1. GEK1544 The Mathematics of Games
Suggested Solutions to Tutorial 5
1. In Craps, a field bet is a bet that a total of 2, 3, 4, 9, 10, 11, or 12 will come up
on the roll (refer to the craps table in the lecture note). Double 2 and double 6 pay 2 : 1,
all others paying even odds (1:1). Compute the bettor’s expectation on a $1 field bet.
Suggested Solution. As in Q. 1 (T. 4),
Total Number of combinations.
2 1
3 2
4 (but not double 2) 2
9 4
10 3
11 2
· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ··
Double 2 1
Double 6 1
Sum= 16
Hence
1 16 18 − 20 1
X= [1 + 2 + 2 + 4 + 3 + 2 + 1 ∗ 2 + 1 ∗ 2] + 1 − (−1) = = − ≈ −5.56% .
36 36 36 18
2. A casino in which the “no result” craps roll for Don’t Pass bettors is “total of 3”
(instead of the usual “total of 12”). That is, when the initial roll has total equal to 3, it
results a loss to a Pass bettor, and neither win nor loss for a Don’t Pass bettor. Compute
the casino’s edge (in percent) on a Don’t Pass bet. Would you prefer to bet Pass or Don’t
Pass in a craps game at this casino?
Suggested Solution. From the lecture notes, with the “natural” craps,
X(Pass) ≈ −0.0143 (≈ −1.4%) ,
X(Don’t Pass) ≈ +0.0143 (≈ +1.4%) .
Under this casino craps’ rule, it changes nothing for Pass bettors, but decreases the
2
expectation of Don’t Pass bettors by 36 . Hence, in the new situation,

2. X(Pass) ≈ −0.0143 (≈ −1.4% , same as before) ,
2
X(Don’t Pass) ≈ +0.0143 − ≈ 0.0143 − 2 × 0.0278 = −0.0413 (≈ −4.1%)
36
↑
(precise sum can be found in lecture notes).
Without “free-odds”, it becomes clear that you better bet on Pass in this casino.
3. Consider the binomial expansions :
(a + b)2 = a2 + 2ab + b2 ;
(a + b)3 = a3 + 3a2 b + 3ab2 + b3 ;
(a + b)4 = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 ;
(a + b)5 = a5 + 5a4 b + 10a3 b2 + 10a2 b3 + 5ab4 + b5 ;
·
·
·
(a + b)n = an + C(n, 1)an−1 b + C(n, 2)an−2 b2 + · · · + C(n, r)an−r br + · · · + bn .
Here
n!
C(n, r) = for n ≥ 1 , r ≥ 1.
r ! (n − r) !
Show that when n is even,
n
C(n, 1) < C(n, 2) < · · · < C(n, ) > · · · > C(n, n − 1) ;
2
and when n is odd,
n−1 n+1
C(n, 1) < C(n, 2) < · · · < C(n, ) = C(n, ) > · · · > C(n, n − 1).
2 2
Suggested Solution. The relations are verified for n = 2, 3, 4, & 5 . Let’s start with
n ≥ 6.
n! n!
C(n, 1) < C(n, 2) ⇐⇒ < ⇐⇒ 2 ! (n − 2) ! < (n − 1) !
1 (n − 1) ! 2 ! (n − 2) !
⇐⇒ 2 < (n − 1) , which is the case as n ≥ 6 .
Likewise,
n! n!
C(n, r) < C(n, r + 1) ⇐⇒ <
r ! (n − r) ! (r + 1) ! [n − (r + 1)] !
⇐⇒ (r + 1) ! [n − (r + 1)] ! < r ! (n − r) !
⇐⇒ r+1<n−r
⇐⇒ 2r + 1 < n
n−1
⇐⇒ r< .
2

3. Moreover,
n−1
C(n, r) = C(n, r + 1) ⇐⇒ 2r + 1 = n ⇐⇒ r =
2
n−1 n+1
⇐⇒ n is odd , that is , C(n, ) = C(n, ).
2 2
When n is even,
n n−1 n n
r= −1< =⇒ C(n, r) < C(n, r + 1) =⇒ C(n, , − 1) < C(n, , ) .
2 2 2 2
The other direction follows from the symmetry
C(n, r) = C(n, n − r) ,
as
n! n!
C(n, n − r) = = = C(n, r) .
(n − r)! (n − (n − r))! (n − r)! r!