Introduction

Lecture 1, DSCI 220, 2026W1

🎪 Sages & Tricksters

Setup

• Draw one card from the envelope. Keep it hidden.
• Black = Sage. Always tell the truth.
• Red = Trickster. Always lie.
• The six cards are printed on the envelope.
• The fact on the slip is true.

Goal

Find the tricksters.

🔄 How to play

1. Discovery — two rounds

• Each round, every player asks one person a yes/no question about their card.
• Sages answer truthfully. Tricksters lie.

2. Debate

• Compare answers. Look for contradictions.
• You may not say which card you hold, or which role you are.

3. Conclusion

• Agree on everyone’s roles, then reveal.

🧿 Asking well

• Ask about a card’s suit or rank.
• Asking about a card’s colour is allowed. Try it and see what happens.

Click “Asking well” bottom right when you are ready, or leave it shut.

“Is your card red?” gets no from everyone; “is your card black?” gets yes from everyone. Say only that colour questions are a trap, not why — let them predict it. Hold the real method back: if a group stalls, hint that two properties which cannot both be true of one card will do it.

What did you just do?

Write down the step that convinced you:

“I knew __________ was a Trickster because __________.”

Collect two or three. Nearly all take the form: if they were a Sage, then X and not-X. Do not name it yet.

Lying is an operator

Let \(1\) mean yes and \(0\) mean no.

Their card has it	They lie	They say
0	0	0
0	1	1
1	0	1
1	1	0

They say 1.

Two rows fit. Nothing so far separates them.

To eliminate one, you must fix one of the input columns.

You cannot fix they lie — that is the unknown.

So fix their card has it.

What could you ask that would do that?

Drive the discussion from the table. Walk down the right column one reveal at a time and let them supply the last step.

The answer they should reach: ask about a property whose value you already know for that person — then the “has it” column is fixed, only two rows remain, and those two differ in the output. So the answer they give reports the lie bit directly.

Either value works. Fix the column to 0 and the top two rows survive, so a “yes” means lying. Fix it to 1 and the bottom two survive, so a “no” means lying. If a group offers one direction, ask for the other.

In practice known-false is much easier to find. After a “yes” the surviving cards are (black with the property) together with (red without it) — a set spanning both colours. Naming a suit or rank none of them has is easy; finding something a mixed bag all share is rare.

Careful with the “true of at least one card” rule. It bans a question that is false of every card in the packet. It does not ban one that is false of every card this person could still be holding, which is what the strategy actually needs — and that is usually still legal. Packet 9, after a yes to “spade”: club, heart and even-number all work, and club is legal because 2C is in the packet even though no candidate is a club.

The clean two-question kill is available after 192 of 219 possible first answers, about 88%. In the remaining cases no single follow-up fixes the column, and the group has to combine what they heard across several people instead – which is what the debate round is for.

Do not name XOR. The Truth slide introduces the same four-row shape for OR, and naming it there lands better.

Later in the term, the direction is the point: a truth table is normally read left to right, inputs to output. All hour they read it right to left. That is what you do with data — you see what came out and argue about what went in.

Two claims

A. ________________ is a Trickster.

B. Students who come to tutorials do better on examlets.

What would it take to settle each of these?

A: ________________

B: ________________

Deliberately from different domains. If both came from the card game it would look like a fact about card games.

Two claims

	A. Seat 3 is a Trickster	B. Tutorials help
The claim is about	six people	all students
Can you examine every case?	yes	no
So the method is	check them all	sample, and generalise
Verdict	settled	supported
Being wrong means	you made an error	you were unlucky

Both are inference. The method follows from the information.

The discriminator is whether the evidence reaches as far as the claim does. Exact inference is not a different mental operation; it is what inference becomes when the domain is small enough to exhaust.

This course is the left-hand column. Most of their degree is the right. Do not oversell it yet — lecture 3 earns it, when the same cereal claim is asked first of a dataset and then of every cereal ever manufactured.

Instructional team

• Prof: Cinda Heeren 📧 cheeren@cs.ubc.ca ⏰ Mon 12–1pm · 🏢 ICCS 233

• TA: Marko Ciric 📧 mciric01@student.ubc.ca

• TA: Noah Hynds 📧 nhynds@student.ubc.ca

• TA: Perrie Soleimani 📧 perriesl@student.ubc.ca

Term structure

Week	Mon	M	W	F	HW	Tut	EX
1	Sep 7		🎉	🎉
2	Sep 14	🎉	🎉	📹	🏎	🪄
3	Sep 21	🎉	🎉	📹		🪄	🤩
4	Sep 28	🎉		📹	🏎	🪄
5	Oct 5	🎉	🎉	📹		🪄	🤩
6	Oct 12		📹	📹
7	Oct 19	🎉	🎉	📹	🏎	🪄
8	Oct 26	🎉	🎉	📹		🪄	🤩
9	Nov 2	🎉	🎉	📹	🏎	🪄
10	Nov 9			📹
11	Nov 16	🎉	🎉	📹		🪄	🤩
12	Nov 23	🎉	🎉	📹	🏎	🪄
13	Nov 30	🎉	🎉	📹		🪄	🤩
14	Dec 7	🎉

Legend: 🎉 in person   📹 async video   🪄 tutorial   🤩 examlet   🏎 HW due

Grading

Component	Weight
🎉 Class Meetings	0%
🪄 Tutorials	10%
🏎 Homework	10%
🤩 Examlets	50%
🤩 Final Exam	30%

Communications

• 📧 Email cheeren@cs.ubc.ca — put “DSCI220” in the subject
• 💬 Piazza https://piazza.com/ubc.ca/winterterm12026/dsci220
• 🌐 Website https://ubc-cs.github.io/dsci220
• 🎓 Class Mon/Wed/Fri 11:00–12:00, BIOL Rm 2200

Logic

Truth

Make at least 3 observations about the following table:

\(p\)	\(q\)	\(p \lor q\)
F	F	F
F	T	T
T	F	T
T	T	T

Summary

\(p\)	\(q\)	\(p \lor q\)
F	F	F
F	T	T
T	F	T
T	T	T

• The diagram is a Truth Table
• \(p\), \(q\) are variables taking one of two Boolean values (T/F, 1/0)
• \(\lor\) is a binary operator: a function from 2 Boolean inputs to 1 Boolean output, characterized by its output column
• \(\lor\) corresponds to the English or
• \(p \lor q\) is a proposition

Cool down

• How many rows for a proposition on 3 variables? On \(k\)?

• Name an operator taking 1 variable: __________

• Name an operator taking 3 variables: __________

• How many different binary operators could there be?

\(2^k\) rows. \(2^{2^k}\) operators on \(k\) variables; 16 binary ones. Half the binary operators are commutative (8).

Before Friday

You counted how many binary operators there could be.

Activity 1 introduces the ones that have names.

It will be posted on PrairieLearn. Watch Piazza for the link.

PrairieLearn is not published yet, so do not send them to the site today — the QR code and link are removed for that reason. Announce Activity 1 on Piazza once the course instance is live, and set the due date then.

Words and Ideas

Truth Table

Proposition

Binary operator

Argument