2025-01-24 09:27:23 -08:00

118 lines
2.6 KiB
Typst

#import "@local/handout:0.1.0": *
#import "@preview/cetz:0.3.1"
#show: doc => handout(
doc,
quarter: link(
"https://betalupi.com/handouts",
"betalupi.com/handouts",
),
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)