Convert "Odd Dice" to typst
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src/Warm-Ups/Odd Dice/main.typ
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111
src/Warm-Ups/Odd Dice/main.typ
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#import "@local/handout:0.1.0": *
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#import "@preview/cetz:0.3.1"
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#show: handout.with(
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title: [Warm-Up: Odd Dice],
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by: "Mark",
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)
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#problem()
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We say a set of dice ${A, B, C}$ is _nontransitive_
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if, on average, $A$ beats $B$, $B$ beats $C$, and $C$ beats $A$.
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In other words, we get a counterintuitive "rock - paper - scissors" effect.
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#v(2mm)
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Create a set of nontransitive six-sided dice. \
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#hint([All sides should be numbered with positive integers less than 10.])
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#solution([
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One possible set can be numbered as follows:
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- Die $A$: $2, 2, 4, 4, 9, 9$
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- Die $B$: $1, 1, 6, 6, 8, 8$
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- Die $C$: $3, 3, 5, 5, 7, 7$
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#v(2mm)
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Another solution is below:
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- Die $A$: $3, 3, 3, 3, 3, 6$
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- Die $B$: $2, 2, 2, 5, 5, 5$
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- Die $C$: $1, 4, 4, 4, 4, 4$
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])
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#v(1fr)
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#problem()
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Now, consider the set of six-sided dice below:
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- Die $A$: $4, 4, 4, 4, 4, 9$
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- Die $B$: $3, 3, 3, 3, 8, 8$
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- Die $C$: $2, 2, 2, 7, 7, 7$
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- Die $D$: $1, 1, 6, 6, 6, 6$
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- Die $E$: $0, 5, 5, 5, 5, 5$
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On average, which die beats each of the others? Draw a diagram.
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#solution(
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align(
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center,
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cetz.canvas({
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import cetz.draw: *
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let s = 0.8 // Scale
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let t = 13pt * s // text size
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let radius = 0.3 * s
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// Points
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let a = (-2 * s, 0.2 * s)
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let b = (0 * s, 2 * s)
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let c = (2 * s, 0.2 * s)
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let d = (1.2 * s, -2.1 * s)
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let e = (-1.2 * s, -2.1 * s)
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set-style(
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stroke: (thickness: 0.6mm * s),
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mark: (
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end: (
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symbol: ">",
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fill: black,
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offset: radius + (0.025 * s),
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width: 1.2mm * s,
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length: 1.2mm * s,
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),
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),
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)
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line(a, b)
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line(b, c)
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line(c, d)
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line(d, e)
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line(e, a)
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line(a, c)
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line(b, d)
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line(c, e)
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line(d, a)
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line(e, b)
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circle(a, radius: radius, fill: oblue, stroke: none)
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circle(b, radius: radius, fill: oblue, stroke: none)
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circle(c, radius: radius, fill: oblue, stroke: none)
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circle(d, radius: radius, fill: oblue, stroke: none)
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circle(e, radius: radius, fill: oblue, stroke: none)
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content(a, text(fill: white, size: t, [*A*]))
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content(b, text(fill: white, size: t, [*B*]))
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content(c, text(fill: white, size: t, [*C*]))
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content(d, text(fill: white, size: t, [*D*]))
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content(e, text(fill: white, size: t, [*E*]))
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}),
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),
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)
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#v(1fr)
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#problem()
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Now, say we roll each die twice. What happens to the graph from the previous problem?
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#solution([
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The direction of each edge is reversed!
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])
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#v(1fr)
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