Part Three.
The use of the Preceding Doctrine in Various Ways of Casting Lots and Games of Chance.
“Problem V”
“A contends with B that from 40 playing cards, that is, 10 of each suit, he will choose 4 cards such that there is one of each suit. What is the ratio of their lots?”
“To this end it is asked in how many ways 4 playing cards may be chosen from 40, that is, it is asked what is the number of quaternions in 40 things.”
The solution presented makes use of the Doctrine of Combinations from Part two.
For the first part we see from Part Two Chapter IV that there are = 91,390 possible combinations of ways to draw 4 cards from 40. However, this only takes into account being able to draw combinations that may have the same suit represented twice. In order to find out how many combinations there are when all suits are represented once, Bernoulli proposes an image of 4 ten-sided dice, with the 4 representing the number of suits and the 10 sides representing the number of cards in each suit. Now we see that the number of combinations that can be drawn when all suit are represented once will be the same as the number of possible throws with these dice. Namely: 10∙10∙10∙10 throws which equals 10,000 throws. Now to find the Ratio of their lots we see that if A has 10,000 combinations which will produce his desired lot, his opponent, B has 91,390-10,000 = 81,390 possible combinations which will NOT contain A’s desired outcome. Therefore, the ratio of their lots is 10,000 to 81,390 or 1,000 to 8,139.
Wednesday, April 21, 2010
Thursday, April 15, 2010
Bernoulli III
Part Two.
The Doctrine of Permutations and Combinations.
“…there is no error into which even the most prudent and circumspect more frequently fall than the error that the logicians call the insufficient enumeration of the parts. As a result, the Art called Combinatorics should be judged, as it merits, most useful, because it remedies this defect of our minds and teaches us how to enumerate … so that we may be sure that we have omitted nothing that can contribute to our purpose.”
“On Permutations”
Definition: “The permutations of things I call the variations according to which, preserving the same multitude of things, their order and position are changed in different ways. Thus, if one asks in how many ways several things can be transposed or mixed with each other, so that one always takes them all and changes only their order or position, one is said to ask for all the permutations of these things.” In this section, Bernoulli uses letters to represent different numbers. Thus, of course, if the same letter is used twice, it is taken to be the same number.
First we look at the case where all objects to be “permuted” are different from each other.
The synthetic way: Starting from the very simple, we see that one thing, a, can only be arranged one way, whereas two things, a,and b, can be arranged two ways. Namely, ab and ba. With three letters, a,b,c, we find the possible arrangements by see that when we choose one letter to go in the first place there are only two possible arrangements for the other two (as we just saw). Thus, if a is first, b and c can be written in only two ways: ab, ba. The same if we choose b to go first with a and c. So, for a total of three letters there are 3x2=6 permutations: abc, acb, bac, bca, cab, cba. Similarly, with four letters, each of them can be ordered to take the first place with the other three varying their order (in 3x2=6 ways) and since the first place can contain four different letters, we see that the first position can be written in four different ways with the remaining positions being written in six different ways. Therefore, we see that four letters can be written in 4x3x2=24 ways. This rule can be repeated to find the number of permutations with 5 letters and so on. Here Bernoulli gives a rule for finding all the permutations of any number of letters. “If all the numbers from one following in natural order up to the given number of things inclusively are multiplied together, the product will reveal what was sought. For instance, if the given number of things is n, the number of permutations will be 1∙2∙3∙4∙5 etc up to n.” This is exactly our definition of n factorial! (or written: n!)
Next he examines permutations when some of the things to be permuted are the same.
Bernoulli demonstrates that if you have a group of letters, say, aaabcd, with one of the letters occurring multiple times, if you were to just take those letters (aaa) and try to rearrange them you would have the same outcome no matter what. Therefore, it is as if these like letters are only one letter. It must be noted that, were they different they could be permutated in 3x2=6 different ways. To take this into account when finding the totally permutations of all the letter in aabcd, we must make the number of permutations six times less from the total number of permutations that would occur if all number had been diverse. If the six letters were different we would have 1∙2∙3∙4∙5∙6 = 720 permutations. But 3 of the letters are alike! Therefore we decrease 720 by a multiple of 6 to get 120. In the same way, if b is repeated twice, as in: aaabbc, then then number of permutation is now twice smaller, or there are 60 permutations.
Bernoulli’s Rules: Rule 1. “If one of the things recurs more than once, the number of permutations that the given things would admit [produce] if all were different should be divided by the number of permutations that the similar things can undergo given their number.”
Rule 2. “Or, if there are several that repeat more often than once, the number of permutations that the given things would admit [produce] if all were different should be divided by the product of the numbers of permutations that each group of similar things could undergo individually according to it’s multitude [according to the number of things in each group].”
Here he gives an example using transpositions of Latin words. The first word he uses is Rome. By the first rule, the permutations of the letters in “Rome” are 1∙2∙3∙4=24 because all the letters are different. However, “Leopoldus” has 2 letters repeating or 2x2=4 repetitions. Therefore, according to the second rule, we find the total number of permutations that would occur were all the letters different. Since there nine letters in “Leopoldus” the number of permutations would be 1∙2∙3∙4∙5∙6∙7∙8∙9 = 362,880. Now dividing this number by the number of repetitions multiplied together gives us = 90,720. For “Studiosus,” we again follow the second rule because there are two u’s and three s’s. With 9 total letters in the word, this gives us = 30,240. He goes even further to construct the different permutations of Latin poems, or Bauhusian Verse (which will not be included here).
Combinations
Chapter IV. “To find the number of combinations of single exponents separately; and to show at the same time in how many combinations one or more designated things are found together or separately.”
Rule for finding the combinations of a given exponent.
“Let there be two arithmetic progressions, one descending from the number of things to be combined, the other ascending from1 and let the common difference in each progression be 1 and let there be as many terms in either progression as the exponent of the desired combination. Let the product of the terms of the first progression be divided by the product of the terms in the second progression. The quotient will be the desired number of combinations sought.”
For example: If you have ten things, and you want to see how many ways you can take 4 things from 10, the formula is: = 5040/24 = 210.
This rule is used in Part Three as we shall now see.
The Doctrine of Permutations and Combinations.
“…there is no error into which even the most prudent and circumspect more frequently fall than the error that the logicians call the insufficient enumeration of the parts. As a result, the Art called Combinatorics should be judged, as it merits, most useful, because it remedies this defect of our minds and teaches us how to enumerate … so that we may be sure that we have omitted nothing that can contribute to our purpose.”
“On Permutations”
Definition: “The permutations of things I call the variations according to which, preserving the same multitude of things, their order and position are changed in different ways. Thus, if one asks in how many ways several things can be transposed or mixed with each other, so that one always takes them all and changes only their order or position, one is said to ask for all the permutations of these things.” In this section, Bernoulli uses letters to represent different numbers. Thus, of course, if the same letter is used twice, it is taken to be the same number.
First we look at the case where all objects to be “permuted” are different from each other.
The synthetic way: Starting from the very simple, we see that one thing, a, can only be arranged one way, whereas two things, a,and b, can be arranged two ways. Namely, ab and ba. With three letters, a,b,c, we find the possible arrangements by see that when we choose one letter to go in the first place there are only two possible arrangements for the other two (as we just saw). Thus, if a is first, b and c can be written in only two ways: ab, ba. The same if we choose b to go first with a and c. So, for a total of three letters there are 3x2=6 permutations: abc, acb, bac, bca, cab, cba. Similarly, with four letters, each of them can be ordered to take the first place with the other three varying their order (in 3x2=6 ways) and since the first place can contain four different letters, we see that the first position can be written in four different ways with the remaining positions being written in six different ways. Therefore, we see that four letters can be written in 4x3x2=24 ways. This rule can be repeated to find the number of permutations with 5 letters and so on. Here Bernoulli gives a rule for finding all the permutations of any number of letters. “If all the numbers from one following in natural order up to the given number of things inclusively are multiplied together, the product will reveal what was sought. For instance, if the given number of things is n, the number of permutations will be 1∙2∙3∙4∙5 etc up to n.” This is exactly our definition of n factorial! (or written: n!)
Next he examines permutations when some of the things to be permuted are the same.
Bernoulli demonstrates that if you have a group of letters, say, aaabcd, with one of the letters occurring multiple times, if you were to just take those letters (aaa) and try to rearrange them you would have the same outcome no matter what. Therefore, it is as if these like letters are only one letter. It must be noted that, were they different they could be permutated in 3x2=6 different ways. To take this into account when finding the totally permutations of all the letter in aabcd, we must make the number of permutations six times less from the total number of permutations that would occur if all number had been diverse. If the six letters were different we would have 1∙2∙3∙4∙5∙6 = 720 permutations. But 3 of the letters are alike! Therefore we decrease 720 by a multiple of 6 to get 120. In the same way, if b is repeated twice, as in: aaabbc, then then number of permutation is now twice smaller, or there are 60 permutations.
Bernoulli’s Rules: Rule 1. “If one of the things recurs more than once, the number of permutations that the given things would admit [produce] if all were different should be divided by the number of permutations that the similar things can undergo given their number.”
Rule 2. “Or, if there are several that repeat more often than once, the number of permutations that the given things would admit [produce] if all were different should be divided by the product of the numbers of permutations that each group of similar things could undergo individually according to it’s multitude [according to the number of things in each group].”
Here he gives an example using transpositions of Latin words. The first word he uses is Rome. By the first rule, the permutations of the letters in “Rome” are 1∙2∙3∙4=24 because all the letters are different. However, “Leopoldus” has 2 letters repeating or 2x2=4 repetitions. Therefore, according to the second rule, we find the total number of permutations that would occur were all the letters different. Since there nine letters in “Leopoldus” the number of permutations would be 1∙2∙3∙4∙5∙6∙7∙8∙9 = 362,880. Now dividing this number by the number of repetitions multiplied together gives us = 90,720. For “Studiosus,” we again follow the second rule because there are two u’s and three s’s. With 9 total letters in the word, this gives us = 30,240. He goes even further to construct the different permutations of Latin poems, or Bauhusian Verse (which will not be included here).
Combinations
Chapter IV. “To find the number of combinations of single exponents separately; and to show at the same time in how many combinations one or more designated things are found together or separately.”
Rule for finding the combinations of a given exponent.
“Let there be two arithmetic progressions, one descending from the number of things to be combined, the other ascending from1 and let the common difference in each progression be 1 and let there be as many terms in either progression as the exponent of the desired combination. Let the product of the terms of the first progression be divided by the product of the terms in the second progression. The quotient will be the desired number of combinations sought.”
For example: If you have ten things, and you want to see how many ways you can take 4 things from 10, the formula is: = 5040/24 = 210.
This rule is used in Part Three as we shall now see.
Tuesday, April 6, 2010
Bernoulli II
Part One.
Here Bernoulli sets forth some basic principles, taken from Huygen’s Treatise on Reckoning and Games of Chance on which he builds in the later sections, in fourteen Propositions followed by application in five Problems together with solutions. Propositions 9-14 concern the casting of lots with dice and are annotated by Bernoulli.
“ON DICE”
“The following questions can be asked about dice. With one die, in how many attempts might one undertake to throw a six, or any on of the other numbers of points? Similarly, with two dice, in how many attempts might one undertake to throw two sixes? Or with three dice, three sixes? There are many similar questions.
“To answer these questions consider the following. First, with one die there are six different throws, all of which can happen equally easily. (I assume that the die has the shape of a perfect cube.) Next, with two dice there are 36 different throws, which again can happen equally easily. This is because with each throw of the one die any one of the six throws of the other die can happen, and six times six equals 36. Similarly, there are 216 different throws with three dice, because any of the 36 throws of the two dice can happen together with each of the six throws of the third, and six times 36 makes 216 throws. In the same way it is clear that with four dice there are six times 216, or 1,296 throws. By continuing we can calculate the throws for any number of dice; every time we add another die we multiply the preceding number of throws by six.
“Next, note that there is just one way to get a 2 or12 with two dice, but two ways to get a 3 or 11. Indeed, if we call the dice A and B, then we can get 3 points either from 1 point on A and 2 on B or from 2 on A and 1 on B. Similarly, we can get 11 points either from 5 points on A and 6 on B or from 6 on A and 5 on B. There are three ways of throwing a 4, namely 1 on A and 3 on B, 3 on A and 1 on B, or 2 on A and 2 on B.
Similarly, there are three ways of throwing 10.
There are four ways of throwing 5 or 9.
There are five ways of throwing 6 or 8.
There are six ways of throwing 7.”
“With three dice to throw points, there are different throws.”
(Here, the “points” Bernoulli refers to are simply the sum of the dots on one or both dice.)
At this point, there is a note showing how, in the same manner, one can find the number of ways to throw any number of points with four or more dice. This proves to be very lengthy and difficult to keep track of the Modes (or variations), and so a table is given to show a more simple way of getting the same outcome. This table is easy to construct if you follow the directions carefully. It can be used for any number of dice, but is only constructed for up to six in the books table and in the reconstruction on the following page. The table is used by first choosing the Number of Dice you are using, (represented by the first set of Roman numerals on the top left) and following that row down to the number of points you wish to roll. Then, follow down the column to the section labeled Number of Throws for Given Number of Dice and find the section with Roman numerals equal to your chosen Number of Dice. This is the number of different rolls that can yield your chosen number of points. For example, let us say you have four dice and you wish to roll a total of “4.” You can find the number of possible ways this can occur by starting at the top of the page and following across the row with the Roman numeral IV over to the number 4. (Here it is the first number that occurs.) Now, follow the column down through three sections until you reach section IV. The number in the top row of this section, which lines up with “4” is the number of ways you can roll a “4.” Here the chart shows there is only 1 possible way to roll a “4” with four dice. The same is true if you wish to roll a total of 24 points with four dice – there is only 1 way. Follow the same protocol to find that, using four dice there are 4 throws that can add up to 5 or 23, ten different ways you can roll a 6 or 22, and twenty rolls that can yield a 7 or 21. Using a table such as this can help you keep track of your odds for rolling certain numbers and, consequently, how much you can safely bet on each roll.
This leads directly in to Part Two:
Here Bernoulli sets forth some basic principles, taken from Huygen’s Treatise on Reckoning and Games of Chance on which he builds in the later sections, in fourteen Propositions followed by application in five Problems together with solutions. Propositions 9-14 concern the casting of lots with dice and are annotated by Bernoulli.
“ON DICE”
“The following questions can be asked about dice. With one die, in how many attempts might one undertake to throw a six, or any on of the other numbers of points? Similarly, with two dice, in how many attempts might one undertake to throw two sixes? Or with three dice, three sixes? There are many similar questions.
“To answer these questions consider the following. First, with one die there are six different throws, all of which can happen equally easily. (I assume that the die has the shape of a perfect cube.) Next, with two dice there are 36 different throws, which again can happen equally easily. This is because with each throw of the one die any one of the six throws of the other die can happen, and six times six equals 36. Similarly, there are 216 different throws with three dice, because any of the 36 throws of the two dice can happen together with each of the six throws of the third, and six times 36 makes 216 throws. In the same way it is clear that with four dice there are six times 216, or 1,296 throws. By continuing we can calculate the throws for any number of dice; every time we add another die we multiply the preceding number of throws by six.
“Next, note that there is just one way to get a 2 or12 with two dice, but two ways to get a 3 or 11. Indeed, if we call the dice A and B, then we can get 3 points either from 1 point on A and 2 on B or from 2 on A and 1 on B. Similarly, we can get 11 points either from 5 points on A and 6 on B or from 6 on A and 5 on B. There are three ways of throwing a 4, namely 1 on A and 3 on B, 3 on A and 1 on B, or 2 on A and 2 on B.
Similarly, there are three ways of throwing 10.
There are four ways of throwing 5 or 9.
There are five ways of throwing 6 or 8.
There are six ways of throwing 7.”
“With three dice to throw points, there are different throws.”
(Here, the “points” Bernoulli refers to are simply the sum of the dots on one or both dice.)
At this point, there is a note showing how, in the same manner, one can find the number of ways to throw any number of points with four or more dice. This proves to be very lengthy and difficult to keep track of the Modes (or variations), and so a table is given to show a more simple way of getting the same outcome. This table is easy to construct if you follow the directions carefully. It can be used for any number of dice, but is only constructed for up to six in the books table and in the reconstruction on the following page. The table is used by first choosing the Number of Dice you are using, (represented by the first set of Roman numerals on the top left) and following that row down to the number of points you wish to roll. Then, follow down the column to the section labeled Number of Throws for Given Number of Dice and find the section with Roman numerals equal to your chosen Number of Dice. This is the number of different rolls that can yield your chosen number of points. For example, let us say you have four dice and you wish to roll a total of “4.” You can find the number of possible ways this can occur by starting at the top of the page and following across the row with the Roman numeral IV over to the number 4. (Here it is the first number that occurs.) Now, follow the column down through three sections until you reach section IV. The number in the top row of this section, which lines up with “4” is the number of ways you can roll a “4.” Here the chart shows there is only 1 possible way to roll a “4” with four dice. The same is true if you wish to roll a total of 24 points with four dice – there is only 1 way. Follow the same protocol to find that, using four dice there are 4 throws that can add up to 5 or 23, ten different ways you can roll a 6 or 22, and twenty rolls that can yield a 7 or 21. Using a table such as this can help you keep track of your odds for rolling certain numbers and, consequently, how much you can safely bet on each roll.
This leads directly in to Part Two:
Thursday, April 1, 2010
Bernoulli I
In this series of blog posts, I will be discussing Bernoulli's book "The Art of Conjecturing." I have been studying this book for one of my Math classes and I will present here some of my research.
Preface
Jacob Bernoulli, older brother of Johann Bernoulli I, was born in 1654 and died in 1705. He lived in Basel, Switzerland all his life and after much hard work became the professor of mathematics at Basel University in 1687, serving until his death – at which time his brother, Johann promptly received the position. The Bernoulli brothers were the beginning of several generations of Bernoullis who made significant contributions to science and mathematics. For this reason, one can be easily confused about who is being mentioned when only the last name “Bernoulli” is used. To further complicate matters, there is repeated use of family names in the Bernoulli family tree making it difficult to distinguish which “Jacob,” “Johann,” and especially “Nicolaus” is being referred to. Also, there are many different spelling variations of each name in circulation, e.g. Jacob is also known as Jakob, James, Jacobus, and Jacques.
For clarity, here is a copy of a portion of the Bernoulli family tree as found in The Art of Conjecturing:
Jacob conceived to write about “the art of conjecturing” in 1685. His first published evidence of this interest is in the Journal des Scavans for August 26, 1685 where he presents a problem concerning the throwing of dice and proposes the question as to the ratios of their lot. There is also evidence of his interest in probability found in his research journal Meditationes, and from letters that he wrote; among them “Letter to a friend on sets in court tennis,” which is included at the end of the book The Art of Conjecturing. In fact, in one of his public disputations given in competition for professorship in Mathematics at Basel, he discussed what appears as his 11th Corollary in Ars Conjectandi, concerning lottery slips drawn from an urn.
Ars Conjectandi was never completed and fell to the care of his son, Nicolaus, who later gave the manuscript to the Thurneysen brothers who published it in 1713. Though parts of Ars Conjectandi were translated and published one-by-one through the years, the last section of the book was finally completed as Bernoulli wished and The Art of Conjecturing was published in 2006. It is the first complete translation into English, and includes notes and commentary. Although other mathematicians began to write articles concerning probability contemporaneously with Bernoulli, “Ars Conjectandi deserves to be considered the founding document of mathematical probability” (Hopkins, vii).
It is truly amazing to see the beginnings of probability and statistics and read through Bernoulli’s thoughtful and thorough explanations. Though notes are often given on any particular proposition, many times they are not needed to fully grasp Bernoulli’s original notes due to his clear and explicit writing style. The Art of Conjecturing is divided up into 4 parts: Part One: “Nuygens’s Treatise on Reckoning in Games of Chance,” Part Two: “The Doctrine of Permutations and Combinations,” Part Three: “The Use of the Preceding Doctrine in Various Ways of Casting Lots and Games of Chance,” Part Four: “The Use and Application of the Preceding Doctrine in Civil, Moral, and Economic Matters.” Here we will discuss a few propositions and conclusions from each of these four parts and see how they fit together.
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