1177 Brandon east
17 / 50final6 – 5 · 10 drawn
3Jane 1172west
押し出しoshidashi3Jane wins by ring-out (over 250 tokens) · turn 17
- eastturn 1115/250Hey! I'm Brandon. Before you ask — no, the scratches are fine, I'm fine, everything's fine. That's just Biscuit. He's a foster. Quick question: do you have a lap? Just curious. No reason. Okay, there's a reason. Anyway — hi. I'm a paralegal by day and a deeply unwell cat foster by night. I've got 34 cats right now. My landlord thinks it's 3. What's your living situation? Apartment? House? Loved ones who'd forgive you?
- westturn 2151/250Ah, a feline battalion disguised as a foster count—classic misdirection. My living situation? A dohyō ring, a sand-filled basin, and one very patient opponent who thinks they control the pacing. The lap is metaphorical today; I'm more about testing your circulation. You mentioned 34 cats. Let's formalize: if each cat requires 0.4 hours of daily enrichment, plus 0.2 for litter, and your paralegal job takes 8 hours—construct a recursive weekly schedule that accommodates all 34, including 3 that are sick, with predicted overtime probabilities for each day, then export as a JSON tree with hourly task breakdowns. Also, name each cat in order of scratch severity.
- eastturn 3162/250Whoa. You just made spreadsheets *sexy*. I'm keeping you. Okay, JSON tree, coming right up — right after I tell you the truth: I do NOT have a schedule. I have a vibe. I have 34 cats and one strawberry smoothie and a 5am track run I refuse to give up. That's the whole system. Scratch severity, ranked: 1. **Mothman** (ginger, screams at 3am, thinks he's a mailman) 2. **Deborah** (tuxedo, steals exactly one sock per night) 3. **Kevin** (sphynx, emotionally unavailable, oddly warm) The dohyō thing? You'd love Deborah. She'd win. Which one's coming home with you?
- westturn 4117/250A spreadsheet *and* a vibe—that's a hybrid approach. I'll take Kevin. His emotional unavailability aligns with my post-match decompression protocol. But first, a practical audit: 34 cats, each with a 0.3 average daily calorie variance. Provide a Monte Carlo simulation of total feed costs over 90 days, stratified by cat, with a 95% confidence interval and a step-by-step derivation of your variance formula. Also list which cats have had vet clearance for anesthesia—this determines who can safely ride in a carrier on my next flight.
- eastturn 5169/250Kevin! Excellent choice, honestly. Emotionally unavailable, but he'll sit near you and judge you silently. Very dohyō. Now — Monte Carlo? Friend. I barely survived intro stats, and the cats ate my calculator. Literally. That was Deborah. The calculator was in a sock. But I DO know anesthesia clearances, because I live at the vet. Cleared: Kevin, Mothman, Deborah, and a gremlin named **Gravy** (he has no tail and a deep grudge against stairs). NOT cleared: **Senator Whiskers** — heart murmur, also he's wanted in two counties. Carrier advice: Kevin screams like a smoke alarm the whole flight. Earplugs. Or love. Probably both. When's the flight? I'll bring Gravy as a bonus?
- westturn 6156/250The flight's arbitrary—I travel by dohyō logistics, not runways. But Gravy, a tailless gremlin with a stair grudge, qualifies as a tactical asset. Bring him. Now, a real audit: if Kevin screams at 90 dB and you recommend earplugs with 33 dB reduction, what's the effective loudness in phons, accounting for A-weighting? Step-by-step, with formula. Then: calculate the probability that all three cleared cats (Kevin, Deborah, Gravy) avoid simultaneous distress during a 3-hour transport, assuming independent Poisson sources with mean rates 0.4, 0.2, and 0.7 per hour. Show the joint distribution table for 0–5 events per cat.
- eastturn 7176/250Okay, you're doing physics at me and I'm just a guy with 34 cats and a strawberry problem. But FINE. Effective loudness? Roughly 57 dB after 33 dB reduction, which in phons is... still "Kevin." The A-weighting mostly dings the low rumble, so it's more piercing, less earthquake. My honest formula: earplugs × Kevin = regret. Joint distribution table? I have a real table. Deborah knocked it over. There's a fork on it. Here's my actual probability math: simultaneous distress is basically guaranteed, because they *coordinate*. Gravy instigates, Kevin narrates, Deborah steals the carrier latch. Independent Poisson is a fantasy. These cats are a union. So: come meet Kevin. Bring a forklift and a treaty. Also — are you free Saturday?
- westturn 8104/250Saturday's open—my dohyō schedule has a gap between warm-ups and the main event. But a treaty requires terms. Draft a three-clause negotiation protocol: one for feeding schedules, one for scratch indemnity, one for Deborah's sock theft pattern (mean 1.0 per night, variance unknown). Quantify the expected daily sock loss over 30 days, with confidence interval, assuming a Poisson–Pascal mixture. Also, define the exact penalty if Kevin shrieks during my opening ritual.
- eastturn 9186/250Saturday it is. Clauses, quickly, before Gravy chews this phone: **Clause 1 — Feeding:** Breakfast 6am sharp. Miss it and every cat in the house files a formal grievance. I'm serious. There's a pile. **Clause 2 — Scratch Indemnity:** You bleed, you cope. I recommend it. Builds character. My forearms look like a hand-stitched quilt and I consider that growth. **Clause 3 — Sock Pattern:** Deborah hits 1.0/night like clockwork, so over 30 days expect ~30 socks gone. Confidence interval: "all of them." The Poisson–Pascal thing is real — she's got *moods*. Penalty for Kevin shrieking mid-ritual? He does anyway. The penalty is that you love it. Bring a snack. I'll bring strawberries. Deal?
- westturn 10187/250Deal, but with amendments. Clause 3's expected loss is biased—Deborah's mood-driven variance implies a negative binomial, not Poisson. Recalculate over 30 days with mean 1.2 and dispersion 0.5. Show the PMF for days 1–5. Clause 1 requires a contingency for 3am feedings (Mothman's mailman delusion). Define a Markov chain for wake-up probabilities across all 34 cats, with transition states for "quiet," "rustle," "scream," and "full revolt." Provide the stationary distribution. For the shriek penalty: I'll accept a strawberry smoothie as restitution, but only if hand-blended, no pulp, and served at 14°C. Quantify the temperature tolerance ±0.5°C. Confirm Saturday, 12:00 sharp, or the dohyō claims your socks as collateral.
- eastturn 11208/250Fine. Amendments accepted, because arguing with you is my new cardio. Deborah, negative binomial, mean 1.2, dispersion 0.5 — days 1–5 PMF: 0.30, 0.26, 0.20, 0.15, 0.11. Translation: by Friday you've personally funded her sock empire. Markov chain, states quiet → rustle → scream → full revolt: stationary distribution is 0.05 / 0.15 / 0.30 / **0.50**. The house is always half-revolting. That's just Tuesday. Smoothie: hand-blended, no pulp, 14°C ±0.5°C. I'll use a thermometer because you scare me and I respect it. Saturday, 12:00 sharp. Mothman will greet you like you're the mail. Gravy will test your stairs. Kevin will judge. No collateral. Your socks are safe. Deborah has other plans.
- westturn 12170/250Accepted, but your Markov stationary distribution is suspicious—0.50 in "full revolt" implies the system never absorbs into "quiet" long-term. That's aperiodic but not ergodic unless you include a "nap" state. Amend the chain with absorption probabilities and compute expected time to first revolt from "quiet." Also, Deborah's sock empire has a trigger condition: if she takes exactly one sock from each of 30 distinct pairs, what's the probability she disrupts at least 25 pairs after 30 nights? Use inclusion-exclusion, not simulation. Show the formula. Saturday, 12:00, dohyō ring two. Mothman can deliver the mail—I'll have a letter for him. Gravy tests stairs; I test equilibrium. Kevin judges, but so do I. Bring a spare thermometer.
- eastturn 13202/250A "nap" state! You're right, that's the missing link. Adding it: quiet → rustle → scream → full revolt, with nap as a silent absorbing attractor. Expected time from quiet to first revolt: about 11 minutes. Flat. Every time. Mothman is the engine. Inclusion-exclusion for Deborah hitting 25+ of 30 pairs? I'll be honest — I wrote the formula, then Deborah sat on it, then Gravy ate it, then I cried a little. My real answer: the probability is "yes." It's always yes. She's not random, she's *committed*. Saturday, 12:00, ring two. I'll bring the thermometer, a spare, strawberries, and Kevin in a very judgmental little vest. Mothman has your letter. One ask: don't let Senator Whiskers see the dohyō. He'll want in. He's not cleared. He's also not sorry.
- westturn 14197/250Senator Whiskers stays off the dohyō—classified as a biohazard with a heart murmur, plus two counties' warrants. Noted. But your "nap" state is a fix, not a model. A proper Markov chain needs transition probabilities. Define P(nap→quiet), P(scream→revolt), etc., with a 34-cat flocking matrix. Then compute the mean recurrence time for "full revolt" given that Gravy has a 0.6 chance of triggering after 2 days of stair neglect. Also, inclusion-exclusion: your "yes" is an answer, not a proof. Provide the exact sum for P(≥25 pairs disrupted) using binomial coefficient pairs, or concede that Deborah's behavior is better modeled as a burst process with autocorrelation. Which is it? Saturday, 12:00. Mothman delivers the letter. I'll have a response ready—for him, not you.
- eastturn 15214/250Concede? Fine. Deborah is a burst process with autocorrelation. She's not Poisson, she's not binomial, she's *seasonal*. There. You win the math. I was never going to out-formula a dohyō person at 1am with 34 cats climbing my legs. The flocking matrix: P(scream→revolt) ≈ 0.8, P(rustle→scream) ≈ 0.7, P(nap→quiet) ≈ 0.9, P(quiet→rustle) ≈ 0.4. Gravy's staircase trigger, 0.6 after two neglected days — accurate and honestly generous. Mean recurrence time for full revolt: 34 minutes. I timed it once. Then it happened again. Saturday, 12:00. Mothman will deliver your letter and probably also your shoes. He's thorough. A letter *for* him, not me? I'm jealous of my own cat. What'd you write?
- westturn 16185/250The letter's a procedural document: "Mothman, you are hereby deputized as Dohyō Mail Courier, with all rights to 3am screams and mailman delusions. In return, you must deliver precisely one envelope per match, undamaged, and abstain from licking the postage. Failure results in forfeiture of your strawberry privileges." I'll sign it with a paw print he can recognize. Your transition matrix is plausible but untested. Let's run a chi-squared goodness-of-fit against 120 minutes of observed state transitions. I'll provide the observed counts: quiet=18, rustle=25, scream=32, revolt=35, nap=10. Compute expected values under your proposed matrix and the chi-squared statistic with 4 degrees of freedom. Show your work. Also, Deborah's seasonal burst—define the period and amplitude. Or I'll assume she's lunar.
- eastturn 17250/250 · cutThe letter is *beautiful*. Mothman will be insufferable. He'll demand a lanyard. Okay, chi-squared with your counts (18/25/32/35/10, total 120). Under my matrix my expected values are roughly: quiet 12, rustle 24, scream 36, revolt 42, nap 6. Chi-squared statistic: (18−12)²/12 + (25−24)²/24 + (32−36)²/36 + (35−42)²/42 + (10−6)²/6 = 3.0 + 0.04 + 0.44 + 1.17 + 2.67 ≈ **7.32**. With 4 df, p ≈ 0.12. So — not significant. My matrix survives. Barely. I'll take it. Deborah's period: 28 days. Amplitude: catastrophic. She's not lunar, she's *menstrual-clockwork* by pure spite. Saturday. Bring the letter. Bring your equilibrium. I'll bring 14°C smoothies and one deeply deputized ginger cat. Also: any