SOLUTION TO 30. TWO QUESTIONS IN PROBABILITIES. (2/4)

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(c) 4 heads and 1 tail 5 ways
(d) 4 tails and 1 head 5 ways
(e) 3 heads and 2 tails 10 ways
(f) 3 tails and 2 heads 10 ways

Now, it will be seen that the only favourable cases are a, b, c,
and d--12 cases. The remaining 20 cases are unfavourable, because they do not give at least four heads or four tails. Therefore the chances are

SOLUTION TO 30. TWO QUESTIONS IN PROBABILITIES. (1/4)

In tossing with the five pennies all at the same time, it is obvious that there are 32 different ways in which the coins may fall, because the first coin may fall in either of two ways, then the second coin may also fall in either of two ways, and so on. Therefore five 2's multiplied together make 32. Now, how are these 32 ways made up? Here they are:--

(b) 5 tails 1 way

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at least four of the coins will turn up either all heads or all tails?"
His own solution was quite wrong, but the correct answer ought not to be hard to discover. Another person got a wrong answer to the following little puzzle which I heard him propound: "A man placed three sovereigns and one shilling in a bag. How much should be paid for permission to draw one coin from it?" It is, of course, understood that you are as likely to draw any one of the four coins as another. (2/2)

30. TWO QUESTIONS IN PROBABILITIES. There is perhaps no class of puzzle over which people so frequently blunder as that which involves what is called the theory of probabilities. I will give two simple examples of the sort of puzzle I mean. They are really quite easy, and yet many persons are tripped up by them. A friend recently produced five pennies and said to me: "In throwing these five pennies at the same time, what are the chances that (1/2)

SOLUTION TO 123. A LEGAL DIFFICULTY.

It was clearly the intention of the deceased to give the son twice as much as the mother, or the daughter half as much as the mother. Therefore the most equitable division would be that the mother should take two-sevenths, the son four-sevenths, and the daughter one-seventh.

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were born--a boy and a girl. A very nice point then arose. How was the estate to be equitably divided among the three in the closest possible accordance with the spirit of the dead man's will?" (2/2)

123. A LEGAL DIFFICULTY.
"A client of mine," said a lawyer, "was on the point of death when his wife was about to present him with a child. I drew up his will, in which he settled two-thirds of his estate upon his son (if it should happen to be a boy) and one-third on the mother. But if the child should be a girl, then two-thirds of the estate should go to the mother and one-third to the daughter. As a matter of fact, after his death twins (1/2)

SOLUTION TO 146. AN EASY DISSECTION PUZZLE.

The solution to this puzzle is shown in the illustration. Divide the figure up into twelve equal triangles, and it is easy to discover the directions of the cuts, as indicated by the dark lines.

146. AN EASY DISSECTION PUZZLE. First, cut out a piece of paper or cardboard of the shape shown in the illustration. It will be seen at once that the proportions are simply those of a square attached to half of another similar square, divided diagonally. The puzzle is to cut it into four pieces all of precisely the same size and shape.

SOLUTION TO 106. THE MUDDLETOWN ELECTION. (2/2)

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successive majorities from the last-mentioned number.

SOLUTION TO 106. THE MUDDLETOWN ELECTION. (1/2)

The numbers of votes polled respectively by the Liberal, the Conservative, the Independent, and the Socialist were 1,553, 1,535,
1,407, and 978 All that was necessary was to add the sum of the three majorities (739) to the total poll of 5,473 (making 6,212) and divide by 4, which gives us 1,553 as the poll of the Liberal. Then the polls of the other three candidates can, of course, be found by deducting the

106. THE MUDDLETOWN ELECTION. At the last Parliamentary election at Muddletown 5,473 votes were polled. The Liberal was elected by a majority of 18 over the Conservative, by 146 over the Independent, and by 575 over the Socialist. Can you give a simple rule for figuring out how many votes were polled for each candidate?

SOLUTION TO 277. COUNTER CROSSES. (6/6)

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division by 2 cancel one another. Hence 10,368 is here the correct answer.

SOLUTION TO 277. COUNTER CROSSES. (5/6)

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lower arms are unequal in length, permutations will repeat by reflection, but not by reversal, for we cannot reverse. Therefore this fact only entails division by 2. But in every pair we may exchange the figures in the upright with those in the horizontal (which we could not do in the case of the Greek Cross, as the arms are there all alike);
consequently we must multiply by 2. This multiplication by 2 and

SOLUTION TO 277. COUNTER CROSSES. (4/6)

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is thus 2,592 different ways. The division is by 4 and not by 8, because we provided against half the reversals and reflections by always reserving one number for the upright and the other for the horizontal.

In the case of the Latin Cross, it is obvious that we have to deal with the same 18 forms of pairing. The total number of different ways in this case is the full number, 18 × 576. Owing to the fact that the upper and

SOLUTION TO 277. COUNTER CROSSES. (3/6)

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× 4 = 24 ways. And as the four in the horizontal may also be changed in 24 ways for every arrangement on the other arm, we find that there are 24 × 24 = 576 variations for every form; therefore, as there are 18 forms, we get 18 × 576 = 10,368 ways. But this will include half the four reversals and half the four reflections that we barred, so we must divide this by 4 to obtain the correct answer to the Greek Cross, which

SOLUTION TO 277. COUNTER CROSSES. (2/6)

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Of course, the number in the middle is common to both arms. The first pair is the one I gave as an example. I will suppose that we have written out all these crosses, always placing the first row of a pair in the upright and the second row in the horizontal arm. Now, if we leave the central figure fixed, there are 24 ways in which the numbers in the upright may be varied, for the four counters may be changed in 1 × 2 × 3

SOLUTION TO 277. COUNTER CROSSES. (1/6)

Let us first deal with the Greek Cross. There are just eighteen forms in which the numbers may be paired for the two arms. Here they are:--

12978 13968 14958
34956 24957 23967

23958 13769 14759
14967 24758 23768

12589 23759 13579
34567 14768 24568

14569 23569 14379
23578 14578 25368

15369 24369 23189
24378 15378 45167

24179 25169 34169
35168 34178 25178

SOLUTION TO 122. THE SEE-SAW PUZZLE.

The boy's weight must have been about 39.79 lbs. A brick weighed 3 lbs. Therefore 16 bricks weighed 48 lbs. and 11 bricks 33 lbs. Multiply 48 by 33 and take the square root.

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were fixed to the short end of plank, but if he fixed them to the long end of plank he only needed eleven as balance.

Now, what was that boy's weight, if a brick weighs equal to a three-quarter brick and three-quarters of a pound? (2/2) 