Background
Probability Theory
Pascal's Triangle & Probability
Application of Probability Theory
Pascal's wager
Objections
Homework
Joyce Lam Nga Ching
2001714828
Phil1007
12-4-2002
27-4-2002
Probability
Theory
Use of Pascal's Triangle
The triangle is easily compiled. Each line is formed by adding together each pair of adjacent numbers in the line above. The first thing to notice about the triangle is how neatly line 5 summarises the five tosses of a coin (there are a total of 32 possible results of which one contains no heads, five contain 1 head, ten contain 2 heads, ten contain 3 heads, five contain 4 heads and one contains 5 heads). The Triangle is of great interest to gamblers, and provides the answer to questions relating to equipartition and combinations.
1.Combinations and permutations
First, look at figure 1.
Looking at row five of Pascal's triangle because it represents a combination of five items. Adding up the sum of the numbers on this row, one should arrive at 32. This is the total possible number of combinations that can be made with five items.
| 1 | ||||||||||
| 1 | 1 | |||||||||
| 1 | 2 | 1 | ||||||||
| 1 | 3 | 3 | 1 | |||||||
| 1 | 4 | 6 | 4 | 1 | ||||||
| 1 | 5 | 10 | 10 | 5 | 1 |
Figure 1
The binomial expansion of Pascal's triangle shows the possible ways of getting each combination of items.
| (a+b)0 | = | 1 | ||||||||||
| (a+b)1 | = | 1a | + | 1b | ||||||||
| (a+b)2 | = | 1a2 | + | 2ab | + | 1b2 | ||||||
| (a+b)3 | = | 1a3 | + | 3a2b | + | 3ab2 | + | 1b3 | ||||
| (a+b)4 | = | 1a4 | + | 4a3b | + | 6a2b2 | + | 4ab3 | + | 1b4 | ||
| (a+b)5 | = | 1a5 | + | 5a4b | + | 10a3b2 | + | 10a2b3 | + | 5ab4 | + | 1b5 |
(a) Looking at row five of Pascal’s triangle, one sees that there are 32 possible ways to choose five a and b, but there is only one way to choose five a. Therefore, the probability is 1/32.
(b) Looking
at row five of Pascal’s triangle, one sees that there are 32 possible ways to
choose five a and b, but there are five ways to choose four a and one b.
Therefore, the probability is 5/32.
(c) Looking
at row five of Pascal’s triangle, one sees that there are 32 possible ways to
choose five a and b, but there are ten ways to choose three a and two b.
Therefore, the probability is 10/32, which reduces to 5/16.
(d)
Looking at row five of
Pascal’s triangle, one sees that there are 32 possible ways to choose five a
and b, but there are ten ways to choose two a and three b.
Therefore, the probability is 10/32, which reduces to 5/16.
(e) Looking
at row five of Pascal’s triangle, one sees that there are 32 possible ways to
choose five a and b, but there are five ways to choose one a and four b.
Therefore, the probability is 5/32.
A
practical problem for gamblers is the calculation of combinations and
permutations. A gambler frequently wants to know how many different ways a
smaller number of items can be taken from a larger. This occurs in horse
racing, where the number of four-horse accumulators which can be taken from a
total of eight selections might be required, or in selecting draws for a
treble-chance football pool, where perhaps the number of combinations of eight
draws from ten selections are needed.
Each
of these problems is an exercise in calculating combinations, although British
football pools companies and pools journalists always refer for some reason to
the second example, as a permutation.
The
answer to both problems is contained in Pascal's Triangle. Modern calculators
and spreadsheets embed a simpler way of working out these calculations. In Excel
use the combin(n,r) function, where n is the number of articles and r is the
repetitions (combinations).
The
full formula is =(n!/((n-r)!*r!))
2.Equipartition
Another
glance at Pascal's Triangle will answer the following question. When
tossing a coin, what is the probability that after n events, the number of heads
will equal the number of tails? Clearly this can happen only with an even
number of events. Those who believe the 'law of averages' fallacy maintain that
the probability of equipartition, as it is called, increases with the number of
events. Pascal's Triangle proves the opposite. Line 4 shows that if
we toss four times, there are 16 Possible outcomes, of which six contain two
heads and two tails. The Probability of equipartition is 6/16. Line
6 shows that with six tosses, the probability of equipartition is 20/64. Line 10
shows that with ten tosses, equipartition is a 252/1024 chance. The
probabilities are becoming smaller as the tosses increase. The formula to
discover the probability of equipartition in n events is to divide the number of
combinations which give equipartition by the number of possible outcomes.
Example
:
What
is the probability, when tossing six dice, of throwing each number once i.e.
achieving equipartition?
The
total number of ways equipartition can occur is 6! The first die can clearly be
any of the six numbers. the second die one of the five remaining, and so on,
giving a total number of ways of 6 x 5 x 4 x 3 x 2 x 1 = 720.
The
number of possible outcomes is power(6,6) since there are six ways each of the
six dice can fall. Therefore this is equal to 46656.
So the
probability of equipartition is:-
720/46656
= 0.0154 or 1.5432%
Most
people would be surprised to discover that if you threw six dice, on about 98.5%
of occasions, at least one number will appear more than once.
If
there are seven children in a family, what is the probability that they were
born on different days of the week?
This
is a question of equipartition, the answer being =fact(7)/power(7,7). The answer
is 0.6120%.
Some
gamblers might he tempted to base staking plants on the theory that in any
series of even-money events there must come a time sooner or later when the
outcomes reach equipartition. This theory is not true, and if a gambler
backs black at roulette, for example, and the first winner is a red, there is a
calculable chance that black will never catch up.
Go to Application of Probability Theory
Reference:1.http//curry.edschool.virginia.edu/teacherlink/content/math/interactive/probability/history/briefhistory/home.html