Three spinners in a cycle

Question

Three spinners each have three equal sections.

Spinner A shows $2$ , $4$ , or $9$ .
Spinner B shows $1$ , $6$ , or $8$ .
Spinner C shows $3$ , $5$ , or $7$ .

Two players each spin one spinner and the higher number wins.

Show that A beats B, B beats C, and C beats A, and give the common winning probability in each of these matchups.

Accepted Answer

Each matchup has 3	imes 3 = 9 equally likely outcomes. Count the wins. A versus B: A wins when 2 beats 1; 4 beats 1; and 9 beats 1,6,8. That is 1+1+3 = 5 of 9. So A beats B with probability 	frac59. B versus C: B wins when 6 beats 3,5; and 8 beats 3,5,7 (the value 1 beats nothing). That is 2+3 = 5 of 9. So B beats C with probability 	frac59. C versus A: C wins when 3 beats 2; 5 beats 2,4; and 7 beats 2,4. That is 1+2+2 = 5 of 9. So C beats A with probability 	frac59. Every arrow goes the same strength: A beats B beats C beats A, each with probability \frac{5}{9} \approx 55.6\%. No spinner…

Three spinners in a cycle

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