Solutions Frequently Asked

Version I: It is not possible to connect the three utilities to all three houses without crossing pipes in the same plane. However, the usual quibble is to solve the puzzle by running one of the pipes underneath one of houses on its way to another house; the puzzle's instructions forbid crossing other *pipes*, but not crossing other *houses*.

Version II: Without using tricks, it can't be done. One trick is to draw a line to one corner of the paper, raise the corner, with the pen tip still touching it, and then lower it on the final point to complete the puzzle.

Another trick is to set the problem on a toroid (doughnut-shaped object).

One sample.
Start by drawing from A/O box. Whatever comes out, the box contains nothing else (since all labels are incorrect). Label this box correctly. Move the A/O label to the box originally marked the opposite of what you drew. Now put the last label on the last box.

There are only two ways to rearrange three things so that none of them stays the same, which can be described as shifting all labels one to the left or one to the right. By drawing from the A/O box, you can immediately tell in which direction the labels have been shifted.

Trip 1 - 2 cannibals.
Trip 2 - 1 cannibal comes back.
Trip 3 - 2 cannibals (leaving 3 missionaries on this bank).
Trip 4 - 1 cannibal comes back.
Trip 5 - 2 missionaries - leaves one of each on this side and 2 of each on the other side.
Trip 6 - 1 cannibal and 1 missionary come back (there are now 2 of each on this side and 1 of each across).
Trip 7 - the last 2 missionaries cross, leaving 2 cannibals on this side and three missionaries and one cannibal across.
The missionaries can now walk along while the cannibal goes back to get his mates.

The problem is confusing for two reasons: first, there are hidden assumptions about Monty's motivation that cloud the issue; and second, novice probability students do not see that the opening of the door gave them any new information.

Monty can have one of three basic motives:

1. He randomly opens doors.
2. He always opens the door he knows contains nothing.
3. He only opens a door when the contestant has picked the grand prize.

1. No improvement when switching. He has a 1/3 chance of winning.
2. Double your odds by switching. The contestant who does not switch doors (i.e. who keeps the originally chosen door) has only a 1/3 chance of winning. The contestant who switches doors, however, has a 2/3 chance of winning.
3. Don't switch! You win.

A variation on the above puzzle, again based on verbal sleight of hand. The answer is "what". No question is asked in the third sentence, just a statement that WHAT is the third word. However invariably a question mark is appended to the third line, making the puzzle unsolvable.

The answer to this bit of verbal trickery is "language". There are three words in the phrase "the English language", and the third word is "language". There is no meantion of a third word ending in "gry", but that is what most people tend to think. Infact there are other words ending in "gry"; unangry is at least one of them.

Divide the balls into three sets of three. Balance two of the three sets. Take the set that has the heavy ball (i.e. the set that weighs more on the balance, or if they weigh the same then the set you didn't weigh). Now balance two of the three balls from that set, and the heavy one should be obvious (same method as before).

Divide the coins into three sets of four coins each. For a first weight trial place a group of four coins in each pan. There are two possibilities, which we shall investigate separately:

(1) The pans balance.
(2) One pan outweighs the other.

WHEN THE PANS BALANCE.
In the event the counterfeit is in the unweighed set, all eight coins on the scale are genuine. Number the coins in the doubtful group 1, 2, 3, 4. Carry out a second weight trial by placing coins 1, 2 and 3 in one pan and placing three genuine coins in the other pan. There are two possibilities:

(A) The pans balance. In this case, coin 4 is counterfeit. A third weighing, comparing coin 4 with a genuine one, will tell whether it is lighter or heavier.

(B) One pan is heavier. In this case, the counterfeit is one of the coins 1, 2 or 3. If the genuine coins are heavier, then the counterfeit is a light coin, and vice versa. One more weight trial will identify which of coins 1, 2 or 3 is counterfeit. (1 in one pan, 2 in the other; if they balance, then 3 is counterfeit; if one side is heavy and it was determined that the counterfeit is heavy, then that is your coin; if it was determined that the counterfeit is light, then the other coin is your coin.)


WHEN ONE PAN OUTWEIGHS THE OTHER.
In this case, all other coins are genuine. Designate the coins in the heavy pan as 1, 2, 3, 4 (if one of these coins is false, then it is heavier than the others) and the coins in the lighter pan by 1', 2', 3', 4' (if one of these coins is false, then it is lighter than the others). A second weight trial will be made by placing coins 1, 2, 1' in one pan and coins 3, 4, 2' in the other. Again, there are several possibilities:

(A) The pans balance. In this case, the counterfeit coin is either 3' or 4' (and is lighter than a genuine coin). A third weight trial is made by placing coin 3' in one pan and coin 4' in the other; the lighter coin will be counterfeit.

(B) The pan containing coins 1, 2 and 1' is heavier. In this case, coins 3, 4 and 1' are genuine; were either coin 3 or coin 4 heavier than the other, or were coin 1' light, then in the second weight trial the pan containing coins 3, 4 and 2' would have been heavy, which was not the result in this case. Therefore, the counterfeit coin is either coin 1 or 2 (and it is a heavier coin), or else it is coin 2' (and it is a lighter coin). A third weight trial is made, placing coin 1 in one pan and coin 2 in the other. If the pans balance, then the counterfeit coin is 2'. If the pans fail to balance, then the counterfeit coin is in the heavier pan.

(C) The pan containing coins 3, 4 and 2' is heavier. Reasoning as before, we conclude that coins 1, 2 and 2' are genuine and that if coin 3 or 4 is counterfeit, then it is a heavier coin than the others, and if coin 1' is the counterfeit, then it is lighter. A third weighing is made by placing coin 3 in one pan and coin 4 in the other. If the pans balance, then the counterfeit is 1'; if, on the other hand, one pan is heavier, then it contains the counterfeit coin.