1. GEK1544 The Mathematics of Games
Suggested Solutions to Tutorial 4
1. In Backgammon a player’s count on a given roll of two dice is determined as follows :
If doubles are rolled the count is twice the total on the two dice (e.g., a double 5 would
give a count of 2 × 10 = 20). Otherwise, the count is simply the total on the two dice
(e.g. 5 + 4 = 9 counts). Compute the “expected” count on a Backgammon roll. Your
answer shoud be 8 1 .
6
Suggested solution. We have
Count Number of combinations.
3 2
4 3 [ they are (1, 3), (3, 1) & (1, 1)]
5 4
6 4
7 6
8 5 [ they are (2, 2), (2, 6), (6, 2), (3, 5) (5, 3)]
9 4
10 2
11 2
12 1 [ that is (3, 3)]
16 1
20 1
24 1
Expected Count
1
= · [3 · 2 + 4 · 3 + 5 · 4 + 6 · 4 + 7 · 6 + 8 · 5 + 9 · 4 + 10 · 2 + 11 · 2
36
294 1
+ 12 · 1 + 16 · 1 + 20 · 1 + 24 · 1] = =8 .
36 6
2. Refer to the Backgammon in the tutorial. You (black) have one checker on the bar.
White controls 2, 4 and 6 points and has a blot on 5 point. It is your turn.
(a) What is the probability of being able to hit the blot on your next roll ?
(b) What is the probability of your having to enter from the bar without hitting the
blot on your next roll ?

2. Suggested solution. (To simplify situation, we assume that Black has only one checker
left, i.e., the one on the bar.)
(a) To hit the blot, a 5 is present, or one of the rolls : (2, 3), (3, 2), (1, 4), (4, 1).
Hence the probability is
11 4 15
+ = .
36 35 36
(b) Refer to the following table.
→ (1, 1) (2, 1) (3, 1) (4, 1) (5, 1) (6, 1)
(1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2)
→ (1, 3) (2, 3) (3, 3) (4, 3) (5, 3) (6, 3)
(1, 4) (2, 4) (3, 4) (4, 4) (5, 4) (6, 4)
(1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5)
(1, 6) (2, 6) (3, 6) (4, 6) (5, 6) (6, 6)
↑ ↑
Once you have a 1 or 3, you can enter. Thus consider the rows and columns indicated
by the arrows in the table. We have 4 · 6 − 4 = 20 pairs to consider (4 over-counts). The
pairs (1, 5), (5, 1), (3, 5), (5, 3) will force you to hit the blot, as the rules say that you
should use the bigger number first whenever it is possible. Likewise, (2, 3), (3, 2), (1, 4)
and (4, 1) will force you to hit the blot. All other combination will at least one 5 with
also force yo to hit the blot. Hence the probability is
20 − 4 − 4 12 1
= = .
36 36 3
3. Refer to the Backgammon in the tutorial. Imagine the doubling cube is re-
placed by a “tripling cube” (i.e., with faces of 3, 9, 27, 81, 243, 729). Following the
analysis given in lecture, compute the expectations on X(triple ; accept retriple) and
X(triple ; refuse retriple) .
Suggested solution. As in the lecture, set
A = white wins on the first roll (P (A) = 19/36) ;
17 3
B = white does not win on the first roll, black wins on the next roll (P (B) = 36
· 4) ;
C = white does not win on the first roll, black does not win on the next roll, white ‘almost
surely’ wins on the second roll (P (C) = 17 · 4 · 1) .
36
1
P (triple , accept retriple) = P (A) · (3) + P (B) · (−9) + P (C) · (9)
19 17 3 17 1
= ·3+ · · (−9) + · ·9
36 36 4 36 4
≈ −0.54 .
P (triple , black retriple, white refuses) = P (A) · (3) + [1 − P (A)] · (−3)
19 17
= ·3+ · (−3)
36 36
1
= .
6