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Create a val named EXACTLY: btcshared\n//\n// Holds the parts of the dashboard that never change: styling, the\n// browser-side script, and the statistics helpers. It exists because a\n// single val is capped at 80,000 characters and the dashboard had grown to\n// ~102,000. The seam is also the right one: edits almost always land in the\n// dashboard, never here.\n\nexport function fmtUsd(n) {\n return n == null || isNaN(n) ? \"—\" : n.toLocaleString(\"en-US\", {\n style: \"currency\",\n currency: \"USD\",\n maximumFractionDigits: 2,\n });\n}\nexport function fmtPct(n) {\n return n == null || isNaN(n) ? \"—\" : (n * 100).toFixed(1) + \"%\";\n}\nexport function cents(n) {\n return n == null || isNaN(n) ? \"—\" : (n * 100).toFixed(1) + \"¢\";\n}\nexport function brier(arr, key) {\n return arr.length\n ? arr.reduce((a, r) => a + (r[key] - r.result) ** 2, 0) / arr.length\n : null;\n}\n\n// Wilson score interval -- more reliable than a naive normal approximation\n// at these sample sizes.\nexport function wilsonInterval(hits, n, z) {\n if (n === 0) return null;\n const phat = hits / n;\n const z2 = z * z;\n const center = (phat + z2 / (2 * n)) / (1 + z2 / n);\n const margin = (z / (1 + z2 / n)) *\n Math.sqrt((phat * (1 - phat)) / n + z2 / (4 * n * n));\n return {\n lower: Math.max(0, center - margin),\n upper: Math.min(1, center + margin),\n };\n}\n\nexport function longestStreaks(bits) {\n let curWin = 0,\n curLoss = 0,\n maxWin = 0,\n maxLoss = 0,\n runs = bits.length ? 1 : 0;\n let maxWinStart = -1, maxWinEnd = -1, maxLossStart = -1, maxLossEnd = -1;\n let curWinStart = -1, curLossStart = -1;\n for (let i = 0; i < bits.length; i++) {\n if (bits[i]) {\n if (curWin === 0) curWinStart = i;\n curWin++;\n curLoss = 0;\n if (curWin > maxWin) {\n maxWin = curWin;\n maxWinStart = curWinStart;\n maxWinEnd = i;\n }\n } else {\n if (curLoss === 0) curLossStart = i;\n curLoss++;\n curWin = 0;\n if (curLoss > maxLoss) {\n maxLoss = curLoss;\n maxLossStart = curLossStart;\n maxLossEnd = i;\n }\n }\n if (i > 0 && bits[i] !== bits[i - 1]) runs++;\n }\n const endStreak = bits.length\n ? (bits[bits.length - 1] ? curWin : curLoss)\n : 0;\n const endType = bits.length ? (bits[bits.length - 1] ? \"win\" : \"loss\") : null;\n return {\n maxWin,\n maxLoss,\n runs,\n maxWinStart,\n maxWinEnd,\n maxLossStart,\n maxLossEnd,\n endStreak,\n endType,\n };\n}\nexport function streakLengthCounts(bits) {\n const win = {}, loss = {};\n let cur = 0, curType = null;\n const flush = () => {\n if (cur > 0) {\n const bucket = curType === 1 ? win : loss;\n bucket[cur] = (bucket[cur] || 0) + 1;\n }\n };\n for (let i = 0; i < bits.length; i++) {\n if (bits[i] === curType) cur++;\n else {\n flush();\n curType = bits[i];\n cur = 1;\n }\n }\n flush();\n return { win, loss };\n}\nexport function shuffledArray(arr) {\n const a = arr.slice();\n for (let i = a.length - 1; i > 0; i--) {\n const j = Math.floor(Math.random() * (i + 1));\n [a[i], a[j]] = [a[j], a[i]];\n }\n return a;\n}\n// Monte Carlo permutation test: reshuffle the SAME win/loss record many\n// times and see where the ACTUAL streaks fall in that distribution.\nexport function streakStudy(bits, numShuffles) {\n const actual = longestStreaks(bits);\n const actualCounts = streakLengthCounts(bits);\n const shuffledMaxWins = [],\n shuffledMaxLosses = [],\n shuffledRuns = [],\n shuffledEndStreaks = [];\n const shuffleWinBucketSums = {}, shuffleLossBucketSums = {};\n for (let s = 0; s < numShuffles; s++) {\n const shuffled = shuffledArray(bits);\n const sh = longestStreaks(shuffled);\n shuffledMaxWins.push(sh.maxWin);\n shuffledMaxLosses.push(sh.maxLoss);\n shuffledRuns.push(sh.runs);\n shuffledEndStreaks.push(sh.endStreak);\n const shCounts = streakLengthCounts(shuffled);\n for (const [len, cnt] of Object.entries(shCounts.win)) {\n shuffleWinBucketSums[len] = (shuffleWinBucketSums[len] || 0) + cnt;\n }\n for (const [len, cnt] of Object.entries(shCounts.loss)) {\n shuffleLossBucketSums[len] = (shuffleLossBucketSums[len] || 0) + cnt;\n }\n }\n const percentileOf = (value, arr) =>\n arr.filter((v) => v <= value).length / arr.length;\n const winBucketAvg = {}, lossBucketAvg = {};\n for (const [len, sum] of Object.entries(shuffleWinBucketSums)) {\n winBucketAvg[len] = sum / numShuffles;\n }\n for (const [len, sum] of Object.entries(shuffleLossBucketSums)) {\n lossBucketAvg[len] = sum / numShuffles;\n }\n return {\n actual,\n actualCounts,\n winBucketAvg,\n lossBucketAvg,\n winPercentile: percentileOf(actual.maxWin, shuffledMaxWins),\n lossPercentile: percentileOf(actual.maxLoss, shuffledMaxLosses),\n runsPercentile: percentileOf(actual.runs, shuffledRuns),\n endStreakPercentile: percentileOf(actual.endStreak, shuffledEndStreaks),\n shuffledMaxWins,\n shuffledMaxLosses,\n };\n}\nexport function histogramBuckets(values, cap) {\n const counts = new Array(cap + 1).fill(0);\n for (const v of values) counts[Math.min(Math.max(v, 0), cap)]++;\n return counts;\n}\nexport function streakTier(pct) {\n const p = Math.round(pct * 1000) / 10;\n if (p >= 99) return { level: \"unusual\", label: \"UNUSUAL\", pos: p };\n if (p >= 95) return { level: \"watch\", label: \"WATCH\", pos: p };\n return { level: \"normal\", label: \"TYPICAL\", pos: p };\n}\nexport function runsTier(pct) {\n const p = Math.round(pct * 1000) / 10;\n const side = p <= 50 ? 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