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of a deviation of less than ¬ve. The area from 0 to .1 under the normal
curve is .040 and hence the probability of a deviation from 5000 of less
than ¬ve is approximately .08. Thus, while a deviation of 100 is very
unlikely, it is also very unlikely that a deviation of less than ¬ve will

Example 4.33 The normal approximation can be used to estimate
the individual probabilities f (n, x; p) for large n. For example, let us
estimate f (200, 65; .3). The graph of the probabilities f (200, x; .3) was
given in Figure 4.28 together with the normal approximation. The
desired probability is the area of the bar corresponding to x = 65. An
inspection of the graph suggests that we should take the area under the
normal curve between 64.5 and 65.5 as an estimate for this probability.
In normalized units this is the area between



or between .6944 and .8487. Our table is not ¬ne enough to ¬nd this
area, but from more complete tables, or by machine computation, this
area may be found to be .046 to three decimal places. The exact value
to three decimal places is .045. This procedure gives us a good estimate.
If we check all of the values of f (200, x; .3) we ¬nd in each case
that we would make an error of at most .001 by using the normal
approximation. There is unfortunately no simple way to estimate the
error caused by the use of the Central Limit Theorem. The error will
clearly depend upon how large n is, but it also depends upon how near
p is to 0 or 1. The greatest accuracy occurs when p is near 2 .

Example 4.34 Suppose that a drug has been administered to a num-
ber of patients and found to be e¬ective a fraction p of the time. Assum-
ing an independent trials process, it is natural to take p as an estimate
for the unknown probability p for success on any one trial. It is useful
to have a method of estimating the reliability of this estimate. One
method is the following. Let x be the number of successes for the drug
given to n patients. Then by the Central Limit Theorem
x ’ np
| ¤ 2] ≈ .95.
Pr[| √

This is the same as saying

x/n ’ p
| ¤ 2] ≈ .95.

Putting p = x/n, we have

Pr[|¯ ’ p| ¤ 2 pq/n] ≈ .95.

Using the fact that pq < 4 (see Exercise 12) we have
Pr[|¯ ’ p| ¤ √ ] ≥ .95.

This says that no matter what p is, with probability ≥ .95, the true
value will not deviate from the estimate p by more than √n It is cus-
tomary then to say that
1 1
p’ √ ¤p¤p+ √
¯ ¯
n n

with con¬dence .95. The interval
1 1
[¯ ’ √ , p + √ ]
p ¯
n n
is called a 95 per cent con¬dence interval. Had we started with
x ’ np
| ¤ 3] ≈ .99,
Pr[| √
we would have obtained the 99 per cent con¬dence interval
3 3
[¯ ’ √ , p + √ ]
p ¯
2n 2n
For example, if in 400 trials the drug is found e¬ective 124 times,
or .31 of the times, the 95 per cent con¬dence interval for p is
1 1
[.31 ’ , .31 + ] = [.26, .36]
20 20
and the 99 per cent con¬dence interval is
3 3
[.31 ’ , .31 + ] = [.235, .385].
40 40

1. Let x be the number of successes in n trials of an independent
trials process with probability p for success. Let x = x’np For

large n estimate the following probabilities.
(a) Pr[x < ’2.5].
[Ans. .006.]
(b) Pr[x < 2.5].
(c) Pr[x ≥ ’.5].
(d) Pr[’1.5 < x < 1].
[Ans. .774.]
2. A coin is biased in such a way that a head comes up with proba-
bility .8 on a single toss. Use the normal approximation to esti-
mate the probability that in a million tosses there are more than
800,400 heads.

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
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.

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.]

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.

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
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
2 2
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.

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.

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 .

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 ),
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 .

And then we solve for x4 :
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.

Figure 4.31: ™¦

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.

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

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