Nobody owes your queue a service-level agreement#

Game Theory and Asynchronous Communication

SLOP1276 · Week 07 · Alex’s story

Fictional Alex’s event trace records a read receipt. It supplies no explanation for the silence. The dashboard has an empty field and too much confidence.

Investigate: What can the timed observations actually establish?

The engineering output#

Build a Communication state machine from Week 6’s tree.

Calculate arrival counts and separate them from a recommended reply time.

Solve utilities, best responses and changed payoffs.

Trace timeouts, duplicates and refusal without emitting a message.

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.

Reconnect the versions after the break#

Alex: board games, terrible puns, a possible relationship.

Friday 5–7 pm · bus · $20 total.

Carry the supported bio, frozen photo, checked Library invitation and G1’s quote.

The constructor calls Week 6’s actual classifier. A pending or closed record cannot start a fresh opener.

Four events, four observations#

Exercise · repair the receipt inference#

5 minutes. Send at 0 h, read at 0.5 h, no reply observed by 2 h.

“They read it, so they want a reminder.”

Separate recorded events from the invented conclusion.

What does a missing delivery receipt establish?

Answer · keep the observations#

Keep the send time, read observation and absence of an observed reply.

Remove the motive and reminder inference.

A missing delivery receipt does not establish non-delivery.

Neither this trace nor the earlier feasible invitation establishes acceptance.

Name the clock you actually have#

at is observation time in hours, from one synthetic origin.

It need not equal a server’s underlying event time.

The reducer retains read and delivery observations separately. It does not fabricate a missing timestamp.

These are authored logs, not claims about a real app’s receipt behaviour.

Count a declared event#

The toy arrival model counts inbound messages, including declines.

Send, delivery and read receipts are not inbound-message arrivals.

Assume a homogeneous Poisson process with λ = 0.4 events/hour.

The state traces are separate fixtures; six authored examples cannot validate this rate.

Rate times duration gives a mean count#

μ = λt = 0.4 × 2 = 0.8 events.

P(N = k) = exp(−μ) × μᵏ / k!

P(N = 0) = exp(−0.8) ≈ 0.449329.

SciPy: Poisson count distribution

Change the assumed rate#

A count is not a reply deadline#

Expected count 0.8 does not mean 80% chance of at least one arrival.

That probability is 1 − exp(−0.8) ≈ 55.07%.

A count may exceed one. A probability cannot.

Neither number identifies when someone should reply or when another message is appropriate.

Exercise · compute three quantities#

6 minutes. At λ = 0.4/hour and t = 2 hours, calculate:

Zero arrivals, at least one arrival, and exactly one arrival.

Then use λ = 0.2 and 0.8 for zero arrivals.

What happens at zero rate or zero duration?

Answer · keep the event definitions#

Zero: 0.449329. At least one: 0.550671.

Exactly one: 0.8 × exp(−0.8) ≈ 0.359463.

At rates 0.2 and 0.8, zero-count probabilities are 0.670320 and 0.201897.

At zero rate or duration, zero arrivals has probability 1. Negative or non-finite inputs are invalid.

The process assumptions do work#

Constant rate; independent counts in disjoint intervals.

Work, sleep, notifications and clustered turns are omitted structure.

After zero in the first two hours, the next two still have zero probability exp(−0.8) under this process.

MIT: Poisson processes, §§4-5–4-6

Same mean, different silence probability#

Pause for the full route · 10 minutes#

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

Keep λ, duration and the event definition beside every probability.

Next: your optimiser has three options and only checked two.

That is a small implementation detail with a large conclusion attached.

One decision maker, three options#

Clarify: 7q − 5. Wait: 2q − 1. Stop: 0.

q is an assumed probability that clarification would be welcome.

Costs and coefficients are arbitrary utility units.

This is a decision against uncertainty; the other person is not yet a strategic player in this model.

Positive utility is not enough#

Clarify beats stop when q > 5/7.

Clarify beats wait when 5q > 4, or q > 0.8.

It must beat both.

At q = 0.75, clarify scores 0.25 but wait scores 0.50. A positive score can still lose.

Solve the whole decision policy#

Exercise · change the cost#

7 minutes. Replace clarification’s cost 5 with 6.

Compare clarify, wait and stop at q = 0.9 and 1.

Can clarification strictly win anywhere on [0,1]?

Now use cost 4. Find the crossover with waiting and check stop too.

Answer · a threshold can leave the domain#

Cost 6, q = 0.9: clarify 0.3, wait 0.8, stop 0.

At q = 1: clarify and wait tie at 1.

Clarification never strictly beats waiting on [0,1].

Cost 4: clarify and wait tie at q = 0.6, both scoring 0.2.

Declare the double-text penalty#

Let d ≥ 0 be an assumed extra cost.

U(C) = 7q − 5 − d.

Clarify beats wait only if q > 0.8 + d/5.

At d = 1, strict preference is impossible. This is sensitivity to a parameter, not a measured social law.

q is not the Poisson complement#

“Would welcome clarification” differs from “an inbound message arrives.”

Do not set q = 1 − P(zero) without a model and evidence connecting those events.

A receipt also supplies no q.

The workbook assumes q; it does not estimate feelings, attention or consent.

Review admissibility before ranking#

The base arithmetic assumes a separately reviewed admissible clarification.

clarificationAllowed: false excludes it, even at q = 0.9.

Week 6’s pending branch supplies no automatic clarification opportunity.

The state reducer has no second-send transition. Closed selects stop before any utility comparison.

Now give both players a choice#

A and B each choose Ask or Wait, simultaneously, once.

Each payoff depends on the pair of choices.

Rows belong to A; columns belong to B; each cell is (A, B).

This is a separate invented game with admissible actions, not a reconstruction of a conversation.

Read the correct payoff coordinate#

Find both players’ best responses#

Fix B = Ask: A receives 1 for Ask, 0 for Wait.

Fix B = Wait: A receives 2 for Ask, 0 for Wait.

Ask is better against either opposing action: strict dominance.

Use B’s second coordinate to repeat the check. B also strictly prefers Ask against either action.

Equilibrium is a joint check#

A Nash equilibrium puts both players on best responses.

Neither improves by changing alone while the other choice stays fixed.

Here the unique equilibrium is (Ask, Ask), because Ask strictly dominates for both.

MIT: best responses and Nash equilibrium

Change one cell, change the game#

Exercise · repair the dominance claim#

7 minutes. In the changed matrix, mark A’s and B’s best responses.

Does Ask still strictly dominate Wait?

Check all four cells for a profitable unilateral change.

List every pure equilibrium. Keep the two payoff coordinates separate.

Answer · two pure equilibria remain#

Against Ask, choose Wait: 0 beats −1.

Against Wait, choose Ask: 2 beats 0.

The equilibria are (Ask, Wait) and (Wait, Ask).

At Ask/Ask, either player can improve by waiting. At Wait/Wait, either can improve by asking. Neither action strictly dominates.

Mix only after calculating indifference#

Let r be the other player’s Ask probability in the changed game.

Expected Ask payoff: −r + 2(1 − r) = 2 − 3r.

Wait payoff: 0. Equalise them: r = 2/3.

By symmetry, both choosing Ask with probability 2/3 form a mixed equilibrium. Both actions then yield zero expected utility.

A game result is not a texting script#

r is a strategy probability; q is the separate welcome assumption; λ is an arrival rate.

The mixed calculation does not remove the two pure equilibria.

Payoffs are invented. Equilibrium does not establish consent or predict behaviour.

Do not turn 2/3 into a rule for randomising real messages.

Turn the tree into event transitions#

Keep phase separate from observation#

Write the contract before the timer#

Each event has ID, type and observation hours. Retain the returned snapshot.

The window is [send time, send time + 2 h).

Two hours is this trace fixture’s convention; Week 6’s aggregate fixture used 48 hours.

This reducer has no real clock, persistence, network or background scheduler.

Replay an observation timeout#

Exercise · substitute a refusal#

Send at 0, wait at 0.1, timeout at 2, reply at 3.

Now replace the timeout with decline at 1.

Trace phase, observation outcome and local action for both histories.

Explain why the same late reply produces different results. Include this in the tutorial’s implementation review.

Answer · one ends a window, one closes contact#

Timeout trace: replied / no-reply, with late-content review.

Refusal trace: closed / reply, with closure reason decline.

The decline itself counts as an inbound reply before the deadline. Later events leave closed unchanged.

A local close before the deadline without a reply gives ended, not a completed no-reply window.

Duplicate records need an identity rule#

Same event ID, type and time → no-op.

Same ID with different content → conflict for review.

A new second-send ID is rejected. Old timestamps on new events are rejected too.

This works only with retained state and IDs. It is not an exactly-once delivery guarantee.

Check expiry before classifying the reply#

If an open observation reaches time ≥ deadline, freeze no-reply.

Then process the supplied event.

A late reply is late even when no explicit timeout event came first.

Nothing wakes up automatically: replay progresses only when the caller supplies an event.

Exercise · test the exact endpoint#

Compare replies at 1.999 h and 2 h in a window starting at 0.

At 2 h, process timeout then reply; reverse their order in a second trace.

Should the window result change with that order?

Predict the literal outcomes before running the workbook.

Answer · equality is outside this window#

At 1.999 h: observation reply.

At 2 h: observation no-reply, with late-content review.

Both equal-time orders end replied / no-reply.

The ≥ expiry check enforces the half-open convention. Changing it to > breaks the endpoint test.

A refusal must survive a competing event#

Break the rule to test the test#

Temporarily remove the closed-state guard. Replay refusal, then late reply.

The expected closed phase must fail if it becomes replied.

Restore the guard and run again. Check input snapshots remain unchanged.

A literal state assertion tests behaviour; searching the code for “closed” does not.

Run five files offline#

Save the five files linked in the lecture notes together.

Run node week-07-worked-examples.mjs.

Inspect arrival counts, changed costs, both games and six event traces.

The download reuses Week 6’s classifier and the existing Poisson module. It emits no messages.

Hand over evidence, not a verdict#

Prepare the reasoning for paper#

Midterm: 23 April, 11 am–12.30 pm.

Question 4: Poisson zero count, utility threshold and best responses in supplied payoffs.

Use a non-programmable calculator and your permitted double-sided A4 notes. No executable code or AI tools during the exam.

Show assumptions and working. Practice values are not promised exam inputs.

Exit ticket · three broken conclusions#

“0.8 expected arrivals means 80% chance.” → Use 1 − exp(−0.8).

“At q = 0.75, positive clarification utility wins.” → Waiting is larger.

“A late reply reopens refusal.” → Closed stays closed.

Alex’s timeout supplies no motive. Week 8 compares silence with evidence that warrants review.