181

4.14. THE CENTRAL LIMIT THEOREM

3. Plot a graph of the probabilities f (10, x, .5). Plot a graph also of

the normalized probabilities as in Figures 4.27 and 4.28.

4. An ordinary coin is tossed one million times. Let x be the number

of heads which turn up. Estimate the following probabilities.

(a) Pr[499, 500 < x < 500, 500]

[Ans. Approximately .682.]

(b) Pr[499, 000 < x < 501, 000],

[Ans. Approximately .954.]

(c) Pr[498, 500 < x < 501, 500],

[Ans. Approximately .997.]

5. Assume that a baseball player has probability .37 of getting a hit

each time he or she comes to bat. Find the probability of getting

an average of .388 or better if he or she comes to bat 300 times

during the season. (In 1957 Ted Williams had a batting average

of .388 and Mickey Mantle had an average of .353. If we assume

this di¬erence is due to chance, we may estimate the probability

of a hit as the combined average, which is about .37.)

[Ans. .242.]

6. A true-false examination has 48 questions. Assume that the prob-

ability that a given student knows the answer to any one question

3

is 4 . A passing score is 30 or better. Estimate the probability

that the student will fail the exam.

7. In Example 4.21 of Section 4.10, assume that the school decides

to admit 1296 students. Estimate the probability that they will

have to have additional dormitory space.

[Ans. Approximately .115.]

8. Peter and Paul each have 20 pennies. They agree to match pen-

nies 400 times, keeping score but not paying until the 400 matches

are over. What is the probability that one of the players will not

be able to pay? Answer the same question for the case that Peter

has 10 pennies and Paul has 30.

182 CHAPTER 4. PROBABILITY THEORY

9. In tossing a coin 100 times, the probability of getting 50 heads

is, to three decimal places, .080. Estimate this same probability

using the Central Limit Theorem.

[Ans. .080.]

10. A standard medicine has been found to be e¬ective in 80 per

cent of the cases where it is used. A new medicine for the same

purpose is found to be e¬ective in 90 of the ¬rst 100 patients on

which the medicine is used. Could this be taken as good evidence

that the new medication is better than the old?

11. In the Weldon dice experiment, 12 dice were thrown 26,306 times

and the appearance of a 5 or a 6 was considered to be a suc-

cess. The mean number of successes observed was, to four deci-

mal places, 4.0524. Is this result signi¬cantly di¬erent from the

expected average number of 4?

[Ans. Yes.]

1 1

12. Prove that pq ¤ 4 . [Hint: write p = + x.]

2

13. Suppose that out of 1000 persons interviewed 650 said that they

would vote for Mr. Big for mayor. Construct the 99 per cent

con¬dence interval for p, the proportion in the city that would

vote for Mr. Big.

14. Opinion pollsters in election years usually poll about 3000 voters.

Suppose that in an election year 51 per cent favor candidate A and

49 per cent favor candidate B. Construct 95 per cent con¬dence

limits for candidate A winning.

[Ans. [.492, .528].]

15. In an experiment with independent trials we are going to estimate

p by the fraction p of successes. We wish our estimate to be within

.02 of the correct value with probability .95. Show that 2500

observations will always su¬ce. Show that if it is known that p

is approximately .1, then 900 observations would be su¬cient.

183

4.15. GAMBLER™S RUIN

16. An experimenter has an independent trials process and he or she

has a hypothesis that the true value of p is p0 . He decides to carry

out a number of trials, and from the observed r calculate the 95

per cent con¬dence interval for p. He will reject p0 if it does not

fall within these limits. What is the probability that he or she

will reject p0 when in fact it is correct? Should he or she accept

p0 if it does fall within the con¬dence interval?

17. A coin is tossed 100 times and turns up heads 61 times. Using

the method of Exercise 16 test the hypothesis that the coin is a

fair coin.

[Ans. Reject.]

18. Two railroads are competing for the passenger tra¬c of 1000 pas-

sengers by operating similar trains at the same hour. If a given

passenger is equally likely to choose one train as the other, how

many seats should the railroad provide if it wants to be sure that

its seating capacity is su¬cient in 99 out of 100 cases?

[Ans. 537.]

4.15 Gambler™s ruin

In this section we will study a particular Markov chain, which is inter-

esting in itself and has far-reaching applications. Its name, “gambler™s

ruin”, derives from one of its many applications. In the text we will

describe the chain from the gambling point of view, but in the exercises

we will present several other applications.

Let us suppose that you are gambling against a professional gambler,

or gambling house. You have selected a speci¬c game to play, on which

you have probability p of winning. The gambler has made sure that the

1

game is in his or her favor, so that p < 2 . However, in most situations

p will be close to 1 . (The cases p = 1 and p > 2 are considered in the

1

2 2

exercises.)

At the start of the game you have A dollars, and the gambler has B

dollars. You bet $1 on each game, and play until one of you is ruined.

What is the probability that you will be ruined? Of course, the answer

depends on the exact values of p, A, and B. We will develop a formula

for the ruin-probability in terms of these three given numbers.

184 CHAPTER 4. PROBABILITY THEORY

Figure 4.30: ™¦

First we will set the problem up as a Markov chain. Let N = A+B,

the total amount of money in the game. As states for the chain we

choose the numbers 0, 1, 2, . . . , N . At any one moment the position of

the chain is the amount of money you have. The initial position is

shown in Figure 4.30.

If you win a game, your money increases by $1, and the gambler™s

fortune decreases by $1. Thus the new position is one state to the right

of the previous one. If you lose a game, the chain moves one step to

the left. Thus at any step there is probability p of moving one step

to the right, and probability q = 1 ’ p of one step to the left. Since

the probabilities for the next position are determined by the present

position, it is a Markov chain.

If the chain reaches 0 or N , we stop. When 0 is reached, you are

ruined. When N is reached, you have all the money, and you have

ruined the gambler. We will be interested in the probability of your

ruin, i.e., the probability of reaching 0.

Let us suppose that p and N are ¬xed. We actually want the prob-

ability of ruin when we start at A. However, it turns out to be easier to

solve a problem that appears much harder: Find the ruin-probability

for every possible starting position. For this reason we introduce the

notation xi , to stand for the probability of your ruin if you start in

position i (that is, if you have i dollars).

Let us ¬rst solve the problem for the case N = 5. We have the

unknowns x0 , x1 , x2 , x3 , x4 , x5 . Suppose that we start at position 2.

The chain moves to 3, with probability p, or to 1, with probability q.

Thus

Pr[ruin|start at 2] = Pr[ruin|start at 3] · p + Pr[ruin|start at 1] · q.

using the conditional probability formula, with a set of two alternatives.

But once it has reached state 3, a Markov chain behaves just as if it

had been started there. Thus

Pr[ruin|start at 3] = x3 .

185

4.15. GAMBLER™S RUIN

And, similarly,

Pr[ruin|start at 1] = x1 .

We obtain the key relation

x2 = px3 + qx1 .

We can modify this as follows:

(p + q)x2 = px3 + qx1 ,

p(x2 ’ x3 ) = q(x1 ’ x2 ),

x1 ’ x2 = r(x2 ’ x3 ),

where r = p/q and hence r < 1. When we write such an equation for

each of the four “ordinary” positions, we obtain

x0 ’ x 1 r(x1 ’ x2 ),

=

x1 ’ x 2 r(x2 ’ x3 ),

=

(4.4)

x2 ’ x 3 r(x3 ’ x4 ),

=

x3 ’ x 4 r(x4 ’ x5 )

=

We must still consider the two extreme positions. Suppose that the

chain reaches 0. Then you are ruined, hence the probability of your

ruin is 1. While if the chain reaches N = 5, the gambler drops out of

the game, and you can™t be ruined. Thus

x0 = 1, x5 = 0. (4.5)

If we substitute the value of x5 in the last equation of 4.4, we have

x3 ’x4 = rx4 . This in turn may be substituted in the previous equation,

etc. We thus have the simpler equations

1 · x4 ,

x4 =

x3 ’ x 4 = rx4 .

r 2 x4 .

x2 ’ x 3 = (4.6)

r 3 x4 .

x1 ’ x 2 =

r 4 x4

x0 ’ x 1 =

Let us add all the equations. We obtain

x0 = (1 + r + r 2 + r 3 + r 4 )x4 .

From 4.5 we have that x0 = 1. We also use the simple identity

(1 ’ r)(1 + r + r 2 + r 3 + r 4 ) = 1 ’ r 5 .

186 CHAPTER 4. PROBABILITY THEORY

And then we solve for x4 :

1’r

x4 = .

1 ’ r5

If we add the ¬rst two equations in 4.6, we have that x3 = (1 + r)x4 .

Similarly, adding the ¬rst three equations, we solve for x2 , and adding

the ¬rst four equations we obtain x1 . We now have our entire solution,

1 ’ r4 1 ’ r3 1 ’ r2 1 ’ r1

x1 = , x2 = , x3 = , x4 = . (4.7)

1 ’ r5 1 ’ r5 1 ’ r5 1 ’ r5

The same method will work for any value of N . And it is easy to guess

from 4.7 what the general solution looks like. If we want xA , the answer

is a fraction like those in 4.7. In the denominator the exponent of r is

always N . In the numerator the exponent is N ’ A, or B. Thus the

ruin-probability is

1 ’ rB

xA = . (4.8)

1 ’ rN

We recall that A is the amount of money you have, B is the gambler™s

stake, N = A + B, p is your probability of winning a game, and r =

p/(1 ’ p).

In Figure 4.31 we show some typical values of the ruin-probability.

Some of these are quite startling. If the probability of p is as low as .45

(odds against you on each game 11: 9) and the gambler has 20 dollars

to put up, you are almost sure to be ruined. Even in a nearly fair game,

say p = .495, with each of you having $50 to start with, there is a .731

chance for your ruin.

It is worth examining the ruin-probability formula 4.8 more closely.

Since the denominator is always less than 1, your probability of ruin is

at least 1 ’ r B . This estimate does not depend on how much money

you have, only on p and B. Since r is less than 1, by making B large

enough, we can make r B practically 0, and hence make it almost certain

that you will be ruined.

Suppose, for example, that a gambler wants to have probability .999

of ruining you. (You can hardly call him or her a gambler under those

circumstances!) The gambler must make sure that r B < .001. For

example, if p = .495, the gambler needs $346 to have probability .999

of ruining you, even if you are a millionaire. If p = .48, the gambler

needs only $87. And even for the almost fair game with p = .499, $1727

will su¬ce.

187

4.15. GAMBLER™S RUIN

Figure 4.31: ™¦

188 CHAPTER 4. PROBABILITY THEORY

There are two ways that gamblers achieve this goal. Small gambling

houses will ¬x the odds quite a bit in their favor, making r much less

than 1. Then even a relatively small bank of B dollars su¬ces to assure

them of winning. Larger houses, with B quite sizable, can a¬ord to let

you play nearly fair games.

Exercises

1. An urn has nine white balls and 11 black balls. A ball is drawn,

and replaced. If it is white, you win ¬ve cents, if black, you lose

¬ve cents. You have a dollar to gamble with, and your opponent

has ¬fty cents. If you keep on playing till one of you loses all

his or her money, what is the probability that you will lose your

dollar?

[Ans. .868.]

2. Suppose that you are shooting craps, and you always hold the

dice. You have $20, your opponent has $10, and $1 is bet on each

game; estimate your probability of ruin.

3. Two government agencies, A and B, are competing for the same

task. A has 50 positions, and B has 20. Each year one position

is taken away from one of the agencies, and given to the other.

If 52 per cent of the time the shift is from A to B, what do you

predict for the future of the two agencies?

[Ans. One agency will be abolished. B survives with probability

.8, A with probability .2.]

4. What is the approximate value of xA if you are rich, and the

gambler starts with $1?

5. Consider a simple model for evolution. On a small island there is

room for 1000 members of a certain species. One year a favorable

mutant appears. We assume that in each subsequent generation

either the mutants take one place from the regular members of

the species, with probability .6, or the reverse happens. Thus,

for example, the mutation disappears in the very ¬rst generation

with probability .4. What is the probability that the mutants

eventually take over? [Hint: See Exercise 4.]

189

4.15. GAMBLER™S RUIN

1

[Ans. .]

3

6. Verify that the proof of formula 4.8 in the text is still correct

1

when p > 2 . Interpret formula 4.8 for this case.

1

7. Show that if p > 2 , and both parties have a substantial amount

of money, your probability of ruin is approximately 1/r A .

1

8. Modify the proof in the text to apply to the case p = 2 . What is

the probability of your ruin?

[Ans. B/N .]

9. You are matching pennies. You have 25 pennies to start with,

and your opponent has 35. What is the probability that you will

win all his or her pennies?

10. Jones lives on a short street, about 100 steps long. At one end of

the street is Jones™s home, at the other a lake, and in the middle

a bar. One evening Jones leaves the bar in a state of intoxication,

and starts to walk at random. What is the probability that Jones

will fall into the lake if

(a) Jones is just as likely to take a step to the right as to the

left?

1

[Ans. .]

2

(b) Jones has probability .51 of taking a step towards home?

[Ans. .119.]

11. You are in the following hopeless situation: You are playing a

1

game in which you have only 3 chance of winning. You have

$1, and your opponent has $7. What is the probability of your

winning all his or her money if

(a) You bet $1 each time?

1

[Ans. .]

255

(b) You bet all your money each time?

1

[Ans. .]

27

190 CHAPTER 4. PROBABILITY THEORY

12. Repeat Exercise 11 for the case of a fair game, where you have

probability 1 of winning.

2

13. Modify the proof in the text to compute yi , the probability of

reaching state N = 5.

14. Verify, in Exercise 13, that xi + yi = 1 for every state. Interpret.

Note: The following exercises deal with the following ruin prob-

lem: A and B play a game in which A has probability W of

winning. They keep playing until either A has won six times or

B has won three times.

15. Set up the process as a Markov chain whose states are (a, b),

where a is the number of times A won, and b the number of B

wins.

16. For each state compute the probability of A winning from that

position. [Hint: Work from higher a- and b-values to lower ones.]

17. What is the probability that A reaches his or her goal ¬rst?

1024

[Ans. .]

2187

18. Suppose that payments are made as follows: If A wins six games,

A receives $1, if B wins three games then A pays $1. What is the

expected value of the payment, to the nearest penny?

Suggested reading.

Cramer, Harald, The Elements of Probability Theory, Part I, 1955.

Feller, W., An Introduction to Probability Theory and its Applications,

1950.

Goldberg, S., Probability: An Introduction, 1960.

Mosteller, F., Fifty Challenging Problems in Probability with Solutions,

1965.

Neyman, J., First Course in Probability and Statistics, 1950.

Parzen, E., Modern Probability Theory and Its Applications, 1960.

Whitworth, W. A., Choice and Chance, with 1000 Exercises, 1934.