1. The document discusses using the binomial expansion and Stirling's formula to estimate the value of r that maximizes a binomial coefficient expression as n becomes large. Taking the limit as n approaches infinity, the optimal value of r is shown to be nq. 2. A example is given of estimating the probability of winning $40 or more by betting $1 on number 8 in roulette 500 times. Using the normal approximation, this probability is estimated to be about 25.8%.

1. GEK1544 The Mathematics of Games
Tutorial 7
We consider Q. 2 first.
2. Let p and q be positive numbers , p + q = 1 . Consider the binomial expansion
1 = (p + q)n = pn + C(n, 1)pn−1 q + · · · + C(n, r) pn−r q r + · · · + q n
r n
n q q q
= p 1 + C(n, 1) + · · · + C(n, r) +···+ .
p p p
Using the Stirling formula
k
√ k
k ! ≈ 2πk for k 1,
e
estimate the value of r (in terms of n when it is large) so that the expression
r r
q n! q
C(n, r) = (2.1)
p r ! (n − r) ! p
is at maximum. (Suggestion. Express (2.1) using the Stirling formula, and treat it as
a function of r. Then differentiate the expression with respect to r , and let n → ∞ . )
Compare in the roulette example when we use the normal distribution, we let
1
r − nq + 2
Z= √ .
np q
Suggested solution. To locate the maximum, we may ignore the term n !, as it is
independent on r . Let a = q/p and consider
1 ar
ar ≈ √ √ r √ √ n−r
.
r ! (n − r) ! 2π r r
· 2π n − r n−r
e e
Here we apply the Stirling formula. Similarly, we ignore the terms involving 2π and e .
Set
ar
f (r) = 1 1
[rr+ 2 ] [(n − r)(n−r)+ 2 ]
1 1
=⇒ ln [f (r)] = r ln a − r + ln r − (n − r) + ln (n − r)
2 2
1 1 1 1
=⇒ (ln [f (r)]) = ln a − ln r − r + · + ln (n − r) + (n − r) + ·
2 r 2 n−r
f (r) 1 1
=⇒ = [ln a − ln r + ln (n − r)] − 1 + + 1+ .
f (r) 2r 2(n − r)

2. At maximum,
1 1
f (r) = 0 =⇒ [ln a − ln r + ln (n − r)] = 1 +
− 1+
2r 2(n − r)
a(n − r) 1 1 (n − r) − r n − 2r
=⇒ ln = − = =
r 2r 2(n − r) 2r(n − r) 2r (n − r)
When n → ∞, r → c n, where 0 < c < 1 is a fixed number to be determined. Hence
n − 2r n − 2cn 1 − 2c
= = →0 as n → ∞ .
2r (n − r) 4cn (n − cn) 4cn (1 − c)
It follows that
a(n − r) a(n − r)
ln → 0 =⇒ → 1 =⇒ an → r(1 + a) .
r r
That is,
q
an ×n q×n
r→ = p q = =⇒ r = nq (p + q = 1) .
1+a 1+ p p+q
1. A person decides to bet $ 1 on the (fixed) number 8 in Las Vegas roulette, and
continue to do so for a total of 500 times. The pay-off on betting on a number is 35 : 1 .
Using the normal distribution table, estimate the person’s probability on winning $ 40 or
more.
Suggested solution. If the number 8 comes up x times in 500 spins, the profit is
35 · x + (500 − x) · (−1) = 36 · x − 500
Solve for x so that
36 · x − 500 ≥ 40 ⇐⇒ 36 · x ≥ 540 =⇒ x ≥ 15 .
That is, one needs to win 15 times or more in order to profit $ 40 or more. The chance
1
of winning is p = 38 , whereas
37
q =1−p= .
38
Consider the binomial expansion
1 = (p + q)500
= p500 + C(500, 1) p499 q + · · · + C(n, s) p500−s q s + · · · + C(500, 485) p15 q 485 + · · · + q n .
It follows that the probability on winning $ 40 or more is
C(500, 0) p500 + C(500, 1) p499 q + · · · + C(500, 485) p15 q 485 .
In the above, p500 is the probability of winning 500 times number 8, etc. There are a lot
of terms to sum, all the way from 0 to r = 485 . Instead, we apply the normal distribution
to approximate the number.

3. From question 2, we let
1
r − nq + 2
Z= √ ,
np q
1 37
where n = 500 , r = 485 , p= , and q = .
38 38
Therefore
37 1 1 1
485 − 500 · 38
+ 2
485 − 500 · 1 − 38
+ 2 485 − 500 + 500 × 1
− 1
38 2
Z = = =
1 37 1 37 1 37
500 · 38 38
500 · 38 38
500 · 38 38
1
15.5 − 500 × 38 2.3421
= − ≈− ≈ − 0.65 (← observe the negative sign) .
500 · 1 37 3.5793
38 38
From the normal distribution table in the lecture notes, we conclude that
P ( winning $ 40 or more ) ≈ 0.258.