Part 20

1. To multiply this number by itself 36 2. So take 1 from the number thought of 5 3. To multiply this by itself 25 4. To tell you the difference between this product and the former 11 You must then add 1 to it 12 And halve this number 6

Which will be the number thought of.

_Fifth Method._

Desire a person to think of a number, say 6. He must then proceed as follows:

EXAMPLE.

1. Add 1 to it 7 2. Multiply by 3 21 3. Add 1 again 22 4. Add the number thought of 28

Let him tell you the figures produced (28):

5. You then subtract 4 from it 24 6. And divide by 4 6

Which you can say is the number thought of.

_Sixth Method._

EXAMPLE.

Suppose the number thought of 6 1. Let him double it 12 2. Desire him to add to this any number you tell him, say 4 16 3. To halve it 8

You can then tell him that if he will subtract from this the number he thought of, the remainder will be, in the case supposed, 2.

_Note._--The remainder is always half of the number you tell him to add.

TO DISCOVER TWO OR MORE NUMBERS THAT A PERSON HAS THOUGHT OF.

_1st Case._--Where each of the numbers is less than 10. Suppose the numbers thought of were 2, 3, 5.

EXAMPLE.

1. Desire him to double the first number making 4 2. To add 1 to it 5 3. To multiply by 5 25 4. To add the second number 28

There being a third number, repeat this process--

5. To double it 56 6. To add 1 to it 57 7. To multiply by 5 285 8. To add the third number 290

And to proceed in the same manner for as many numbers as were thought of. Let him tell you the last sum produced (in this case 290). Then, if there were two numbers thought of, you must subtract 5; if three, 55; if four, 555. You must here subtract 55, leaving a remainder of 235, which are the numbers thought of, 2, 3 and 5.

_2d Case._--Where one or more of the numbers are 10, or more than 10, and where there is an _odd_ number of numbers thought of.

Suppose he fixes upon five numbers, viz. 4, 6, 9, 15, 16.

He must add together the numbers as follows, and tell you the various sums:

1. The sum of the 1st and 2d 10 2. The sum of the 2d and 3d 15 3. The sum of the 3d and 4th 24 4. The sum of the 4th and 5th 31 5. The sum of the 1st and last 20

You must then add together the 1st, 3d and 5th sums, viz. 10 + 24 + 20 = 54, and the 2d and 4th, 15 + 31 = 46; take one from the other, leaving 8. The half of this is the 1st number, 4; if you take this from the sum of the 1st and 2d you will have the 2d number, 6; this taken from the sum of the 2d and 3d will give you the 3d, 9; and so on for the other numbers.

_3d Case._--Where one or more of the numbers are 10, or more than 10, and where an _even_ number of numbers has been thought of.

Suppose he fixes on six numbers, viz. 2, 6, 7, 15, 16, 18. He must add together the numbers as follows, and tell you the sum in each case:--

1. The sum of the 1st and 2d 8 2. The sum of the 2d and 3d 13 3. The sum of the 3d and 4th 22 4. The sum of the 4th and 5th 31 5. The sum of the 5th and 6th 34 6. The sum of the 2d and last 24

You must then add together the 2d, 4th and 6th sums, 13 + 31 + 24 = 68, and the 3d and 5th sums, 22 + 34 = 56. Subtract one from the other, leaving 12; the 2d number will be 6, the half of this; take the 2d from the sum of the 1st and 2d you will get the 1st; take the 2nd from the sum of the 2d and 3d, and you will have the 3d, and so on.

HOW MANY COUNTERS HAVE I IN MY HANDS?

A person having an equal number of counters in each hand, it is required to find how many he has altogether.

Suppose he has 16 counters, or 8 in each hand. Desire him to transfer from one hand to the other a certain number of them, and to tell you the number so transferred. Suppose it be 4, the hands now contain 4 and 12. Ask him how many times the smaller number is contained in the larger; in this case it is 3 times. You must then multiply the number transferred, 4, by the 3, making 12, and add the 4, making 16; then divide 16 by the 3 _minus_ 1; this will bring 8, the number in each hand.

In most cases fractions will occur in the process: when 10 counters are in each hand, and if 4 be transferred, the hands will contain 6 and 14.

He will divide 14 by 6 and inform you that the quotient is 2 3/6 or 2⅓.

You multiply 4 by 2⅓, which is 9⅓.

Add 4 to this, making 13⅓, equal to 40/3.

Subtract 1 from 2⅓, leaving 1⅓ or 4/3.

Divide 40/3 by 4/3 giving 10, the number in each hand.

THE MYSTERIOUS HALVINGS.

_To tell the number a person has thought of._

One of the company must fix upon any one of the numbers from 1 to 15; this he keeps secret, as well as the numbers produced by the succeeding operations:

Suppose he fixes on 8 He must add 1 to it, making 9 Triple it 27 Halve it[11]--1st _halving_--(_larger half_) 14 Triple it 42 Halve it--2d _halving_ 21 Triple it 63 Halve it--3d _halving_--(_larger half_) 32 Triple it 96 Halve it--4th _halving_. 48

He need not inform you that 48 is the figure produced, but he must let you know in which of the four halvings he was obliged to take a "larger half;" having ascertained this point, you discover the number fixed upon in the following manner. Carry in your mind, or on a slip of paper, the following list of names in which the letter a occurs in one or more of the three syllables of all except the last.

The three syllables are intended to represent the 1st, 2d, and 3d halvings, and the occurrence of the letter A corresponds to the occurrence of a "larger half" in one or more of these three halvings. Having been informed where the _larger half_ was taken, refer to the word which has A in the corresponding syllable, and against it stand two numbers, one of which was the number thought of; and of these two, the right hand number is the correct one _if a larger half was taken in the 4th stage_, and the left hand one _if the, 4th halving was exact_.

In the example given, a _larger half_ occurred in the 1st and 3d stage; this points us to _Car-ro-way_, and the halving in the 4th stage being exact, shows us that 8 was the number fixed upon.

If the 4th halving If a _larger half_ occurs is _exact_. in the 4th halving. WAsh-ing-ton 4 12 LA-fAy-ette 2 10 CAr-row-wAy 8 0 MAn-hAt-tAn 6 14 Ger-mA-ny 13 5 Tel-e-grAph 3 11 Bo-nA-pArte 1 9 Long-fel-low 15 7

It will be observed that there is always a difference of 8 between the numbers of the columns, so that it is necessary to recollect only one of them. Perhaps some of our readers who wish to be adepts in this game, would prefer recollecting the above table if put in this form:

2-3 1-2 3 1-2-3 1-3 2 none -------------------------------- 1 2 3 4 8 13 15

where the upper line denotes the cases in which the "larger half" was taken, and the lower line the numbers of the left hand column above given.

_Another Method._

The person having chosen any number from one to fifteen, he is to add twenty-one to that number, and triple the amount. Then,

1st. He is to take half of that triple, and triple that half.

2nd. To take the half of the last triple, and triple that half.

3rd. To take the half of the last triple.

4th. To take the half of the last half.

In this operation there are four distinct cases or stages where the half is to be taken. The three first are denoted by one of the eight following Latin words, each word being composed of three syllables, and the syllables containing the letter i corresponding in numerical order with the cases where the half cannot be taken without a fraction; consequently, in those cases the person who makes the deduction is to add one to the number to be divided. The fourth case shows which of the two numbers corresponding to each word has been chosen. For if the fourth half can be taken without adding one, the number chosen is in the first, or left-hand column; but if not, it is in the second column to the right.

The words. The numbers denoted. Mi-ser-is 8 0 Ob-tin-git 1 9 Ni-mi-um 2 10 No-tar-i 3 11 In-fer-nos 4 12 Or-di-nes 13 5 Ti-mi-di 6 14 Te-ne-ant 15 7

_Example._--Suppose the number chosen to be nine, to which is to be added one, making ten, and which last, being tripled, gives thirty. Then:

1st case. The half of the triple is 15 which tripled, makes 45 2nd case. The half of that triple, 1 being added to make an even number, is 23 and that tripled, makes 69 3rd case. The half of the last triple, 1 being added, is 35 4th case. The half of the last half, 1 being again added, is 18

Here we see, that in the second and third case, one had to be added, and, looking at the table, we find that the only corresponding word having an i in its second and third syllables is _Ob-tin-git_, which represents the figures one and nine. Then, as one had to be added in the fourth case, we know by the rule, that the figure in the second column, 9, is the one required. Observe, that if no addition be required at any of the four stages, the number thought of will be fifteen; and if one addition only be required at the fourth stage, the number will be seven.

WHO WEARS THE RING?

This is an elegant application of the principles involved in discovering a number fixed upon. The number of persons participating in the game should not exceed nine. One of them puts a ring on one of his fingers, and it is your object to discover--1st. The wearer of the ring. 2d. The hand. 3d. The finger. 4th. The joint.

The company being seated in order the persons must be numbered 1, 2, 3, &c.; the thumb must be termed the first finger, the fore finger being the second; the joint nearest the extremity must be called the first joint; the right hand is one, and the left hand two.

These preliminaries having been arranged, leave the room in order that the ring may be placed unobserved by you. We will suppose that the third person has the ring on the right hand, third finger, and first joint; your object is to discover the figures 3131.

Desire one of the company to perform secretly the following arithmetical operations:

1. Double the number of the person who has the ring; in the case supposed, this will produce 6 2. Add 5 11 3. Multiply by 5 55 4. Add 10 65 5. Add the number denoting the hand 66 6. Multiply by 10 660 7. Add the number of the finger 663 8. Multiply by 10 6630 9. Add the number of the joint 6631 10. Add 35 6666

He must apprise you of the figures now produced, 6666; you will then in all cases subtract from it 3535; in the present instance there will remain 3131, denoting the person No. 3, the hand No. 1, the finger No. 3, and the joint No. 1.

PROBABILITIES.[12]

When we look around us at results happening daily, of the causes of which we are ignorant, we are led to regard them as isolated incidents subject to no law or rule; but could we see and understand the secret workings and connection existing between cause and effect, we might frequently discover that all works by rule. As it is, we may readily mark the boundaries, within which events must happen in very many instances; and do much to estimate their probability. We speak of _Chance_, as something without plan or design, but taking in a large range, our calculations will approximate closely to the truth. When we throw a copper into the air, the chances of "heads or tails," as the boys say, are equal, and though one or the other may occur most frequently for a few throws, in a large number, say a thousand, the results will be about equally divided. In this case the sides of the coin must be equal in weight, else it will be like the grumbler's bread and butter:

"I never had a piece of bread, Particularly good and wide, But fell upon the sanded floor, And always on the buttered side."

Had he put on less butter, perhaps the sides would have been more equal in weight, and the probability of the buttered side being uppermost would have been increased. Disturbing causes unknown to us, may often shape the result; but in the absence of these, we may pretty accurately estimate our chances.

We see accidents from fire and flood, happening at times and points least expected; but the insurer has learned by observation to estimate probabilities, and by taking a wide range of country and a period of years, he does a comparatively safe business. Death takes the young and the old; but the life insurer has conned the bills of mortality and studied the ages of those who have died, until he can estimate at once the probability of duration of life, and determine what he can afford to pay for an annuity contingent on life, or engage for a present sum, or an annual sum paid for life, to pay the heirs at the death of the insured. In one instance his estimate may fall short, and in another exceed, but the average will be about right.

So, too, the man who deals in lotteries and games of chance, knows the data and calculates carefully the probabilities, and though "luck" may sometimes be against him, his estimates of probabilities are based on mathematical principles, and he is secure in being ultimately the gaining party.

How these chances are calculated, depends on the data in each case, and it is not within the range of our present plan to attempt more than giving a general idea of the subject; and this with any one of ordinary prudence, will be sufficient to prevent all intermeddling with lotteries and every other species of gambling. The probabilities are always against the casual operator, even if all be conducted fairly; what then must they be when fraud and dishonesty are superadded? It is downright swindling!

In lottery schemes generally, fifteen per cent. is reserved as profit, but this is a small part of what may be secured; yet even this amounts to a great deal. If a man were to draw a prize nominally of $100,000, fifteen thousand would be deducted at once, and he would be entitled to only $85,000. It is true that in his good fortune he would not probably regard the abatement, but that does not change the principle.

VARIATIONS.

It is obvious that if we have a number of single things arranged in any order, we may change the arrangement into a variety of forms, and in doing so, we may take all together, or we may take only part at once. For instance, we may arrange the six vowels, a e, i, o, u, y, in a great number of ways, as a e i o u y, a i e o u y, e a i o u y, &c., &c.; or we may form them into groups, as ae, io, uy, ai, eu, oy, &c.; or, we may take three, four, five, or, as above, all at a time; and it is reasonable to suppose that the number of possible changes may, in all cases, be calculated.

When all are taken together, the operation is called _Permutation_; but if a part only be taken, it is called either a _Variation_ or a _Combination_; a e, i o, u y, are distinct combinations, and are also considered one of the variations of two of which those six letters are susceptible; e a, o i, y u, are three other variations, but they are the same combinations; for a change of order will constitute a new variation but not a new combination; hence the number of variations will always exceed the number of combinations.

The doctrine of variations and combinations forms the basis of many forms of lotteries, and of other calculations used in practical life.

COMBINATIONS AND PERMUTATIONS.

"Combinations" are the different ways in which a certain number of things can be selected out of a larger number, when taken 1 at a time, 2 at a time, or any other number each time, but without regard to the order in which the selected numbers can be arranged among themselves. The latter is the province of "Permutation," which refers to the different ways in which a number can be selected out of one that is larger, and, _in addition to this_, to the different ways of _grouping_ these selected numbers.

Thus 4 things can be taken 2 at a time in 6 different ways; for instance, the letters a, b, c, d, can be taken 2 at a time thus, a and b, a and c, a and d, b and c, b and d, c and d; if we regard the _order_ of the selected letters we shall find that these 4 letters are capable of 12 different permutations, as ab, ba, ac, ca, ad, da, bc, cb, bd, db, cd, dc.

If we selected 3 letters at a time we could make 4 different selections, and 24 different changes of grouping.

The rule to compute the number of these different ways is very simple, but sometimes involves a multitude of figures.

To determine the number of permutations, commence with unity, and multiply by the successive terms of the natural series 1, 2, 3, &c., until the highest multiplier shall express the number of individual things. The last product will indicate the number of possible changes.

_Example 1._ How many changes can be made in the arrangement of 5 grains of corn, all of different colors, laid in a row?

_Solution._ 1 × 2 × 3 × 4 × 5 = 120, _Ans._

This may seem improbable, the number being so great, but if there were but a single grain more, the possible changes would be 720; and another would extend the limit to 5040; and so onward in a constantly increasing ratio. The reason, however, will be obvious on a little scrutiny. If there were but one thing, as _a_, it would admit of but one position; but if two, as _a b_, it would admit of two positions, _ab_, _ba_. If three things, as _a b c_, then they will admit of 1 × 2 × 3 = 6 changes, for the last two will admit of two variations, as _a b c_, _a c b_, and each of the three may successively be placed first, and two changes made to each of the others, so that 3 × 2 = 6, the number of possible changes. In the same way we may show that if there be four individual things, each one will be first in each of the six changes which the other three will undergo, and consequently, there will be 24 changes in all. In this way we might show that when there are 5 individual things, there will be 5 times as many changes as when there were but 4; and when 6, there will be 6 times as many changes as when there are only 5; and so on _ad infinitum_, according to the same law.

_Example 2._ In how many ways may a family of 10 persons seat themselves differently at dinner? _Ans._ 3,628,800.

When we consider that this would require a period of 9935-55/487 years, the mind is lost in astonishment. The story of the man who bought a horse at a farthing for the first nail in his shoe, a penny for the second, &c., is thrown into the shade; and we incline to doubt whether there is not some mistake; and yet on just such chances as one to all these, do gamblers constantly risk their money!

_Example 3._ I have written the letters contained in the word N I M R O D on 6 cards; being one letter on each, and having thrown them confusedly into a hat, I am offered $10 to draw the cards successively, so as to spell the name correctly. What is my chance of success worth? _Ans._ 1-7/18 cents.

_Example 4._ In order to form a lottery scheme, I have put into the wheel as many cards as I can put 4 letters of the word Charleston on, without having the same letters in the same order upon any two cards. I offer $100 to him who draws the card having on it the first four letters of the said word in their natural order (Char). What is the chance of drawing a prize worth?

There are 10 letters in the word, and the combination is of the 4th class; and, according to the mode of determining combinations with repetitions, we find the whole number of combinations of the 4th class which the word admits of is 210. Then he has one chance in 210 of drawing the letters Char, in _some_ order. The number of permutations of four individual things is 1 × 2 × 3 × 4 = 24, and 210 × 24 = 5040 his chance of drawing them in the right order, and $100 divided by 5040 gives _Ans._ 1 52/63 cents.

Suppose that the numbers from 1 to 78, inclusive, be placed upon 78 cards, and the cards placed in a wheel by which they are thoroughly mixed; and then 13 cards be successively drawn out, by a person who has no means of choosing, and the numbers on them registered. Suppose also that tickets have been issued, containing each three of the 78 numbers, but no two having _all_ the same numbers, and that he who holds the ticket having on it the first three drawn numbers in their regular order, shall be entitled to $100,000; what would the probability of drawing such a ticket be worth?

_Ans._ 21 5183/5858 cents.

_Note._--It is usual also, to give smaller prizes to the holders of tickets having the numbers in any order, or having any two or one of the drawn numbers. Lotteries may be arranged on a great diversity of plans, and in each the probability of drawing prizes will vary.

A speaks the truth 3 times in 4; B 4 times in 5, and C 6 times in 7. What is the probability of an event which A and B assert, and C denies? _Ans._ 140/143

Suppose a coin be thrown up, having two faces; what is the probability that the obverse (heads) side will fall upward, and what the reverse?

Here there are only two possible cases, and one favors each of the contingencies the probability of each will be 1/(1 + 1) = 1/2; there being no reason why one side should fall uppermost rather than the other.

What would be the probability of either side presenting upwards twice in two throws?

Here we have 4 possible cases, viz.:

Obverse and reverse; Obverse both times; Reverse and obverse; Reverse both times.

Of the 4 possibilities there is only one which favors the turning up of the obverse twice in succession, and the same is true of the reverse, hence the probability of either is only 1/4.

In like manner we might show that the probability of the obverse presenting upwards three times in succession will be 1/8, or 1/2 × 1/2 × 1/2; the general principle being to multiply successively together the independent probabilities of an event for the fraction expressing the chance of all the events happening.

THE VISITORS TO THE CRYSTAL PALACE.