Article #183
Subject: Pascal on Probability
Author: Andrew W. Harrell
Posted: 7/30/2012 11:34:28 AM
Pascal on Probability
The game of points studied by Fermat and Pascal consisted of a stake being
made by both players, then a fair coin being flipped a number of times.
The stake goes to whichever player has a given number of heads or the most
heads at the end of the flipping. If, at a certain point in the game [before
the given number of points has been achieved by either player] they decide to
quit and divide up winnings so far, the question comes up, “how is this to be
done?”. Thus the game is an example of what is called nowadays a Bernoulli
trial [the probability of the results of each flip are independent of the
results that have gone before]. It is clear from the fact that their are only
two possibilities for each throw that a given game of n points can only last
for at most 2n throws. And, if, at a certain point in game it m out of 2n
throws have already occured, then in order to figure out the solution to the
problem as stated, it is only necessary to consider the possibilities and
combinations of throws out to 2*(n-m) more throws.
After Fermat’s letter [see Great Books volume 33 reference] Pascal started
thinking about if the game was analyzed during it course of play how to we
define the probability for one player winning, considered what has happened
up to that time.
In order to do this he formulated a tentative definition, later to be made
more precise by Laplace: “The theory of chance consists in reducing all the
events of the same kind to a certain number of cases equally possible, that
is to say, to such as we may be equally undecided about in regard to their
existence, and in determining the number of cases favorable to the event
whose probability is sought. The ratio of this number to that of all the
cases possible is the measure of this probability, which is thus simply a
fraction whose numerator is the number of favorable cases and whose
denominator is the number of all the cases possible.” — Pierre-Simon Laplace,
A Philosophical Essay on Probabilities
Certain rules were postulated around this time by mathematicians studying in
this area (or assumed) as to how to calculate “probabilities p(E) of an event
E”
1) for all events 0 < p(E) < 1
2) p(impossible events) = 0. p(certain events)=1
3) p( not an event happening)= 1- p(E)
4) If two events, A,B are disjoint in occurence: p(A or B)= p(A) + p(B)
5) If two events A,B are with independently determined outcomes, but But
successive results of the same experiments p(A then B) =p(A and B)= p(A)*p(B)
In order to determine the ratio of favorable possible outcomes to unfavorable
at a particular point in the analysis of the partial results of the complete
results
Pascal computed a lot of the values of what we now call, due to Newton I
believe, binomial coefficients[ because they are the coefficients of the
expansion of the powers a binomial function (a + b)^n [(a+b) raised to the
power n] = sum of terms k=1 to n (n k) a^k*b^(n-k) (. If you think about it
in terms of how the powers of a^k*b^(n-l) you will understand that these
coefficients count the number of ways of picking k objects out of a set of n.
In the course of doing this he discovered the important step by step
relationship which allows us to compute a table of the values of these
coefficients for a given n, assuming we already know what the coefficients
are for the case of n-1.
Pascal's algorithm to compute his famous arithmetic triangle which can be
used to compute binomial coefficients First of all, by the definition of the
binomial coefficients in terms of combinations of k objects picked out of n
we know they have the property that the number is (k n)=n* (n-1)..(n-k+1)/k*
(k-1)…1
In Pascal’s paper he wrote his triangle rotated 45 degrees. But, in order to
understand If in relation to the binomial coefficients lets start out with it
in the form
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
It is defined inductively using the relation between the coefficients (which
is easy to verify from by writing out the above definition and getting a
common denominator):
(k n) = ( k n-1) + (k -1 n-1)
(k n)=n* (n-1)..(n-k+1)/k*(k*(k-1)..1= (n-1)*(n-2)…(n-k+1/k*(k-1)*(k-2)…1 +
(n-1)*(n-2)..(n-k)/(k-1)*(k-2)..1
Thus, for example, this relation says that the 2nd coefficient in the third
row (2 3)
Is the sum of the 2nd coefficient in the 2nd row (2 2) and the 1st
coefficient in the 2nd row (1 2).
Or writing it out:
3*2/2*1=6/2=3=2*1/2 +1/1=2+1=3
The formula can be proved algebraically, as above, using the definition of
the binomial coefficients in terms of factorials.
Or, a simple combinatorial argument using the definition in terms of k
subsets of n objects, can be given:
Of the set of n objects, let one be called T. All of the k subsets are
either of two types:
a) those that contain T (which number (k-1 n-1) )
b) those that do not (which number (k n-1)
thus (k n) = (k-1 n-1) + (k n-1) Q.E.D.
For the case n =6 referred to in his letter to Fermat [triangle is rotated
45 degrees here]
cells numbered along the top are called numbered by parallel rows [PLR]
cells numbered from top to bottom are called numbered by perpendicular rows
[PPR]
and cells numbered from the top left to the bottom center are called numbered
by the base [BR]
Thus, the algorithm says the for any cell (PLR=n PPR= k), its value [the
value in the cell]
AT(k n) = AT(k n-1)[the value in cell on its left] + AT(k-1 n)[the value just
on top of it]
1 2 3 4 5 6
1 1 1 1 1 1 1
2 1 2 3 4 5
3 1 3 6 10
4 1 4 10
5 1 5
6 1
In his essay Pascal after defining the triangle and its parameters he proves
some simple theorems about the triangle thus rotated:
1) Each cell is equal to the sum of all the cells of the preceeding row from
its own perpendicular row to the first, inclusive. Eg. the value at cell 3,3
= 6 which equals the values at cells 2,1 +2,2 + 2,3 = 1+ 2+ 6
2) In each such triangle the value at each cell exceeds by unity the sum of
all the cells within its parallel
And perpendicular rows, exclusive. Eg. the value at cell 3,3=6 which equals
the values at cells
3,1+3,2 +1,2+2,2 = 1+3+1+2=7
3) each cell is equal to its reciprocal eg. 3,2=3 is equal to 2,3=3
4) the sum of all the cells of each base is double that of the preceding
base. Eg.
sum of cells of base 3 1+3+3+1=8 ½ sum of cells of base 4 = 1+ 4+6+4+1=16
5) If we compare the sum of the coefficients [ (0 j) +(1 j) + (2 j) +... (j
j)] = X (in the triangle to their interpretation as polynomial coefficients
in the term by term expansion of (1+1)^j we find the fact that X = 2^j.
6) If, on the other hand, we look at the polynomial expansion of (1 -1)^j, we
find the fact that
[(0 j) - (1 j) + (2 j) - ... (-1)^j (j j)] = 0
Those interested can consult the excellent Mathematical Association of
America's (MAA) introductory text "Mathematics of Choice, How to Count
without Counting" by Dr. Ivan Niven and the more advanced one "Combinatorial
Mathematics" by Herbert Ryser for more details on these Theorems . Also, the
MAA's book, "The Mathematics of Games and Gambling" by Edward Packel has a
lot of information about how to use binomial coefficients to compute the
probabilities of results in backgammon, roulette and craps and also bridge
and poker hands. Binomial coefficients and the arithmetric triangle have many
applications. In addition to the theory of games and gambling already
mentioned, then can be used to count out and construct symmetrical
experimental designs, patterns of symmetry and tilings in continous and
discrete transformation groups, fractal image compression transformations
(see my UC Berkeley graduate mathematics department friend Hans Otto Pietgen
and his several Springer books on Fractals).
Another one of the applications that Pascal gives of the arithmetrical
triangle is to compute what he defines as the "expected payoff"
or "probabilistic expectation" or "mathematical expectation" at a given
position in a game of one player versus the other in the game of points.
Let E1, E2,E3,E4...EN be pairwise disjoint events ( no pair can occur
simulataneously) with their respective probabilities P1,P2,P3,P4...PN. Then
what we call the "mathematical expectation" in an experiment in which one of
these events must occur is defined to be equal to:
P1*E1+P2*E2+P3*E3+P4*E4 +...PN*EN
Pascal’s Theorem [used to compute expectation values for the game of points]:
In a game of points
in which each player lacks a certain number of points r[1st player] and s
[2nd player] to win:
In order to find the division of expectations [hopes for a win of the stakes]
Use the arithmetical triangle with base of order n=r+s and add up the
coefficients of the base of
The triangle of that order in that proportion [the sum of those r
coefficients is compared
To the sum of those s]
The sum of all the coefficients in the base row is equal to the sum of the
binomial coefficients of
the ways k objects can be taken from n[computed by means of the formula to be
equal to (k n)= n*(n-1)…(n-k+1)/k*(k-1)…1) where n is the number of the base
And k is the index running diagonally along the base.
In order to prove the validity of the formula for the expectations of each
player:
expectation of 1st player = number of ways of winning in last r+s throws/
total number of possibile throws
expectation of 2nd player = number of ways of winning in last r+s
throws/total number of possible throws
we use the 4th axiom listed above which define the probability of a series of
disjoint events {the throws} being the sum of the probability of each. And,
we also use the 3rd axiom to tell us that to get the probability at each
throw we multiply the independent particular combinations of probability of
separate dice outcomes times the total number of possible outcomes.
In order to prove the validity of the algorithm for all triangles [and hence
all game situations], since we have already identified the expectation as a
ratio of sums of binomial coefficients and we know the base rows in the
triangle are made up of these coefficients, defined inductively and recursely
by Pascal's formula (which already has been proved in two ways) the theorem
follows by induction on the base index of the triangle, by using base index
as the inductive index.
How is this theorem of Pascal used to compute the answer to the earlier
question
About stopping a game of stakes midway in the game?
In the game of stakes for n heads to win, with one player needing k more
and the other n-k, you can sum up the 1st k columns and compare it to the sum
of the last n-k columns and you have their respective "hopes with the odds "
for a final win.
For example, if two players have wagered money on being able to win three
times, and the 1st player has already won twice and the second once. If two
more throws happen there are four possibilities,
1) the 1st player wins both
2) the 1st player wins the next one and the second player the last one
3) the 2nd players win the next one and the the 1st the last one
4) the 2nd player wins both
In the first three cases the 1st player wins the game and in the fourth case
the 2nd player does.
By the accepted reasoning up to that time if you quit before the throws you
would divide up the wagers with ¾ to the 1st player and ¼ to the 2nd
But, according to Pascal’s “method of expectations.” we need to take into
account all the possible ways these things can happen in two different extra
throws. Computing the triangle we have:
1 1 1
1 2
1
. The 1st can win in 1 +2 ways according to Pascal and the 2nd in 1 way. Thus
the odds are 3 to 1. But, according to the argument of Cardano if would be 3
to 4.
By mathematical induction, and using the recursive properties of the binomial
coefficients (n k)(which can also be computed, interpreting them as
combinations as n*(n-1)…(n-k+1)/k*(k-1)…1)
triangle, Pascal was able to solve the problem for any number of points the
players might
lack.
In the case of 5 points to win, stopped when the 1st player has 3 points and
the second 2 points
1 2 3 4 5
1 1 1 1 1 1
2 1 2 3 4
3 1 3 6
4 1 4
5 1
the odds will be (1+4+6)=11 to (4+1)= 5 in the favor of the the 1st player
and so the money should be given to him in the ratio of 11/5.
And, in the case of 6 throws which corresponds to the 1st triangle listed
above, if the 1st player lacks four points and the 2nd lacks two, then their
shares are found by adding the numbers in the base of the triangle in Figure
1 2 3 4 5 6
1 1 1 1 1 1 1
2 1 2 3 4 5
3 1 3 6 10
4 1 4 10
5 1 5
6 1
1: the 1st
share is to the 2nd as 1+5+10+10 to 5+1, or 13 to 3.
Thus, considering this, the question is, how do these different approaches to
computing what "hope" means mathematically translate into differences in
theology?
Not an easy question, is it? Pascal noted in his essays that if we use the
same reasoning as above, eg. same axioms to define probability and same rules
to compute probabilistic expectation and apply it to the question of whether
it is reasonable to believe that God exists and that there is some argument
for holding this belief. Let the probability that God exists be P>0 . If
we believe that God exists and He does not then there will be some negative
payoff due to wasted time and energy and effort. Call this payoff -Z. If we
believe in Him and He does exists there will be some positive payoff. Call
this payoff X. If we do not believe in God and He does not exist there will
be a positive payoff. Call this payoff Y. However, if God does exists and we
do not believe in him the payoff will be a much worse, missing out on all the
benefits of His eternal help, and possibility suffering damnation. Call it -
infinity. This is because the effects of God existing and not are infinitely
positive and zero respectively.
x
So we have two mathematical expectations X(belief) and X(nonbelief):
X(belief)= P*X +(1-P)*(-z)
X(non-belief)= P*(-infinity)+(1-P)* Y= -infinity
Therefore when we sum the expectation formulas, at any given time in our
lives, as in the game of points the positive part is heavily outweighed by
the effect of the negative infinity part in the second of the formulas
summing up the possibilites.