History Of Mathematics Prepared by CHU Lap foo chulapfoo ... · Pascalto Fermat(July 29, 1654) Education Bureau Pascal Mathematics Education Section Your method is very sound and

History Of MathematicsPrepared by CHU Lap‐[email protected]

Problem of Points

Two players, A and B, agree to play a series of fairgames until one of them has won 4 games (4‐point game). For some accidental reason, the playis stopped when A has won 2 and B 1 games. Howshould the stakes be divided? Suppose the stakefrom each player are 24 pistoles.

A 24 pistoles

B 24 pistoles

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1st Attempt

A 24 pistoles

B 24 pistoles

Pacioli (1494): divide the stakes as2 : 1

A should get 32 pistoles and B should get 16pistoles


1st Attempt

How if

A1 : 0 ???

B

A2 : 1 ???

B


Other Attempts

A 24 pistoles

B 24 pistoles

Tartaglia (1556): A should get 24 + [(2 – 1)/4] 24,i.e. 30 pistolesB should get 18 pistolesDivide as 5 : 3


Other Attempts

Cardano (1539), Forestani (1603)[A History of Probability and Statistics and Their Applications before 1750]

Pascal and Fermat (1654)The problem was introduced to them by Chevalier de Mere[A Source Book in Mathematics II]


Fermat to Pascal (1654)

A, B have equal chance in a 4‐point gameA: 2 points; B: 1 point (stakes = 24 + 24 pistoles)

A 24 pistoles

B 24 pistoles

At most 2 + 3 – 1 = 4more games to play


Fermat to Pascal (1654)

Fermat’s method (Combination):the possible arrangements are:


Fermat

No. of outcomes favorable to A = 11, No. of outcomes favorable to B = 5

So A should get 11/16 of the total, i.e. 33 pistolesand B should get 15 pistoles.

a a a a a a a a b b b b b b b b

a a a a b b b b a a a a b b b b

a a b b a a b b a a b b a a b b

a b a b a b a b a b a b a b a b

A A A A A A A B A A A B A B B B

Pascal to Fermat (July 29, 1654)

Education Bureau

PascalMathematics Education Section

Your method is very sound and it is the first one that cameto my mind in these researches, but because the trouble ofthese combinations was excessive, I found an abridgmentand indeed another method that is much shorter and moreneat, which I should like to tell you here in a few words;for I should like to open my heart to you henceforth if Imay, so great is the pleasure I have had in our agreement.I plainly see that the truth is the same at Toulouse and atParis.


A, B have equal chance in a 4‐point gameA: 2 points; B: 1 point (stakes = 24 + 24 pistoles)Pascal’s method (Recursion):

First consider the case whenA: 3 points; B: 2 points

Mathematics Education Section Education Bureau



f(m,n) = # of pistoles A can get when A gets m points and B gets n points

f(2,1)=?f(3,2)=?

f(3,2)

f(4,2)

f(3,3)

=48

=24

=24+(4824)0.5=24+12=36

f(3,1)

f(4,1)=48

=36+(4836)0.5=36+6=42f(2,1)

f(2,2)=24

=24+(4224)0.5=24+9=33

Same as Fermat’s result


A, B have equal chance in a s‐point game



f(m,n+1)

f(m,n)

f(m+1,n)

f(m,n) = [f(m+1,n) + f(m,n+1)] / 2

Pascal attempted to solve this recursion by using Pascal Triangle



42312424

232424

21

21

2424)1,2(

3

44

43

42

42

CCC

Cf

Another way to express f(2,1) (Pascal Triangle)

86427531

21

21

642531

21

21

4231

21

21

88

87

86

85

84

84

66

65

64

63

63

44

43

42

42

CCCCC

C

CCCC

C

CCC

C

At most 4 more games to play



11 1

1 2 11 3 3 1

1 4 6 4 144

43

42

41

40 CCCCC

832424

21

21

2424

21

21

21

24

224

248)1,2(

44

43

42

42

44

43

42

44

43

42

42

3

44

43

42

4

44

43

42

CCC

C

CCC

CCCC

CCC

CCCf

1 4 3 3 4 1

B:24 pistoles A:24 pistoles3/8 from B to A

A will win in the following cases:2a2b, 3a1b and 4a0b




4‐point game

f(3,2)

f(4,2)

f(3,3)

=48

=24

=24+(4824)0.5=24+12=36

f(3,1)

f(4,1)=48

=36+(4836)0.5=36+6=42f(2,1)

f(2,2)=24

=24+(4224)0.5=24+9=33




5‐point game

f(4,3)

f(5,3)

f(4,4)

=48

=24

=24+(4824)0.5=24+12=36

f(4,2)

f(5,2)=48

=36+(4836)0.5=36+6=42f(3,2)

f(3,3)=24

=24+(4224)0.5=24+9=33


One can see thatf(2,1) (in 4‐point game)

= f(3,2) (in 5‐point game)= f(4,3) (in 6‐point game)= ….

The division rule should depend on the number of games each player lacks in winning



Using the Pascal Triangle, a general division rule can be found:Suppose A lacks 3 points and B lacks 4 points,at most 6more games to play, Awill win in the following cases:

3a3b, 4a2b, 5a1b, 6a0bThe fraction of stakes A should get is


6

66

65

64

63

2CCCC

Pascal to Fermat (August 24, 1654)

Pascal mentioned Fermat’s method to M. de Roberval


A lacks 2 pointsB lacks 3 points

At most 4 more games to play

Objection from M. de Roberval:Not necessary to play FOUR games.They may play TWO, THREE or FOUR games.

a a a a a a a a b b b b b b b b

a a a a b b b b a a a a b b b b

a a b b a a b b a a b b a a b b

a b a b a b a b a b a b a b a b

A A A A A A A B A A A B A B B B


Pascal applied Fermat’s method to the problemfor three players, he suspected that in this casethe method failed:

It seems to me that this is the way in which it isnecessary to make the division by combinationsaccording to your method, unless you have somethingelse on the subject which I do not know. But if I am notmistaken, this division is unjust.



3 players A, B and C


At most 3 more games

i.e. 33 = 27 possible arrangements

Solution 1 A : B : C = 19 : 7 : 7

A lacks 1 pointB lacks 2 pointsC lacks 2 points

Count all favorable outcomes

a a a a a a a a a b b b b b b b b b c c c c c c c c c

a a a b b b c c c a a a b b b c c c a a a b b b c c c

a b c a b c a b c a b c a b c a b c a b c a b c a b c

A A A A A A A A A A A A A A A A A A A

B B B B B B B

C C C C C C C






Solution 2 A : B : C = 16 : 5.5 : 5.5


Count half of the outcomes favorable to two players





B B B B B B B

C C C C C C C





B B B B B B B

C C C C C C C







Solution 3 A : B : C = 17 : 5 : 5

Delete some favorable outcomes

Agree with Pascal’s method (Recursion)



I believe you have recognized from this that the theory ofcombinations is good for the case of two players byaccident, as it is also sometimes good in the case of threegamblers, as when one lacks one point, another one, andthe other two, because, in this case, the number of pointsin which the game is finished is not enough to allow two towin, but it is not a general method and it is good only inthe case where it is necessary to play exactly a certainnumber of times.

Fermat to Pascal (September 25, 1654)

3 players A, B and CFermat’s verification:


Chance that A wins in 1 throw (a) = 1/3Chance that A wins in 2 throws (ba or ca) = 2/9Chance that A wins in 3 throws (bca or cba) = 2/27

Total chances = 1/3+2/9+2/27 = 17/27


Fermat to Pascal (September 25, 1654)

Using Fermat’s combinatorial method, another general division rule can be found:Suppose A lacks 3 points and B lacks 4 points,Awill win in the following cases:1. 2a+a 2. 2a1b+a3. 2a2b+a 4. 2a3b+aFraction of stake that A should get is


6

52

5

42

4

32

3

22

2222CCCC

6

66

65

64

63

2CCCC

Equivalent Division Rules

Suppose A lacks m points and B lacks n points, atmost m+n–1 more games to play, the probabilitythat Awill win is:Pascal’s Division Rule (Binomial Distribution):

Fermat’s Division Rule (Negative Binomial Distribution):


1

21

11

11

222

nm

nmm

m

mm

m

mm CCC

1

11

11

1

2

nm

nmnm

nmm

nmm CCC

Summing Up

Pascal and Fermat introduced:

Combinatorial MethodApplication of the Arithmetic Triangle (Pascal

Triangle)Recursion (Difference Equation)Expectation

in Probability Theory


CHU Lap‐foo May [email protected]

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History Of Mathematics Prepared by CHU Lap foo chulapfoo ... · Pascalto Fermat(July 29, 1654) Education Bureau Pascal Mathematics Education Section Your method is very sound and