Your expected date is not a booking#

The Mathematics of the “Match”

SLOP1276 · Week 05 · Alex’s story

The assumed model offers fictional Alex 1.2 expected mutual selections. The calendar accepts no fractional bookings. A Friday invitation still needs a bus and budget check.

Investigate: Which plan is feasible, and what remains unknown?

The engineering output#

Build a Match probability model with explicit denominators.

Separate an expected count from the chance of at least one match.

Check Alex’s travel constraints and audit the Sunday comparison.

Freeze a quality target that never masquerades as a probability.

How this pack fits the week#

60 minutes · Monday’s selected lecture slides.

90 minutes · Thursday’s four practical blocks.

About 210 minutes · independent preparation, remaining examples and consolidation.

Timed selections and tasks.

The complete 180-minute route is a resource guide; its break is for a full sitting.

Start with the frozen inputs#

Week 2: platform audit and exposure denominators.

Week 4: supported candidate text and A-portrait-v1.

Alex: adult CS student, board games, terrible puns; possible relationship.

Friday 5–7 pm · bus travel · $20 total.

Define one opportunity#

One eligible fictional pair, offered both selection directions, with a complete 24-hour outcome window.

A: first participant selects. B: reverse selection.

M = A ∩ B: at most one mutual match per pair.

All fixtures are authored. Week 4’s positive response is a different outcome.

Read the branches, then multiply#

The intersection has six pairs#

Conditional is doing work here#

P(A ∩ B) = P(A) × P(B | A)

Here: 0.30 × 0.20 = 0.06.

The marginal P(B) is 13/100, not 6/30. Multiplying the marginals gives 0.039 and the wrong joint probability.

MIT: conditional product rule, §17.4

Exercise · change the denominator#

6 minutes. Use the 100-pair table: both 6, A only 24, B only 7, neither 63.

Find P(B), P(B | A), P(A | B), and P(A or B).

Decide whether A and B are independent in this fixture.

Explain what happens if nobody selects A’s direction.

Answer · the arrow is not reversible#

P(B) = 13/100. P(B | A) = 6/30 = 20%.

P(A | B) = 6/13 ≈ 46.15%.

P(A or B) = 37/100. Count the six mutual pairs once.

The events are dependent: 0.20 ≠ 0.13. With zero A selections, P(B | A) is undefined; the joint count is zero.

Twenty opportunities: what is X?#

Let Iᵢ = 1 for a mutual match on opportunity i, otherwise 0.

X = I₁ + … + I₂₀ is a count, not a probability.

The repeated model generates new pair opportunities. It does not draw the table’s rows without replacement.

Expectation adds, even with dependence#

E[X] = Σ P(Mᵢ)

Twenty marginal probabilities of 0.06 give E[X] = 1.2.

An expectation can be fractional. The realised count cannot.

Linearity needs no independence between opportunities. The individual probabilities still need justification.

MIT: linearity of expectation, §18.5

At least one: use the complement#

With independent match indicators and the same p = 0.06:

P(X = 0) = 0.94²⁰ ≈ 0.2901.

P(X ≥ 1) = 1 − 0.94²⁰ ≈ 70.99%.

This model leaves about a 29% chance of zero. It does not book 1.2 meetings.

Where the count can land#

NIST: binomial distribution

Exercise · audit the model calls#

7 minutes. Reproduce 1.2 and 70.99% in code or a spreadsheet.

Check zero opportunities, p = 0 and p = 1.

Now give every opportunity the outcome of one shared Bernoulli draw, with success probability 0.06.

Which earlier answer survives?

Answer · same mean, different chance#

Zero opportunities: expected 0, chance of one 0.

Twenty opportunities at p = 0: 0 and 0; at p = 1: 20 and 1.

Shared draw: X = 20 with probability 0.06, otherwise X = 0.

E[X] = 1.2, but P(X ≥ 1) = 6%. Dependence changes the distribution.

The marginal probabilities stayed fixed#

The shared draw is a deliberate stress test, not a claim about real users.

Unequal probabilities need their own product#

For independent events: P(any) = 1 − ∏(1 − pᵢ).

Ten at 0.02 and ten at 0.10 give E[X] = 1.2 and P(any) ≈ 71.51%.

Replacing every pᵢ with the mean 0.06 gives 70.99% instead.

Equal means do not imply equal distributions.

Sensitivity to the reciprocal branch#

Two different independence questions#

Within a pair: is P(B | A) equal to P(B)? The table says no.

Across opportunities: are M₁, …, M₂₀ mutually independent? The repeated toy model assumes yes.

You may use conditional selection within each pair and independent match indicators across pairs. State both choices.

Pause for the full route · 10 minutes#

For an uninterrupted full-pack session; skip this slide in Monday’s selection.

Keep your event definitions and calculation trace.

Next: your model has found a match six kilometres away.

Your bus timetable would like to review the pull request.

A radius is not a journey#

Straight-line distance does not encode routes, waiting or return travel.

Treat the three venue records as invented options, not extra facts about Alex.

Start at 17:00. Include the entire outing before 19:00 and every cost within $20.

Check the whole itinerary#

A hard constraint is a predicate#

minutes = outbound + meeting + return

cost = return transport + activity

Feasible means minutes ≤ 120 AND cost ≤ $20.

Travel legs already include walking and waiting. Never count the return fare twice or omit the return journey.

Exercise · choose a feasible invitation#

8 minutes. Audit all three invented options. Each meeting lasts 60 minutes.

Library: 20 min each way; $6 transport + $8 activity.

Arcade: 35 min each way; $6 + $10.

Riverside: 15 min each way; $6 + $16.

Choose only after checking both constraints.

Time alone rejects the closest option#

Budget supplies a separate rejection#

Passing is not having spare capacity#

Library reserve: 20 minutes and $6.

Exercise · 4 minutes: add 10 minutes to each travel leg. Then add 11 instead. Test an outing costing exactly $20 and one costing $20.01.

Keep Alex’s limits fixed; vary the fictional route inputs.

Answer · boundary conditions count#

Ten extra minutes each way: 120 minutes, passes with no time reserve.

Eleven extra each way: 122 minutes, fails.

Exactly $20 passes; $20.01 fails. Store cents as integers.

A desired buffer is a separate declared decision rule. Passing a deterministic case is no reliability guarantee.

The cap is an input, not an app disclosure#

Course cap: 20 offered pair opportunities in each strategy.

Baseline: P(A) = 0.30, P(B | A) = 0.20.

Broader selection: P(A) = 0.60, P(B | A) = 0.10.

Both assumed models give p = 0.06 and expected count 1.2. More outgoing selections need not improve this objective.

Keep the feasible denominator visible#

If only 10 of 20 offered opportunities satisfy the declared eligibility filter, 0.06 per eligible opportunity gives 0.6 expected eligible matches.

Report offered, eligible and selected counts separately.

Use the same pool and cap for a controlled strategy comparison. Changing eligibility changes the question.

Peak system load: Sunday evening swiping#

This is the week’s synthetic peak-load scenario.

Sunday: 8 mutual matches / 40 exposures.

Tuesday: 3 mutual matches / 10 exposures.

Separate disjoint fictional pools; no random time assignment. These rows are not the earlier 100-pair table.

Sunday wins on counts#

Tuesday wins on observed rate#

Exercise · repair the headline#

6 minutes. “Sunday is 2.67 times better, so move every session there.”

Identify the quantity behind 2.67. Compute the exposure ratio, rate difference and rate ratio.

Write one sentence the table supports and one causal question it cannot answer.

Answer · name what improved#

Count ratio: 8/3 ≈ 2.67. Exposure ratio: 40/10 = 4.

Sunday − Tuesday: −10 percentage points. Rate ratio: 0.20/0.30 ≈ 0.67.

Sunday has more observed matches and a lower observed rate.

The effect of moving the same eligible population to Sunday remains unknown.

A shared cap gives a hypothetical comparison#

Suppose both windows offered 20 eligible exposures, and their observed rates transferred unchanged.

Expected counts would be 4 on Sunday and 6 on Tuesday.

These are plug-in scenarios, not new observations or a randomised result.

With zero exposures, a rate is undefined, not 0%.

Exercise · repair the time experiment#

8 minutes. “Use old bio A on Sunday, new B on Tuesday. Reuse viewers. Stop when Tuesday wins.”

Specify the unit, frozen inputs, assignment, outcome window and stopping rule for a simulation-only proposal.

List what must be supplied before it can run. The existing counts do not supply a generator.

Answer · preserve one contrast#

Freeze the photo, text, pool eligibility and all non-time settings.

Randomly assign 200 distinct simulated pairs per window, once each; finish all 24-hour windows before analysis.

Count mutual matches per assigned pair. Repair missing or duplicate records.

Prespecify the generator, seed and rate difference. The cap is an exercise choice, not a power calculation.

What a simulation would establish#

A synthetic experiment checks behaviour under its generator.

The generator must say whether time changes selection probabilities; assignment alone cannot invent that mechanism.

Tiny observed counts do not justify certainty. Week 4’s normal-test guardrail withholds z here.

No data in this pack establishes a universal best time.

Freeze the right project objective#

Four ratings need four pieces of evidence#

Clarity: what Alex is seeking.

Specificity: a supported detail someone can respond to.

Feasibility: time, bus and total-budget consistency.

Exit: declining or not engaging is acceptable.

Use the toolkit’s 0–4 anchors. Quote the evidence before assigning a score.

Exercise · rank is not score#

7 minutes. Score the illustrative vectors [4, 4, 3, 4] and [4, 4, 4, 4].

Load all 99 Null Island v1 controls and rank each candidate.

Use rank = 1 + count(control ≥ candidate).

Explain why changing ≥ to > would reward a tie.

Answer · the tie occupies a place#

[4, 4, 3, 4] scores 93.75 and ranks 2/100.

[4, 4, 4, 4] scores 100 and ranks 1/100.

One control also scores 93.75. Ties count against the candidate.

These are arithmetic examples. Your candidate’s ratings must follow its evidence.

The benchmark has a designed ceiling#

The 99 controls are invented feature vectors; the maximum is 93.75.

The generator intentionally excludes [4, 4, 4, 4]. Winning this fixture is not external validation.

Only rank 1/100 receives the course’s top-1% label.

A score of 100 predicts neither a match nor a date.

Freeze before evaluation influences revision#

Record candidate text version, photo fingerprint and venue assumptions.

Save feature evidence, primary scoring rule, control file fingerprint, 100-entry denominator and conservative ties.

Prespecify Week 11’s double-feasibility sensitivity and separate availability shift.

Preserve the baseline; label later revisions separately.

Run the workbook offline#

Save the five linked files from the lecture notes in one folder.

Run node week-05-worked-examples.mjs.

Inspect the pair table, independent and shared-draw answers, venue decisions, Sunday rates and control ranks.

Use a spreadsheet or another language if preferred; retain the same definitions.

Exercise · defend the frozen record#

6 minutes. Review your neighbour’s record, or audit your own.

Can you reproduce the text and inputs? Locate each rating’s evidence? Explain every denominator?

Show the 93.75-score example beside the 0.06 match model without converting one into the other.

Return one concrete correction and its reason.

Handoff · a feasible invitation is still optional#

“Library games table Friday 17:20–18:20? My bus trip and activity fit $14 total. Fine to pass.”

An example under the invented route inputs; recipient availability is still unknown.

Week 6 uses your supported bio and checked invitation in its message tree. A match never bypasses the refusal-first branch.

Exit ticket · fix three category errors#

“Expected 1.2” → a model mean, not a booking.

“Sunday wins” → more counts, lower observed rate.

“Top 1%” → rank 1/100 in a named rubric fixture.

Alex’s Library plan fits 100 minutes and $14; acceptance is unknown. Week 6 drafts the opener.

Keep the Match probability model and frozen baseline for Week 11.