**Carnival Dice Game**

**A Classic Puzzle by Sam Loyd**

The following dice game is very popular at fairs and carnivals, but since two persons seldom agree on the chances of a player winning, I offer it as an elementary problem in the theory of probability. On the counter are six squares marked 1, 2, 3, 4, 5, 6. Players are invited to place as much money as they wish on any one square. Three dice are then thrown. If your number appears on one die only, you get your money back plus the same amount. If two dice show your number, you get your money back plus twice the amount you placed on the square. If your number appears on all three dice, you get your money back plus three times the amount. Of course if the number is not on any of the dice, the operator gets your money. To make this clearer with an example, suppose that you bet 1 dollar on No. 6. If one die shows a 6, you get your dollar back plus another dollar. If two dice show 6, you get back your dollar plus two dollars. If three dice show 6, you get your dollar back plus three dollars. A player might reason: the chance of my number showing on one die is 1 / 6, but since there are three dice, the chances must be 3 / 6 or 1 / 2, therefore the game is a fair one. Of course this is the way the operator of the game wants everyone to reason, for it is quite fallacious. Is the game favorable to the operator or the player, and in either case, just how favorable is it?

### Solution

Out of the 216 equally probable ways the dice may be thrown, you will win on only 91 of them, lose on 125. So your chance of winning at least as much as you bet is 91 / 216, your chance of losing 125 / 216. If the dice always showed different numbers, the game would be a fair one. Suppose all the squares covered with a dollar. The operator would, on each roll that showed three different numbers, take in three dollars and pay out three. But on doubles he makes a dollar and on triples he makes two dollars. In the long run, for every dollar wagered by a player, regardless of how he places the money, and in what amounts, he can expect to lose about 7.8 cents. This gives the operator a profit of 7.8 percent on each dollar bet.

**The St. Patrick’s Day Parade**

**A Classic Puzzle by Sam Loyd**

During a recent St. Patrick’s Day parade an interesting and curious puzzle developed. The Grand Marshall issued the usual notice setting forth that “the members of the Honorable and Ancient Order of Hibernians will parade in the afternoon if it rains in the morning, but will parade in the morning if it rains in the afternoon”. This gave rise to the popular impression that rain is to be counted as a sure thing on St. Patrick’s Day. Casey boasted that he “had marched for a quarter of a century in every St. Patrick’s Day parade since he had become a boy”. I will pass over the curious interpretations which may be made of the above remark, and say that old age and pneumonia having overtaken Casey at last, he had marched on with the immortal procession. When the boys met again to do honor to themselves and St. Patrick on the 17th of March, they found that there was a vacancy in their ranks which it was difficult to fill. In fact, it was such an embarrassing vacancy that it broke up the parade and converted it into a panic-stricken funeral procession. The lads, according to custom, arranged themselves ten abreast, and did march a block or two in that order with but nine men in the last row where Casey used to walk on account of an impediment in his left foot. The music of the Hibernian band was so completely drowned out by spectators shouting to ask what had become of “the little fellow with the limp”, that it was deemed best to reorganize on the basis of nine men to each row, as eleven would not do. But again Casey was missed, and the procession halted when it was discovered that the last row came out with but eight men. There was a hurried attempt to form with eight men in each row; again with seven, and then with five, four, three, and even two, but it was found that each and every formation always came out with a vacant space for Casey in the last line. The, although it strikes us as a silly superstition, it became whispered through the lines that every time they started off, Casey’s “dot and carry one” step could be heard. The boys were so firmly convinced that Casey’s ghost was marching that no one was bold enough to bring up the rear. The Grand Marshall, however, was a quick-witted fellow who speedily laid out that ghost by ordering the men to march in single file; so, if Casey did follow in spirit, he brought up the rear of the longest procession that ever did honor to his patron saint. Assuming that the number of the men in the parade did not exceed 7,000, can you determine just how many men marched in the procession?

### Solution

The number of men when Casey was alive must be a multiple of 2, 3, 4, 5, 6, 7, 8, 9 and 10. We take the least common multiple, 2520, then subtract 1 to get the number of members without Casey. This could be the answer were it not for the catch phrase, “as eleven would not do”. Since 2,519 is divisible by 11 we have to go the the next highest multiple, 5040, then subtract 1 to get 5039. Since this is not divisible by 11, and since higher multiples will give answers above 7,000, we conclude that 5039 is the only correct answer.

**Milkman’s Puzzle**

**A Classic Puzzle by Sam Loyd**

Honest John says: “What I don’t know about milk is scarcely worth mentioning,” but he was flabbergasted one day when each of two ladies asked him for two quarts of milk. One lady had a five-quart pail and the other had a four-quart pail. John had only two ten-gallon cans, each full of milk. How did he measure out exactly two quarts of milk for each lady? It is a juggling trick pure and simple, devoid of trick or device, but it calls for much cleverness to get two quarts of milk into those two pails without making use of any receptacles other than the two pails and the two full cans.

### Solution

**Dividing His Flocks**

**A Classic Puzzle by Sam Loyd**

A Western rancher, finding himself well advanced in years, called his boys together and told them that he wished to divide his herds between them while he yet lived. “Now, John,” he said to the eldest, “you may take as many cows as you think you could conveniently care for, and your wife Nancy may have one ninth of all the cows left.” To the second son he said, “Sam, you may take the same number of cows that John took, plus one extra cow because John had the first pick. To your good wife, Sally, I will give one ninth of what will be left.” To the third son he made a similar statement. he was to take one cow more than the second son, and his wife was to have one ninth of those left. The same applied to the other sons. Each took one cow more than his next oldest brother, and each son’s wife took one ninth of the remainder. After the youngest son had taken his cows, there were none left for his wife. Then the rancher said: “Since horses are worth twice as much as cows, we will divide up my seven horses so that each family will own livestock of equal value.” The problem is to tell how many cows the rancher owned and how many sons he had.

### Solution

The rancher had seven sons and fifty-six cows. The eldest son took two cows, and his wife took six. The next son took three cows, and his wife five. the next son took four and his wife four, and so on down to the seventh son who took eight cows, leaving none for his wife. Curiously, each family now has eight cows, so each took one of the seven horses to make their livestock of equal value.

**The Damaged Engine**

**A Classic Puzzle by Henry Ernest Dudeney**

We were going by train from Anglechester to Clinkerton, and an hour after starting an accident happened to the engine. We had to continue the journey at three-fifths of the former speed. It made us two hours late at Clinkerton, and the driver said that if only the accident had happened fifty miles farther on the train would have arrived forty minutes sooner. Can you tell from that statement just how far it is from Anglechester to Clinkerton?

### Solution

The distance from Anglechester to Clinkerton must be 200 miles. The train went 50 miles at 50 m.p.h. and 150 miles at 30 m.p.h. If the accident had occurred 50 miles farther on, it would have gone 100 miles at 50 m.p.h. and 100 miles at 30 m.p.h.

**The Man and the Dog**

**A Classic Puzzle by Henry Ernest Dudeney**

“Yes, when I take my dog for a walk,” said a mathematical friend, “he frequently supplies me with some interesting puzzle to solve. One day, for example, he waited, as I left the door, to see which way I should go, and when I started he raced along to the end of the road, immediately returning to me; again racing to the end of the road and again returning. He did this four times in all, at a uniform speed, and then ran at my side the remaining distance, which according to my paces measured 27 yards. I afterwards measured the distance from my door to the end of the road and found it to be 625 feet. Now, if I walk 4 miles per hour, what is the speed of my dog when racing to and fro?”

### Solution

The dog’s speed was 16 miles per hour. The following facts will give the reader clues to the general solution. The distance remaining to be walked side by side with the dog was 81 feet, the fourth power of 3 (for the dog returned four times), and the distance to the end of the road was 625 feet, the fourth power of 5. Then the difference between the speeds (in miles per hour) of man and dog (that is, 12) and the sum of the speeds (20) must be in the same ratio, 3 to 5, as is the case.

**Crossing the River**

**A Classic Puzzle by Henry Ernest Dudeney**

During the Turkish stampede in Thrace, a small detachment found itself confronted by a wide and deep river. However, they discovered a boat in which two children were rowing about. It was so small that it would only carry the two children, or one grown person. How did the officer get himself and his 357 soldiers across the river and leave the two children finally in joint possession of their boat? And how many times need the boat pass from shore to shore?

### Solution

The two children row to the opposite shore. One gets out and the other brings the boat back. One soldier rows across; soldier gets out, and child returns with boat. Thus it takes four crossings to get one man across and the boat brought back. Hence it takes four times 358, or 1432 journeys, to get the officer and his 357 men across the river and the children left in joint possession of their boat.

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