less (which is why it sank) and the water level will decrease.

• There are 3 ants at 3 corners of a triangle, they randomly start moving towards another corner. What is the probability that they don't collide?

  • Let's mark the corners of the triangle as A,B,C.
    2. There are only two cases under which the ants won't collide : A->B, B->C, C->A and A->C, C->B and B->A
    3. Therefore, probability of ants not colliding : 2/8 = ¼

   

• This puzzle was apparently written by Einstein in the last century. He said that 98% of the people in the world cannot solve the quiz. See if you can. ( Sent by crit crit ).
  Facts:
  1: There are 5 houses in 5 different colors
  2: In each house lives a person with a different nationality.
  3: These 5 owners drink a certain beverage, smoke a certain brand of cigar and keep a certain pet.
  4: No owner has the same pet, smoke the same brand of cigar or drink the same drink.

• HINTS:
  1: The British lives in a red house.
  2: The Swede keeps dogs as pets
  3: The Dane drinks tea
  4: The green house is on the left of the white house (it also means they are next door to each other)
  5: The green house owner drinks coffee
  6: The person who smokes Pall Mall rears birds
  7: The owner of the yellow house smokes Dunhill
  8: The man living in the house right in the center drinks milk
  9: The Norwegian lives in the first house
  10: The man who smokes Blend lives next to the one who keeps cats
  11: The man who keeps horses lives next to the man who smokes Dunhill
  12: The owner who smokes Blue Master drinks beer
  13: The German smokes Prince
  14: The Norwegian lives next to the blue house
  15: The man who smokes Blend has a neighbor who drinks water.
  The question is: WHO KEEPS THE FISH?
  • ONLY for those who want to know the approach for solving such problems
  • Ok first thing to figure out is the order of houses
  • we can use to fact
    1) Green and White are next to each other
    2) blue is the second house.
  • center house is imporatnt becase we have a clue about it
  • So the possiblilites could be:-
  • ____, blue, ____, Green, White,
  • or
    ____, blue, Green, White,____
  • 
    there are 2 blanks and two colors.
  • Red can't be the first house because, firts house should have Norway guy and Red house has British.
  • So we have :-
  • Yellow, blue, ____, Green, White,
  • or
    Yellow, blue, Green, White,____
    (where blank is Red)
  • 2nd option can't be true as we have cluse that center house owner drinks milk, but the green house owner drinks coffee.
  • So we get the order as follow
  • Yellow, blue,Red, Green, White,
    (with last two houses could be switched with no effect)
  • Then make the table as follows
  • house/attributes|Yellow|Blue|Red|Green|White|
    Drink
    Pet
    Smoke
    Nationality
  • Once you have the right order then its only the matter of reading the clues and filling in the table entries.

• There are three wise men in a room. So you decide to give them a challenge. Suspecting that the thing they care about most is money, you give them $100 and tell them they are to divide this money observing the following rule: they are to discuss offers and counter-offers from each other and then take a vote. Majority vote wins. Sounds easy enough... now the question is, assuming each person is motivated to take the largest amount possible, what will the outcome be?

  • 2 probably won't vote the 3rd out, since counter-offers are allowed. Consider: A and B decide to leave C out and split halfway. C offers B to leave A out instead, and as incentive offers B $60 as compared to $50 (Note, for C, $40 is better than nothing at all). Now A will try to raise that figure, but keep no less than $33 for himself. It would also happen that as B starts to get more and more of the share, A and C would decide to keep B out instead and split amongst each other. This would go on and on until they reach equilibrium with each agreeing to take their rightful share of 33 dollars.

• Three men were renting a motel figuring the room cost 30 dollars they would pitch in ten a piece.The room was only 25 so they each gave the bell boy ten,(tip)the bellboy didn"t think that would be fair so he gave them each back 1 dollar and kept 2 for himself.What happened to the other dollar?

  • Oh after giving it some thought actually I was right, there isn't any missing dollar.
  • If you go backwards you come up to this:
    9x3 = 27
    27+2 = 29
    so it had to be 30$ in order to be correct right?
    No, because you have to add 3$ and not 2. You think it would be right if you add 2$ because the tip is that amount. But the tip is already included in the multiplication.
    So the maths aren't right, it should be 9x3+3 and not 9x3+2.

• "You have b boxes and n dollars. If I want any amount of money from 0 to n dollars, you must be able to hand me 0 to b boxes so that I get exactly what I request." The two questions were "What are the restrictions on b and n, and how is money distributed among the boxes?"

  • No one mentioned why one should use binary numbers !!
    Any numner n can be expressed in any base.
    n = a1 * b^0 + a2 * b^1 + a3 * b^2 + ...
    where ai's are from 0 to b-1.
    eg. in decimal system
    123 = 10^2 * 1 + 10 * 2 + 3.
  • In this case you either choose a box or not. So the ai's can be 0 or 1.( 0 means dont choose and 1 means you chose it)
    Hence b = 2;
    So binary base has to be used.
  • max that you can represent using this is
    2^b - 1 >=n;
    It is the relation between b and n.
  • The trick is of Binary Numbers
    2^0=1
    2^1=2
    2^2=4
    2^3=8
    2^4=16
    ....
    2^7=128
  • The formula is [log n to base 2] + 1 = b
    where [] denote integral part of the log base 2 of n
  • Example
    So if we want 0 to 15 dollars
    [log of 15 to base 2] + 1 = [3.90] + 1 = 3 + 1 = 4 boxes
    we need to have 4 boxes each having 1,2,4,8, dollars respectively.Now we know from binary numbers that any amount from 0 to 15 can be formed with this boxes.
  • For 0 to 6 dollars
    [log of 7 to base 2] + 1 = [2.58] + 1 = 2 + 1 = 3 boxes
    we need to have 3 boxes each having 1,2,4, dollars respectively
  • For 0 to 100 dollars
    [log of 100 to base 2] + 1 = [6.64] + 1 = 6 + 1 = 7 boxes
    we need to have 7 boxes each having 1,2,4,8,16,32,64 dollars respectively

• There are 2 rooms. First has 3 switches (on/off) and the other one has 3 bulbs (electric lamps). Initially all the switches are off and all bulbs are turned off. Every bulb is connected to one switch and only one. First you are in the room with switches. You can play with switches however you want and whenever you want, but in the rooms with bulbs you can enter only once. Find the correspondence between switches and bulbs. - (Sent by Oliviu Burlacu)

  • Just turn one light switch on(switch1) and leave it on long enough for the bulb to get warm. Then turn that off and turn another one on(switch2). Then go downstairs and the light that is currently on belongs to (switch2) and then a bulb that is off will be warm since you left it on for awhile so that belongs to (switch1) then the third bulb that is off belongs to the last switch that wasn't touched(switch3)

• There is a prisioner who has two doors infront of him one leads to freedom and one leads to hell, he has no idea which one is which, but there is 2 guards there on tell the truth and one always lies. You can ask them one question that will lead you to your freedom. What is your question?

  • If you are allowed to ask one question per guard then ask them individually "what will the other guard say if I were to ask him the right door ?"
  • When this questioned is posed to the guard who speaks the truth, he will know that the other guard lies so will claim his gurad to be the right door. so the gurad who speaks the truth will point to the gurad who lies.
  • When this question is posed to the bad guard then without hesitation he would say that his door his right. From the above answers one can determine the right door.
  • This answer assumes that One question can be asked to both the guards, if not we need to call both the gurads and pose this question at the same time.
  • This is NOT my answer but found it as a riddle in a greek story (if my memory holds good)
  • Also this answer assumes that the guard who speaks the truth knows that the other guard lies and vice versa

  • If you look at a clock and the time is 3:15, what is the angle between the hour and the minute hands? ( The answer to this is not zero!)

  • Obviously, for digital clock, nothing is meaningful in this problem.
  • For analog clock, the minute pin is exact at "3" and hour pin is a little past "3". Then tha angle is:
    15/60 * (360/12)= 7.5 (degree)
  • where 15/60 = 15 minutes count in an hour(the angle between "3" and "4");
    360/12 = a circle devided by 12 digits.
  • Notice that we have to find the angle swept by the hour hand when it was at 3 o'clock and when it is at 3:15.
  • This is so because the minute hand points to the same location at 3:15 as the hour hand was pointing at 3 o'clock.
  • In other words we have to find the angle swept by the hour hand in 15 minutes.
  • We know hour hand sweeps 360 degrees in 12 hours(12*60 minutes).
  • so, number of degrees it sweeps in 15 minutes = 15 * (360 / (12*60)) = 7.5 degrees.

   

• What is the sum of the numbers from 1 to 1000?

  • In general the sum of numbers from 1 to n is given by: n*(n+1)/2

  • An easier way to remember it might be "times a half, plus a half", which is essentially what that calculation reduces to:
  • (n(n+1))/2 = n*(n/2) + n/2

• You are an employer. You have ten employees. Each month, each one of your ten employees gives you ten bags of gold. Each bag of gold has ten pieces of gold in it. Each piece of gold weighs one pound. One of your employees is cheating you by only putting nine pieces of gold in each of his ten bags of gold. You have a scale (not a balance, a scale), and you can only take one measurement from the scale, only one (1) reading. How can you tell which of the ten employees is cheating you by using this scale and only taking one measurement?

  • When your employees come to pay you at the end of the month, take one bag from employee number one's pile and set it aside. Take two bags from employee number two's pile and add it to the one from employee number one. Take three from employee three, four from employee four, etc...
  • Once you have now compiled a pile of bags, you should have 55 bags (one from one, two from two, etc). Set the whole pile (all 55 bags) on the scale and take a reading. If all the bags were honest, you would have 550 pounds on the scale.
  • If employee one is cheating you, then the reading would be 549, since you only have one bag from employee one. Likewise, if the measurement is 547, then 3 pounds are missing, and employee three is the thief.

   

• How many points are there on the globe where by walking one mile south, one mile east and one mile north you reach the place where you started.

• Imagine that you have 26 constants, labelled A through Z. Each constant is assigned a value in the following way: A = 1; the rest of the values equal their position in the alphabet (B corresponds to the second position so it equals 2, C = 3, etc.) raised to the power of the preceeding constant value. So, B = 2 ^ (A's value), or B = 2^1 = 2. C = 3^2 = 9. D = 4^9, etc., etc. Find the exact numerical value to the following equation:
  (X - A) * (X - B) * (X - C) * ... * (X - Y) * (X - Z)

(X - A) * (X - B) * (X - C) * ... * (X - Y) * (X - Z) equals 0 since (x-x) is zero.

          (X - A) * (X - B) * (X - C) * ... * (X - Y) * (X - Z) = 0 as there is (X-X) which is equal to 0 and anything multiplied by 0 is 0.

• You have two jars, 50 red marbles and 50 blue marbles. A jar will be picked at random, and then a marble will be picked from the jar. Placing all of the marbles in the jars, how can you maximize the chances of a red marble being picked? What are the exact odds of getting a red marble using your scheme?
  • Put 1 red marble in the first jar.
    Put the other 49 reds and the 50 blues in the other jar (99 total marbles!).
  • The probability of selecting the jar with only 1 red marble and getting the only marble in the jar is 50% (1/2 * 1) not 1%!
  • The probability of selecting the other jar is 50% and the odds of getting a red marble from that jar is 49/99 and (1/2 * 49/99) = 24.7475%.
  • The total probability is 74.7475%. Zeephro forgot to correct for the missing red marble in the denominator (49/99, not 49/100) which is why he came out with 74.5%
  • Short answer: In one of the jars put 1 red ball and 0 blue balls, in the second jar put the rest of the balls.
    The probabily of picking a red ball is 0.7474...
  • Long answer:
    Let R1 and B1 be the number of Red and Blue balls in the first jar, and R2 and B2 the number of Red and Blue balls in the second jar.
    Then a) R1 + R2 = 50 and b) B1 + B2 = 50.
    The probabily of picking a Red ball is equal to the probabily of choosing a jar times the probabity of choosing a Red ball in that jar.
  • ProbRed = (prob of jar1) (prob of red in jar1) + (prob of Jar2)(prob of red in jar2)
  • The probability of each jar is 0.5 so:
  • c) ProbRed = 0.5 ( Prob of red in jar1 + Prob of red in Jar2)
    ProbRed1 = (num of red balls in jar1)/(total number of balls in jar1)
    ProbRed1 = R1/(R1+B1) , the same for jar 2:
    ProbRed2 = R2/(R2+B2)
  • If we replace a), b) in c) we get:
  • ProbRed = 0.5( (R1/(R1+B1)) + ((50-R1)/(50-R1+50-R1)) )
  • Now if you are good at math you can find the derivate of this function and equal it to zero, but I am not that good at math so I run a VB program for all posible values of R1 and B1 (except those that cause division by zero):
  • Private Sub Command1_Click()
    Dim R1 As Integer
    Dim B1 As Integer
    Dim ProbRed As Double
    
    For R1 = 1 To 49
    For B1 = 0 To 50
    ProbRed = 0.5 * ((R1 / (R1 + B1)) + ((50 - R1) / (50 - R1 + 50 - B1)))
    List1.AddItem "R1 = " & R1 & " B1 = " & B1 & " ProbRed = " & ProbRed
    Next B1
    Next R1
    End Sub
  • The result confirmed that the maximum probabity is when :
    a) R1 = 1 and B1 = 0 or when
    b) R1 = 49 and B1 = 50
• Imagine you are standing in front of a mirror, facing it. Raise your left hand. Raise your right hand. Look at your reflection. When you raise your left hand your reflection raises what appears to be his right hand. But when you tilt your head up, your reflection does too, and does not appear to tilt his/her head down. Why is it that the mirror appears to reverse left and right, but not up and down?
  • It may be because, our eyes are placed left to right when facing the mirror. So the reflection from mirror crosses like a X in front of us in horizontal direction. As far as vertical reflection goes, the eyes are in same line. So the vertical reflection will be just like a horizontal line without reversals. That's why the image appears reversed horizontally and not vertically.

   

• You have a bucket of jelly beans. Some are red, some are blue, and some green. With your eyes closed, pick out 2 of a like color. How many do you have to grab to be sure you have 2 of the same?
  • It is quite simple. If you only take 4 jelly beans, the possible combos that you can come up with are, in terms or a,b and c:
  • a,b,c,a a,b,c,b a,b,c,c
    a,a,b,c a,a,b,b a,a,c,c
    b,b,a,c b,b,a,a b,b,c,c
    c,c,a,c c,c,a,a c,c,b,b
    a,a,a,b a,a,a,c a,a,a,a
    b,b,b,a b,b,b,c b,b,b,b
    c,c,c,a c,c,c,b c,c,c,c

   

• You are given a scale which you are to use to measure eight balls. Seven of these balls have the same weight: the eigth ball is heavier than the rest. What is the minimum number of weighs you could perform to find the heaviest of the eight balls?. Remmber it's a scale not a balance. (i.e. It can just tell you if one side is heavier than the other it can't give you the exact weight).
  • Good try!..Umm..You can do this in 2 weighs. Put three balls on each side of the scale. That's a total of six balls you're weighing. If the three balls on each side weigh equally, you know that one of the two remaining balls is the heaviest. Weigh those two balls to determine which one is heaviest. If, however, one of the three ball combinations weighs most, remove all balls from the scale, then weigh just two of the three "heavier" balls. If those two balls are equal weight, the third, unweighed ball is the heaviest; otherwise the scale will indicate which of the two balls on the scale is the heavy one.
• There was a sheriff in a town who caught three outlaws. He said he was going to give them all a chance to go free. All they had to do is figure out what color hat they were wearing. The sheriff had 5 hats 3 black and 2 white. The first guy guessed and was wrong so he was put in jail. The second guy also guessed and was also put in jail. finally, a third blind man guessed and he guessed correctly. How did he know?
  • Humm, I puzzled over this for a sec. and then, when checking my answer, I had a hard time understanding what was written (though it is indeed correct) so here is a more lengthy answer that hopefully explains it all.
    
    lets look at it this way; Here are our possibilities:
    1 BBB
    2 BBW
    3 BWB
    4 WBB
    5 WWB
    6 BWW
    7 WBW
  • Ok, now we can eliminate # 6 because in this case convict one would be sure to know that he had on a white hat.
  • # 7 can be eliminated for the same reason for convict 2's guess.
  • In case # 2, the first convict has to see at least 1 black hat (if he saw two whie hats he wouldn't have guessed wrong guessing he had a white hat since he would already have seen 2 white hat). So we know that convict 2 or convict 3 has a black hat (posibaly both). Now convict two has the same delema, but he knows that one or both of convict 2 and 3 has a black hat. He can see that convict 3 has a white hat so in that case he would guess black and be correct, but he didn't (since we know he guessed wrong). given this, we can remove option 2 from consideration.
  • options 1,3,4,5 all have convict 3 wearing a black hat. Thus, assuming that convicts 1 and 2 are as logical as possible, the only options left all have convict 3 wearing a black hat.