1.The sum of the 1st and 2d102.The sum of the 2d and 3d153.The sum of the 3d and 4th244.The sum of the 4th and 5th315.The sum of the 1st and last20
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 anevennumber 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 2d82.The sum of the 2d and 3d133.The sum of the 3d and 4th224.The sum of the 4th and 5th315.The sum of the 5th and 6th346.The sum of the 2d and last24
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 numberof 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 3minus1; 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 22/6or 21/3.
You multiply 4 by 21/3, which is 91/3.
Add 4 to this, making 131/3, equal to40/3.
Subtract 1 from 21/3, leaving 11/3or4/3.
Divide40/3by4/3giving 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 on8He must add 1 to it, making9Triple it27Halve it[11]—1sthalving—(larger half)14Triple it42Halve it—2dhalving21Triple it63Halve it—3dhalving—(larger half)32Triple it96Halve it—4thhalving.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 letterAcorresponds to the occurrence of a "larger half" in one or more of these three halvings. Having been informed where thelarger halfwas taken, refer to the word which hasAin 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 oneif a larger half was taken in the4th stage, and the left hand oneif the, 4th halving was exact.
In the example given, alarger halfoccurred in the 1st and 3d stage; this points us toCar-ro-way, and the halving in the 4th stage being exact, shows us that 8 was the number fixed upon.
If the 4th halving isexact.If alarger halfoccurs in the 4th halving.WAsh-ing-ton412LA-fAy-ette210CAr-row-wAy80MAn-hAt-tAn614Ger-mA-ny135Tel-e-grAph311Bo-nA-pArte19Long-fel-low157
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 2none——————————————123481315
2-3 1-2 3 1-2-3 1-3 2none——————————————123481315
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-is80Ob-tin-git19Ni-mi-um210No-tar-i311In-fer-nos412Or-di-nes135Ti-mi-di614Te-ne-ant157
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 is15which tripled, makes452nd case.The half ofthattriple, 1 being addedto make an even number, is23and that tripled, makes693rd case.The half of the last triple, 1 being added, is354th case.The half of the last half, 1being again added, is18
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 isOb-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 produce62.Add 5113.Multiply by 5554.Add 10655.Add the number denoting the hand666.Multiply by 106607.Add the number of the finger6638.Multiply by 1066309.Add the number of the joint663110.Add 356666
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 ofChance, 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."
"I never had a piece of bread,Particularly good and wide,But fell upon the sanded floor,And always on the buttered side."
"I never had a piece of bread,Particularly good and wide,But fell upon the sanded floor,And always on the buttered side."
"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 calledPermutation; but if a part only be taken, it is called either aVariationor aCombination; 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 thebasis 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 ofgroupingthese 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 theorderof 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.
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 littlescrutiny. If there were but one thing, asa, it would admit of but one position; but if two, asa b, it would admit of two positions,ab,ba. If three things, asa b c, then they will admit of 1 × 2 × 3 = 6 changes, for the last two will admit of two variations, asa b c,a c b, and each of the three may successively be placed first, and two changes made to each ofthe 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 onad 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.17/18cents.
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, insomeorder. 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 givesAns.152/63cents.
Suppose that the numbers from 1 to 78, inclusive, be placed upon 78 cards, and the cards placed in a wheel bywhich 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 havingallthe 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.215183/5858cents.
Ans.215183/5858cents.
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 be1/(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.
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 insuccession, and the same is true of the reverse, hence the probability of either is only1/4.
In like manner we might show that the probability of the obverse presenting upwards three times in succession will be1/8, or1/2×1/2×1/2;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.
In a family consisting of 8 young people, it was agreed that 3 at a time should visit the Crystal Palace, and that the visit should be repeated each day as long as a different trio could be selected. In how many days were the possible combinations of 3 out of 8 completed?
We must multiply 8 × 7 × 6, and also 3 × 2 × 1, and divide the product of the former, 336, by the product of the latter, 6; the result is 56, the number of visits, a different three going each time. So much gratified were they with the results of their agreement, that they wished to be allowed another series of visits, to be continued as many days as they could group 3 together in different order when starting. If Paterfamilias had granted such permission he would have had to wait 56 multiplied by 3 × 2 × 1, or 336 days, before this "new series" of visits would have come to afinis.
HOW MANY CHANGES CAN BE GIVEN TO 7 NOTES OF A PIANO?
That is to say, in how many ways can 7 keys be struck in succession, so that there shall be some difference in the order of the notes each time?
The result of multiplying
7 × 6 × 5 × 4 × 3 × 2 × 1
is 5,040, the number of changes.
THE ARITHMETICAL TRIANGLE.
This name has been given to a contrivance said to have originated with the famous Pascal, or to have been perfected by him.
12 13 3 14 6 4 15 10 10 5 16 15 20 15 6 17 21 35 35 21 7 18 28 56 70 56 28 8 1&c.&c.
12 13 3 14 6 4 15 10 10 5 16 15 20 15 6 17 21 35 35 21 7 18 28 56 70 56 28 8 1&c.&c.
This peculiar series of numbers is thus formed: Write down the numbers 1, 2, 3, &c., as far as you please, in a vertical row. On the right hand of 2 place 1, add them together, and place 3 under the 1; then 3 added to 3 = 6, which placeunder the 3; 4 and 6 are 10, which place under the 6, and so on as far as you wish. This is the second vertical row, and the third is formed from the second in a similar way.
This triangle has the property of informing us, without the trouble of calculation, how many combinations can be made, taking any number at a time out of a larger number.
Suppose the question were that just given; how many selections can be made of 3 at a time out of 8? On the horizontal row commencing with 8, look for the third number; this is 56, which is the answer.
HOW MANY DIFFERENT DEALS CAN BE MADE WITH 13 CARDS OUT OF 52?
To discover this we must make a continued multiplication of 52 × 51 × 50 × 49 × 48 × 47 × 46 × 45 × 44 × 43 × 42 × 41 × 40, being 13 terms for the 13 cards, also a continued multiplication of 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1, and having found the two products, we must divide one by the other, and the quotient is the number of different deals out of 52 cards. This "sum," that looks so formidable with natural figures, is a very short one by logarithms.
THE THREE GRACES.
Three articles, or three names inscribed on cards, having been distributed between three persons, you are to tell which article or card each person has.
Designate the three persons in your own mind, as 1st, 2d, and 3d, and the three articles,A,E,I. Provide 24 counters, and give 1 to the first person, 2 to the 2d, 3 to the 3d. Place the remaining 18 on the table. Request that the three persons will distribute among themselves the three articles, and that, having done so, the person who has the one which you have secretly denoted byA, will take as many counters as he may have already; the holder ofEmust take twice as many as he may have; and the holder ofImust take four times as many. Then leave the room, in order that the distribution of articles and of counters may be made unobserved by you. We will suppose that the three articles are three cards, on which are the words Clara, Rosa, Emily, which you will yourself secretly denote by the lettersA,E,I. Suppose also that in the division the first person has Emily (I), the second has Clara (A), and the third has Rosa (E), then the 1st will take four times as many counters as he has (1), and will therefore take 4; the 2d will take as many ashe has (2), and will therefore take 2; the 3d will take 6, being twice as many as he has (3). On the table will be left 6 counters. The distribution having been made, you will return and observe the number of counters on the table, from which you can find who is the holder of each card by the following method.
It is plain that if the cards held by the 1st and 2d can be told, that held by the 3d will be known. It will be found that only six numbers can remain, viz. 1, 2, 3, 5, 6, 7; never 4, and never more than 7. Now the 6 combinations of a, e, and i, here given, represent the articles held by the 1st and 2d persons.
1234567aeeaai—eiiaie
In the case supposed, 6 counters being on the table, the combinationiaindicates that the first person has the card you have calledI(Emily), the 2d hasA(Clara), so that the 3d hasE(Rosa).
In order to recollect the combinations ofA,E, andI, it will be best to keep in memory some 7 words which form a sentence, and which contain these vowels in the order just given.
Our young friends can amuse themselves in forming a sentence for themselves, but as examples we supply three.
1234567aeeaai—eiiaieJameseasyadmiresnowreigningwith abride.Anger,fear,painmaybe hidwith asmile.GracefulEmma,charmingshereignsin allcircles.
Or, if they prefer Latin, they can use the pentameter made up by the inventor of this beautiful pastime:
123567Salvecertaanimæsemitavitaquies.
ANOTHER METHOD.
The performer must mentally distinguish the articles by the lettersA,B,C, and the persons as 1st, 2d, and 3d. The persons having made their choice, give 12 counters to the 1st, 24 to the 2d, and 36 to the 3d. Then request the 1st person to add together the half of the counters of the person who has chosenA, the 3d of the person who has chosenB,and the 4th of those of the person who has chosenC, and then ask the sum, which must be either 23, 24, 25, 27, 28, or 29, as in the following table:
First.Second.Third.122436ABC23ACB24BAC25CAB27BCA28CBA29
This table shows that if the sum be 25, for example, the 1st person must have chosenB, the 2dA, and the 3dC; or if it be 28, the 1st must have chosenB, the 2dC, and the 3dA.
ANOTHER METHOD.
Three things having been divided between three persons, you are to determine the holder of each.
Call the persons in your own mind 1st, 2d, 3d.
Give to the 1st a card on which you have written the number 12; to the 2d the number 24; to the 3d 36.
The three things you must denote asA,E,I.
To simplify it you may have three cards with a name upon each, of which the initial letters areA,E,I, as Anna, Emma, Isabel.
Request your friends to divide between them the three articles, and then to add together certain parts of the numbers on their cards, as follows:
Whoever hasAmust supply one half of the number on his card;
Whoever hasEmust supply one third;
Whoever hasImust supply one fourth;
This half, third and fourth having been added together, the sum must be announced to you on your return; and from this number you can tell who hasA, who hasE, and who hasI.
If the No. isthe 1st hasthe 2d hasthe 3d has23AEI24AIE25EAI27IAE28EIA29IEA
The sum which will be given to you can be one of six only. There are only six ways in which the articles can be divided, and there is a definite number for each of them.
The number 26 can never occur, and to recollect the six which do occur, and which you perceive are consecutive, you need take note only of what the 1st and 2d persons have.
23242526272829aeaiea—iaeiie
If you make up a line of good (or bad) English, having the vowels in the order here given, you will find it will aid you in their recollection. We give one as a specimen:
aeaiea—iaeiieBravedashingsea,like agiantrevivesitself.
THE FORTUNATE NINTH.
A sharp youth, fresh from school, having gone to visit a good-natured uncle, the latter placed on a table fifteen fine oranges and fifteen apples, and desired his young friend to take half. He, not liking the apples, was about to take the fifteen oranges; but this monopoly of the best fruit being objected to, the old gentleman told him to range all the fruit in a circle, and to take every ninth. The clever fellow ranged them in such a way as that, by taking away every ninth, all the apples were left on the table, and all the oranges were transferred to his capacious pockets. How did he arrange them?
He placed them as in the margin, A representing apples, and O oranges; and it will be found that, by commencing at the four apples, and going round and round the circle, taking away every ninth, all the oranges will be removed, and all the apples will remain.
If we let the vowelsaeioudenote the figures12345,
the arrangement of the figures 4, 5, 2, 1, &c., can be easily recollected by the following line: