#import "@local/handout:0.1.0": * #import "@preview/cetz:0.3.1" #show: handout.with( title: [Warm-Up: Odd Dice], by: "Mark", ) #problem() We say a set of dice ${A, B, C}$ is _nontransitive_ if, on average, $A$ beats $B$, $B$ beats $C$, and $C$ beats $A$. In other words, we get a counterintuitive "rock - paper - scissors" effect. #v(2mm) Create a set of nontransitive six-sided dice. \ #hint([All sides should be numbered with positive integers less than 10.]) #solution([ One possible set can be numbered as follows: - Die $A$: $2, 2, 4, 4, 9, 9$ - Die $B$: $1, 1, 6, 6, 8, 8$ - Die $C$: $3, 3, 5, 5, 7, 7$ #v(2mm) Another solution is below: - Die $A$: $3, 3, 3, 3, 3, 6$ - Die $B$: $2, 2, 2, 5, 5, 5$ - Die $C$: $1, 4, 4, 4, 4, 4$ ]) #v(1fr) #problem() Now, consider the set of six-sided dice below: - Die $A$: $4, 4, 4, 4, 4, 9$ - Die $B$: $3, 3, 3, 3, 8, 8$ - Die $C$: $2, 2, 2, 7, 7, 7$ - Die $D$: $1, 1, 6, 6, 6, 6$ - Die $E$: $0, 5, 5, 5, 5, 5$ On average, which die beats each of the others? Draw a diagram. #solution( align( center, cetz.canvas({ import cetz.draw: * let s = 0.8 // Scale let t = 13pt * s // text size let radius = 0.3 * s // Points let a = (-2 * s, 0.2 * s) let b = (0 * s, 2 * s) let c = (2 * s, 0.2 * s) let d = (1.2 * s, -2.1 * s) let e = (-1.2 * s, -2.1 * s) set-style( stroke: (thickness: 0.6mm * s), mark: ( end: ( symbol: ">", fill: black, offset: radius + (0.025 * s), width: 1.2mm * s, length: 1.2mm * s, ), ), ) line(a, b) line(b, c) line(c, d) line(d, e) line(e, a) line(a, c) line(b, d) line(c, e) line(d, a) line(e, b) circle(a, radius: radius, fill: oblue, stroke: none) circle(b, radius: radius, fill: oblue, stroke: none) circle(c, radius: radius, fill: oblue, stroke: none) circle(d, radius: radius, fill: oblue, stroke: none) circle(e, radius: radius, fill: oblue, stroke: none) content(a, text(fill: white, size: t, [*A*])) content(b, text(fill: white, size: t, [*B*])) content(c, text(fill: white, size: t, [*C*])) content(d, text(fill: white, size: t, [*D*])) content(e, text(fill: white, size: t, [*E*])) }), ), ) #v(1fr) #problem() Now, say we roll each die twice. What happens to the graph from the previous problem? #solution([ The direction of each edge is reversed! ]) #v(1fr)