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 wouldgive 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. Youranswer shoud be 81

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

36· [3 · 2 + 4 · 3 + 5 · 4 + 6 · 4 + 7 · 6 + 8 · 5 + 9 · 4 + 10 · 2 + 11 · 2

+ 12 · 1 + 16 · 1 + 20 · 1 + 24 · 1] =294

36= 8

1

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 theblot on your next roll ?

Suggested solution. (To simplify situation, we assume that Black has only one checkerleft, 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

36+

4

35=

15

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 indicatedby the arrows in the table. We have 4 · 6− 4 = 20 pairs to consider (4 over-counts). Thepairs (1, 5), (5, 1), (3, 5), (5, 3) will force you to hit the blot, as the rules say that youshould 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 withalso force yo to hit the blot. Hence the probability is

20− 4− 4

36=

12

36=

1

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 theanalysis given in lecture, compute the expectations on X(triple ; accept retriple) andX(triple ; refuse retriple) .

Suggested solution. As in the lecture, set

A = white wins on the first roll (P (A) = 19/36) ;

B = white does not win on the first roll, black wins on the next roll (P (B) = 1736· 3

4) ;

C = white does not win on the first roll, black does not win on the next roll, white ‘almostsurely’ wins on the second roll (P (C) = 17

36· 1

4· 1) .

P (triple , accept retriple) = P (A) · (3) + P (B) · (−9) + P (C) · (9)

=19

36· 3 +

17

36· 3

4· (−9) +

17

36· 1

4· 9

≈ −0.54 .

P (triple , black retriple, white refuses) = P (A) · (3) + [1− P (A)] · (−3)

=19

36· 3 +

17

36· (−3)

=1

6.