16.3. Strange Dice 531

1 st A

roll

2 nd A

roll

sum of

A rolls

2

2

7

6

7

7

6

2

2

6

6

7

4

8

9

8

12

13

9

13

14

1 st B

roll

2 nd B

roll

sum of

B rolls

1

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).

We can treat each outcome as a pair.x;y/ 2 SASB, whereAwins iffx > y. If
xD 4 , there is only oney(namelyyD 2 ) for whichx > y. IfxD 8 , there are
three values ofyfor whichx > y. Continuing the count in this way, the number
of pairs for whichx > yis

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 whichx < y, and there are
two pairs (.14;14/,.14;14/) which result in ties. This means thatAlosestoB
with probability42=81 > 1=2and ties with probability2=81. DieAwins with
probability only37=81.
How can it be thatAis more likely thanBto win with one roll, butBis more
likely to win with two rolls? Well, why not? The only reason we’d think otherwise
is our unreliable, untrained intuition. (Even the authors were surprised when they
first learned about this, but at least we didn’t lose $1400 to biker dude. :-) ) In fact,
the die strength reverses no matter which two die we picked. So for 1 roll,

ABCA;

but for two rolls,
ABCA;

where we have used the symbolsandto denote which die is more likely to
result in the larger value.