16.3. Strange Dice 683
1 st A
roll2 nd A
rollsum of
A rolls2
2
7
6
7
7
6
2
2
6
6
7
4
8
9
8
12
13
9
13
14
1 st B
roll2 nd B
rollsum of
B rolls1
1
9
5
9
9
5
1
1
5
5
9
2
6
10
6
10
14
10
14
18
‹
Figure 16.9 Parts of the tree diagram for dieBversus dieAwhere each die is
rolled twice. The first two levels are shown in (a). The last two levels consist of
nine copies of the tree in (b).
the two-roll sums take the values
.4;8;8;9;9;12;13;13;14/:Likewise, two rolls of dieBresult in nine equally likely outcomes in a sample
spaceSBin which the two-roll sums take the values
.2;6;6;10;10;10;14;14;18/:We can treat the outcome of rolling both dice twice as a pair.x;y/ 2 SASB,
whereAwins iff the sum of the twoA-rolls of outcomexis larger the sum of the
twoB-rolls of outcomey. If theA-sum is 4, there is only oneywith a smaller
B-sum, namely, when theB-sum is 2. If theA-sum is 8, there are threey’s with
a smallerB-sum, namely, when theB-sum is 2 or 6. Continuing the count in this
way, the number of pairs.x;y/for which theA-sum is larger than theB-sum is
1 C 3 C 3 C 3 C 3 C 6 C 6 C 6 C 6 D37:A similar count shows that there are 42 pairs for whichB-sum is larger than the
A-sum, and there are two pairs where the sums are equal, namely, when they both
equal 14. This means thatAlosestoBwith probability42=81 > 1=2and ties with
probability2=81. DieAwins with probability only37=81.