˜ ˜

(a) X ∪ Y = (X © Y ) ∪ (X © Y ) ∪ (X © Y ).

˜ ˜

(b) m(X ∪ Y ) = m(X © Y ) + m(X © Y ) + m(X © Y ).

(c) m(X ∪ Y ) = m(X) + m(Y ) ’ m(X © Y ).

23. If X, Y , and Z are any three sets, prove that, for any probability

measure,

m(X ∪ Y ∪ Z) = m(X) + m(Y ) + m(Z)

’m(X © Y ) ’ m(Y © Z) ’ m(X © Z)

+m(X © Y © Z).

24. Translate the result of Exercise 23 into a result concerning three

statements p, q, and r.

25. A man o¬ers to bet “dollars to doughnuts” that a certain event

will take place. Assuming that a doughnut costs a nickel, what

must the probability of the event be for this to be a fair bet?

20

[Ans. .]

21

26. Show that the inclusion-exclusion formula 3.2 is true is n is re-

placed by m. Apply this result to

Pr[p1 ∨ p2 ∨ . . . ∨ pn ].

93

4.3. THE EQUIPROBABLE MEASURE

4.3 The equiprobable measure

We have already seen several examples where it was natural to assign

the same weight to all possibilities in determining the appropriate prob-

ability measure. The probability measure determined in this manner

is called the equiprobable measure. The measure of sets in the case of

the equiprobable measure has a very simple form. In fact, if U has n

elements and if the equiprobable measure has been assigned, then for

any set X, m(X) is r/n, where r is the number of elements in the set

X. This is true since the weight of each element in X is 1/n, and hence

the sum of the weights of elements of X is r/n.

The particularly simple form of the equiprobable measure makes

it easy to work with. In view of this, it is important to observe that

a particular choice for the set of possibilities in a given situation may

lead to the equiprobable measure, while some other choice will not. For

example, consider the case of two throws of an ordinary coin. Suppose

that we are interested in statements about the number of heads which

occur. If we take for the possibility set the set U = {HH, HT, TH, TT}

then it is reasonable to assign the same weight to each outcome, and

we are led to the equiprobable measure. If, on the other hand, we

were to take as possible outcomes the set U = {no H, one H, two H},

it would not be natural to assign the same weight to each outcome,

since one head can occur in two di¬erent ways, while each of the other

possibilities can occur in only one way.

Example 4.4 Suppose that we throw two ordinary dice. Each die can

turn up a number from 1 to 6; hence there are 6 · 6 possibilities. We

1

assign weight 36 to each possibility. A prediction that is true in j cases

will then have probability j/36. For example, “The sum of the dice is

5” will be true if we get 1 + 4, 2 + 3, 3 + 2, or 4 + 1. Hence the

probability that the sum of the dice is 5 is 36 = 1 . The sum can be 12

4

9

1

in only one way, 6 + 6. Hence the probability that the sum is 12 is 36 .

™¦

Example 4.5 Suppose that two cards are drawn successively from a

deck of cards. What is the probability that both are hearts? There

are 52 possibilities for the ¬rst card, and for each of these there are

51 possibilities for the second. Hence there are 52 · 51 possibilities for

the result of the two draws. We assign the equiprobable measure. The

statement “The two cards are hearts” is true in 13 · 12 of the 52 · 51

94 CHAPTER 4. PROBABILITY THEORY

possibilities. Hence the probability of this statement is 13·12/(52·51) =

1

™¦

.

17

Example 4.6 Assume that, on the basis of a predictive index applied

to students A, B, and C when entering college, it is predicted that after

four years of college the scholastic record of A will be the highest, C the

second highest, and B the lowest of the three. Suppose, in fact, that

these predictions turn out to be exactly correct. If the predictive index

has no merit at all and hence the predictions amount simply to guessing,

what is the probability that such a prediction will be correct? There

are 3! = 6 orders in which the men might ¬nish. If the predictions were

really just guessing, then we would assign an equal weight to each of the

six outcomes. In this case the probability that a particular prediction is

1

true is 6 . Since this probability is reasonably large, we would hesitate

to conclude that the predictive index is in fact useful, on the basis

of this one experiment. Suppose, on the other hand, it predicted the

order of six men correctly. Then a similar analysis would show that,

1 1

by guessing, the probability is 6! = 720 that such a prediction would be

correct. Hence, we might conclude here that there is strong evidence

™¦

that the index has some merit.

Exercises

1. A letter is chosen at random from the word “random”. What is

the probability that it is an n? That it is a vowel?

11

[Ans. 6 ; 3 .]

2. An integer between 3 and 12 inclusive is chosen at random. What

is the probability that it is an even number? That it is even and

divisible by three?

3. A card is drawn at random from a pack of playing cards.

(a) What is the probability that it is either a heart or the king

of clubs?

7

[Ans. .]

26

(b) What is the probability that it is either the queen of hearts

or an honor card (i.e., ten, jack, queen, king, or ace)?

95

4.3. THE EQUIPROBABLE MEASURE

5

[Ans. .]

13

4. A word is chosen at random from the set of words

U = {men, bird, ball, ¬eld, book}.

Let p, q, and r be the statements:

p: The word has two vowels.

q: The ¬rst letter of the word is b.

r: The word rhymes with cook.

Find the probability of the following statements.

(a) p.

(b) q.

(c) r.

(d) p § q.

(e) (p ∨ q) § ¬r.

(f) p ’ q.

4

[Ans. .]

5

5. A single die is thrown. Find the probability that

(a) An odd number turns up.

(b) The number which turns up is greater than two.

(c) A seven turns up.

6. In the Primary voting example of Section 2.1, assume that all 36

possibilities in the elections are equally likely. Find

(a) The probability that candidate A wins more states than ei-

ther B or C.

7

[Ans. .]

18

(b) That all the states are won by the same candidate.

1

[Ans. .]

36

(c) That every state is won by a di¬erent candidate.

96 CHAPTER 4. PROBABILITY THEORY

[Ans. 0.]

7. A single die is thrown twice. What value for the sum of the two

outcomes has the highest probability? What value or values of

the sum has the lowest probability of occurring?

8. Two boys and two girls are placed at random in a row for a

picture. What is the probability that the boys and girls alternate

in the picture?

1

[Ans. .]

3

9. A certain college has 500 students and it is known that

300 read French.

200 read German.

50 read Russian.

20 read French and Russian.

30 read German and Russian.

20 read German and French.

10 read all three languages.

If a student is chosen at random from the school, what is the

probability that the student

(a) Reads two and only two languages?

(b) Reads at least one language?

10. Suppose that three people enter a restaurant which has a row

of six seats. If they choose their seats at random, what is the

probability that they sit with no seats between them? What is

the probability that there is at least one empty seat between any

two of them?

11. Find the probability of obtaining each of the following poker

hands. (A poker hand is a set of ¬ve cards chosen at random

from a deck of 52 cards.)

(a) Royal ¬‚ush (ten, jack, queen, king, ace in a single suit).

52

[Ans. 4/ = .0000015.]

5

(b) Straight ¬‚ush (¬ve in a sequence in a single suit, but not a

royal ¬‚ush).

97

4.3. THE EQUIPROBABLE MEASURE

52

[Ans. (40 ’ 4)/ = .000014.]

5

(c) Four of a kind (four cards of the same face value).

52

[Ans. 624/ = .00024.]

5

(d) Full house (one pair and one triple of the same face value).

52

[Ans. 3744/ = .0014.]

5

(e) Flush (¬ve cards in a single suit but not a straight or royal

¬‚ush).

52

[Ans. (5148 ’ 40)/ = .0020.]

5

(f) Straight (¬ve cards in a row, not all of the same suit).

52

[Ans. (10, 240 ’ 40)/ = .0039.]

5

(g) Straight or better.

[Ans. .0076.]

12. If ten people are seated at a circular table at random, what is

the probability that a particular pair of people are seated next to

each other?

2

[Ans. .]

9

13. A room contains a group of n people who are wearing badges

numbered from 1 to n. If two people are selected at random, what

is the probability that the larger badge number is a 3? Answer

this problem assuming that n = 5, 4, 3, 2.

112

[Ans. ; ; ; 0.]

533

14. In Exercise 13, suppose that we observe two men leaving the room

and that the larger of their badge numbers is 3. What might we

guess as to the number of people in the room?

15. Find the probability that a bridge hand will have suits of

(a) 5, 4, 3, and 1 cards.

4!(13)(13)(13)(13)

[Ans. = .129.]

5 4 3 1

(52)

13

98 CHAPTER 4. PROBABILITY THEORY

(b) 6, 4, 2, and 1 cards.

[Ans. .047.]

(c) 4, 4, 3, and 2 cards.

[Ans. .216.]

(d) 4, 3, 3, and 3 cards.

[Ans. .105.]

16. There are 52 = 6.5 — 1011 possible bridge hands. Find the

13

probability that a bridge hand dealt at random will be all of one

suit. Estimate roughly the number of bridge hands dealt in the

entire country in a year. Is it likely that a hand of all one suit

will occur sometime during the year in the United States?

Supplementary exercises.

17. Find the probability of not having a pair in a hand of poker.

18. Find the probability of a “bust” hand in poker. [Hint: A hand

is a “bust” if there is no pair, and it is neither a straight nor a

¬‚ush.]

[Ans. .5012.]

19. In poker, ¬nd the probability of having

(a) Exactly one pair.

[Ans. .4226.]

(b) Two pairs.

[Ans. .0475.]

(c) Three of a kind.

[Ans. .0211.]

20. Verify from Exercises 11, 18, 19 that the probabilities for all pos-

sible poker hands add up to one (within a rounding error).

21. A certain French professor announces that he or she will select

three out of eight pages of text to put on an examination and that

each student can choose one of these three pages to translate.

99

4.4. TWO NONINTUITIVE EXAMPLES

(a) What is the maximum number of pages that a student should

prepare in order to be certain of being able to translate a

page that he or she has studied?

(b) Smith decides to study only four of the eight pages. What

is the probability that one of these four pages will appear on

the examination?

4.4 Two nonintuitive examples

There are occasions in probability theory when one ¬nds a problem for

which the answer, based on probability theory, is not at all in agreement

with one™s intuition. It is usually possible to arrange a few wagers that

will bring one™s intuition into line with the mathematical theory. A

particularly good example of this is provided by the matching birthdays

problem.

Assume that we have a room with r people in it and we propose

the bet that there are at least two people in the room having the same

birthday, i.e., the same month and day of the year. We ask for the

value of r which will make this a fair bet. Few people would be willing

to bet even money on this wager unless there were at least 100 people

in the room. Most people would suggest 150 as a reasonable number.

However, we shall see that with 150 people the odds are approximately

4,100,000,000,000,000 to 1 in favor of two people having the same birth-

day, and that one should be willing to bet even money with as few as

23 people in the room.

Let us ¬rst ¬nd the probability that in a room with r people, no two

have the same birthday. There are 365 possibilities for each person™s

birthday (neglecting February 29). There are then 365r possibilities

for the birthdays of r people. We assume that all these possibilities

are equally likely. To ¬nd the probability that no two have the same

birthday we must ¬nd the number of possibilities for the birthdays

which have no day represented twice. The ¬rst person can have any of

365 days for his or her birthday. For each of these, if the second person

is to have a di¬erent birthday, there are only 364 possibilities for his

or her birthday. For the third person, there are 363 possibilities if he

or she is to have a di¬erent birthday than the ¬rst two, etc. Thus the

probability that no two people have the same birthday in a group of r

people is

365 · 364 · . . . · (365 ’ r + 1)

qr = .

365r

100 CHAPTER 4. PROBABILITY THEORY

Figure 4.1: ™¦

The probability that at least two people have the same birthday is

then pr = 1 ’ qr . In Figure 4.1 the values of pr and the odds for a fair

bet, pr : (1 ’ pr ) are given for several values of r.

We consider now a second problem in which intuition does not lead

to the correct answer. A hat-check clerk has checked n hats, but they

have become hopelessly scrambled. The clerk hands back the hats at

random. What is the probability that at least one head gets its own

hat? For this problem some people™s intuition would lead them to

guess that for a large number of hats this probability should be small,

while others guess that it should be large. Few people guess that the

probability is neither large nor small and essentially independent of the

number of hats involved.

101

4.4. TWO NONINTUITIVE EXAMPLES

Let pj be the statement “the jth head gets its own hat back”. We

wish to ¬nd Pr[p1 ∨ p2 ∨ . . . ∨ pn ]. We know from Exercise 26 that

a probability of this form can be found from the inclusion-exclusion

formula. We must add all probabilities of the form Pr[pi ], then subtract

the sum of all probabilities of the form Pr[pi § pj ], then add the sum of

all probabilities of the form Pr[pi § pj § pk ], etc.

However, each of these probabilities represents the probability that

a particular set of heads get their own hats back. These probabilities

are very easy to compute. Let us ¬nd the probability that out of n

heads some particular m of them get back their own hats. There are n!

ways that the hats can be returned. If a particular m of them are to get