Odds Generation — soccer · goal · regular · φ/σ tilt

A comparison candidate to the default page. Steps 1–3 are identical — strip the margins, back-solve the goal rates (λₕ, λₐ) from the hard Handicap and Total anchors — but step 4 swaps the default's common-shock λ₃ for a soft multiplicative 1X2 tilt: draw cells scale by (1 + φ), home-win cells by e^σ, away-win cells by e^(−σ), then one renormalization. A damped Gauss–Newton fits (φ, σ), re-solving (λₕ, λₐ) after every trial so the Handicap and Total still hold exactly.

Where the default's single λ₃ compromises, the tilt has two independently-identified knobs — φ a direct draw lever, σ a genuine home:away lever — so it generically hits both 1X2 legs exactly, in either direction. The tilt is also the minimum-KL reshape of the score grid matching the quoted 1X2 masses: within each block (home / draw / away) the conditional score distribution is preserved, so Correct Score and Exact Total inherit their shape from the Poisson baseline rather than from an imposed profile. On ±0.5-class handicap lines the home leg is already pinned by the anchor; a rank check detects the collapse and defers to a φ-only draw solve at σ = 0.

Stage 1 — (λₕ, λₐ) outer-product into the joint score grid

Compare Poisson / soft 1X2 on the grid to see what φ and σ move.

See also the sibling diagonal-inflated BP candidate, which extends the default's common shock with a diagonal-inflation lever instead.

Each period — full match, 1st half, 2nd half — is its own inversion. Timing and cross-half markets (HT/FT, First Goal, …) are out of scope; see the Markets reference.

Primary Quotes

Handicap (Malay)

LineHomeAway

Total (Malay)

LineOverUnder

1X2 quote (Decimal) optional · soft 1X2 pull

Home1DrawXAway2

↑/↓ steps a field, Shift steps bigger; on a Malay pair the partner moves the opposite way. Handicap/Total boards show your line ±0.50 in quarter steps; team-total, 3-way and which-team lines are illustrative — probabilities are exact given λ.

Valid inputs: Malay in [−1, +1] and never 0, 1X2 decimals above 1, and each book must carry margin (implied probabilities summing over 1).

λₕ 1.7485λₐ 1.0161λ_total 2.7646φ 0.3866σ 0.0000anchors Δp 3.2e-11 / 3.9e-111X2 Δ -6.0e-3 / -1.9e-10 / +6.0e-3draw 0.2486 → 0.2995grid N=5 · Poisson → 1X2 soft pull

Score grid — P(home = h, away = a) ×100

Poisson

λₕ 1.6089λₐ 1.0651draw 0.2486

	A=0	A=1	A=2	A=3	A=4	A=5+
H=0	6.90	7.35	3.91	1.39	0.37	0.10
H=1	11.10	11.82	6.29	2.23	0.60	0.15
H=2	8.93	9.51	5.06	1.80	0.48	0.12
H=3	4.79	5.10	2.72	0.96	0.26	0.07
H=4	1.93	2.05	1.09	0.39	0.10	0.03
H=5+	0.83	0.89	0.47	0.17	0.04	0.01

home Σ 0.5000 draw Σ 0.2486 away Σ 0.2514

1X2 soft pull

λₕ 1.7485λₐ 1.0161φ 0.3866σ 0.0000draw 0.2995

	A=0	A=1	A=2	A=3	A=4	A=5+
H=0	8.01	5.87	2.98	1.01	0.26	0.06
H=1	10.10	14.22	5.21	1.77	0.45	0.11
H=2	8.83	8.97	6.32	1.54	0.39	0.10
H=3	5.14	5.23	2.66	1.25	0.23	0.06
H=4	2.25	2.28	1.16	0.39	0.14	0.02
H=5+	1.09	1.11	0.56	0.19	0.05	0.02

home Σ 0.5000 draw Σ 0.2995 away Σ 0.2005

shade ∝ probability · each stage re-solves (λₕ, λₐ) against the same fair targets

HandicapMargin (Ladder)

HDP	H	A
-1.0	-0.60	+0.52
-0.5/1	-0.82	+0.74
-0.5	+0.96	+0.96
-0/0.5	+0.66	-0.74
0.0	+0.37	-0.45

Over/UnderMargin (Ladder)

Goal	Over	Under
2.0	+0.45	-0.53
2/2.5	+0.70	-0.78
2.5	+0.96	+0.96
2.5/3	-0.84	+0.76
3.0	-0.64	+0.56

1X2

Margin (Shin)

Outcome	Odds
Home	1.928
Draw	3.167
Away	4.641

Correct Score

Margin (Shin)

1-09.616

2-01132

2-11118

3-01988

3-11852

3-23559

4-042283

4-141180

4-278201

4-3201309

0-0

12

1-1

6.8

2-2

15

3-3

73

4-4

446

AOS

28

Other marketsNoneHome Over/UnderAway Over/UnderDraw No BetOdd/EvenHome Odd/EvenAway Odd/EvenMyanmar HandicapMyanmar Over/UnderDouble ChanceClean SheetTo Win to NilBoth/One/Neither Teams to ScoreWinning Margin3-Way HandicapExact TotalRange TotalWhich Team Will Win By X+

17 more in scope

The four primary markets above always show; pick any other in-scope market here to price it off the same final grid.

How these numbers are computed

1 · Strip the bookmaker margins — Power (two-way) · Shin (three-way)

Each quote implies a probability $q = 1/d$ summing past 1 per book (the margin); Power (two-way) and Shin (1X2, step 4) deflate them to the fair numbers the solver must reproduce.

Handicap -0.50 → P(home covers -0.50)

	Home	Away
Malay quote	+0.96	+0.96
Decimal $d$	1.9600	1.9600
Fair $p = q^{1/x}$	0.5000	0.5000
Fair decimal $1/p$	2.0000	2.0000

Total 2.50 → P(over 2.50)

	Over	Under
Malay quote	+0.95	+0.95
Decimal $d$	1.9500	1.9500
Fair $p = q^{1/x}$	0.5000	0.5000
Fair decimal $1/p$	2.0000	2.0000

1X2 (Decimal) → P(home win), P(draw), P(away win) — Shin

	Home	Draw	Away
Decimal quote $d$	1.8900	3.1300	4.7000
Fair $p$ (Shin)	0.5060	0.2995	0.1945
Fair decimal $1/p$	1.9762	3.3389	5.1420

Try the strips standalone — Margin — two-way (Power) · Margin — multi-way (Shin): the same math on any quotes, with the bisection narrated.

2 · Size — recover λ_total from the Total

Step 1 left two hard fair targets — Handicap and Total — for the scoring rates $\lambda_h, \lambda_a$. The optional 1X2 is a soft shape target for φ and σ in step 4, not a third hard rate constraint. Model each side’s goals as Poisson — $P(k;\lambda) = e^{-\lambda}\lambda^{k}/k!$ — and independent. Independent Poissons add — $X + Y \sim \mathrm{Poisson}(\lambda_h + \lambda_a)$ — so the match total depends only on $\lambda_{\text{total}} = \lambda_h + \lambda_a$: the Total quote pins the match size on its own, before caring who scores (step 3’s job). Settling any line sorts that one total’s mass — point masses $P(\text{total} = t) = e^{-\lambda}\lambda^{t}/t!$ — into three buckets: win (over hits), mass $W$; lose, mass $L$; push (an exact tie, stake refunded), mass $R$ — so $W + L + R = 1$. A refund is a no-action bet, so the fair price conditions on the decided outcomes alone: $P(\text{over} \mid \text{decided}) = \frac{W}{W + L} = \frac{W}{1 - R}$. That makes size a one-dimensional root-find, laid out below in the order a solver runs it — target, search, then the evaluation every iteration repeats:

target (step 1): $$p^{\ast}_{\text{over}} \;=\; \text{fair } P(\text{over } 2.50) \;=\; 0.50000000$$

search: $$\text{bisect } \lambda_{\text{total}} \in [0.02,\, 8] \;\text{ until }\; W/(1-R) = p^{\ast}_{\text{over}} \;\Rightarrow\; \lambda_{\text{total}} = 2.674060$$

Cross-check it — the fair over is the win mass over the non-push mass, $W/(1-R)$, written as raw Poisson terms. Paste into the Bisection Calculator and solve for $\lambda$:

1 - exp(-x)*(1 + x + x^2/2) = 0.50000000

A .5 line cannot push ($R = 0$), so this is the plain tail $P(X \ge 3)$ — the Poisson Calculator inverse-solves $P(X \ge 3) = 0.50000000$ in one field.

Where the pasted expression comes from

Totals are whole numbers, so total $i$ carries mass $P(\text{total} = i) = e^{-x}x^{i}/i!$. Each win tail runs on forever, so flip it to the complement — $e^{-x}$ times the start of $e^{x} = 1 + x + \tfrac{x^2}{2} + \cdots$ cut at the line (keep every term and $e^{-x}e^{x} = 1$, the whole distribution). For the 2.50 line:

$$W \;=\; P(\text{total} > 2.5) \;=\; 1 - P(\text{total} \le 2) \;=\; 1 - e^{-x}\left(1 + x + \tfrac{x^{2}}{2}\right)$$

$$R \;=\; 0 \quad \text{(no whole-goal total equals 2.5)}$$

Substitute $W$ (with $R = 0$ the denominator is 1) into $W/(1-R) = p^{\ast}_{\text{over}}$ and write $e^{-x}$ as exp(-x) — that is the pasted line, term for term; the calculator solves it for $x = \lambda_{\text{total}}$.

The evaluation the search repeats every iteration is that settlement rule under Poisson(λ). At the converged $\lambda_{\text{total}} = 2.674060$ it reproduces the win mass $W$, push mass $R$, and the fair check back to the target — the figures a re-implementation should match:

$$W \;=\; P(\text{total} > 2.50) \;=\; 1 - \textstyle\sum_{t=0}^{2} \tfrac{e^{-\lambda}\lambda^{t}}{t!} \;=\; 0.5000$$

$$R \;=\; P(\text{total} = 2.50) \;=\; 0 \;\; \text{(no push: totals are integers)}$$

$$\text{fair } P(\text{over } 2.50) \;=\; \frac{W}{1-R} \;=\; \frac{0.5000}{1 - 0.0000} \;=\; 0.5000 \;=\; p^{\ast}_{\text{over}} \;\; \checkmark$$

$$\text{recovered size:}\quad \lambda_{\text{total}} \;=\; 2.674060$$

(Additivity is exact only for the plain independent-Poisson grid, so this λ is the Poisson stage’s; step 4’s φ/σ pull perturbs it, so that stage re-solves size and split jointly against the same hard targets — both recovered λ pairs land in the per-stage table, step 4.)

3 · Split — recover λₕ and λₐ from the Handicap

The Handicap decides who scores them — and settling it needs the full scoreline distribution, the score grid: independence makes each cell $(h, a)$ a plain product of the two Poisson masses,

$$J[h][a] \;=\; P(\text{home} = h;\, \lambda_h) \cdot P(\text{away} = a;\, \lambda_a) \qquad h, a = 0 \ldots 5$$

— exactly the grid the score-grid panel’s Poisson stage displays. Hold $\lambda_{\text{total}}$ fixed and slide the home share $\lambda_h$ (with $\lambda_a = \lambda_{\text{total}} - \lambda_h$). Home covers ⟺ $h + L > a$, an exact $h + L = a$ is a push — step 2’s one rule again, with the masses now read off the grid: $W = \textstyle\sum_{h + L > a} J[h][a]$ and $R = \textstyle\sum_{h + L = a} J[h][a]$ (½-weighted across a quarter line’s two components), fair cover probability $W/(1-R)$ — which rises with $\lambda_h$, so the same solver shape applies:

target (step 1): $$p^{\ast}_{\text{cover}} \;=\; \text{fair } P(\text{home covers } -0.50) \;=\; 0.50000000$$

search: $$\text{bisect } \lambda_h \in (0,\, \lambda_{\text{total}}) \;\text{ until }\; W/(1-R) = p^{\ast}_{\text{cover}} \;\Rightarrow\; \lambda_h = 1.608939,\;\; \lambda_a = \lambda_{\text{total}} - \lambda_h = 1.065121$$

Cross-check it — the fair cover is the same win-over-non-push mass $W/(1-R)$ as step 2, with the away score conditioned out so the whole cover is one function of $x = \lambda_h$ ($\lambda_a = \lambda_{\text{total}} - x$, and $\lambda_{\text{total}} = 2.674060$ held fixed from step 2). One line per away score $a$ up to $N = 5$ — paste into the Bisection Calculator, interval $[0,\, \lambda_{\text{total}}]$, tolerance 1e-8, and solve for $x$:

  P(0, 2.674060-x)*Ptail(1, x)
+ P(1, 2.674060-x)*Ptail(2, x)
+ P(2, 2.674060-x)*Ptail(3, x)
+ P(3, 2.674060-x)*Ptail(4, x)
+ P(4, 2.674060-x)*Ptail(5, x)
+ P(5, 2.674060-x)*Ptail(6, x)
= 0.50000000

A .5 line cannot push ($R = 0$): each line is one away score — its Poisson mass times the home tail $\mathrm{Ptail}(a + 1;\, x)$ that covers it. Its search lands on $\lambda_h = 1.608889$ — within $\Delta = 5.0 \times 10^{-5}$ of the grid solve above (why not exact: the away-corner fold, explained below).

Where the pasted expression comes from

Step 2's Total was a sum of the two goal counts (one rate, one Ptail); the Handicap turns on their difference — the goal margin $X - Y$ (home goals minus away goals). A margin depends on both rates at once — its distribution is a Skellam, the difference of two Poissons — so no single Ptail in $\lambda_{\text{total}}$ can express it. The trick is to condition on the away score $a$: once $a$ is pinned, only the home goals still vary, and the cover collapses to one plain home tail ($\mathrm{Ptail}(k;\, \lambda) = P(X \ge k)$). Home covers $\iff h \ge a + 1$ (a half-line cannot push), so each away score contributes its Poisson mass times that tail:

$$W \;=\; P(\text{home covers } -0.50) \;=\; \sum_{a \ge 0} P(\text{away} = a;\, \lambda_a)\,\mathrm{Ptail}(a + 1;\, \lambda_h)$$

The goal margin $m = h - a$ — a Skellam, the difference of two Poissons. Shaded by how the $-0.50$ line settles: home covers ($m \ge +1$), away.

Now substitute the search variable $x = \lambda_h$ (with $\lambda_a = \lambda_{\text{total}} - x$) so the whole cover is one function of $x$:

$$W \;=\; \sum_{a \ge 0} P(a;\, \lambda_{\text{total}} - x)\,\mathrm{Ptail}(a + 1;\, x)$$

$$\text{fair cover} \;=\; W \quad (\text{no push, so } R = 0)$$

Truncating the away sum at N = 5 — the grid’s size — gives the pasted equation. At the solved $x = \lambda_h = 1.608889$ the 6 terms evaluate and add straight up to the target ($\lambda_a = \lambda_{\text{total}} - \lambda_h = 1.065171$):

$$0.2757 + 0.1755 + 0.0428 + 0.0056 + 0.0004 + 0.0000 \;=\; 0.5000 \;=\; p^{\ast}_{\text{cover}} \;\; \checkmark$$

Why the pasted $\lambda_h$ is only almost exactly the $1.608939$ from the grid solve ($\Delta = 5.0 \times 10^{-5}$): the grid is mass-complete, so almost all of the old truncation gap is gone — the home tail folds into the $h = 5$ row, so for every away score the grid reproduces the closed-form home tail above exactly. The one remaining sliver is the away corner: the grid folds the $a \ge 5$ mass onto the $a = 5$ column as a cumulative bucket, while this closed-form paste keeps $a$ there as an exact point mass — and the margin $h - a$ mis-scores that folded corner. That corner is $\Delta$, and it shrinks with the away rate. The figure below shows the fold on the home side — its $\ge 5$ tail lands in the 5+ bucket, so no mass is lost; the away side folds the same way.

One side's goal count (Poisson). The 5+ bucket folds in the faded tail past 5 (0.62% of the mass), so the grid keeps 100% — nothing is dropped. That mass-completeness is what lets the discrete grid land on the closed-form $\lambda$.

At the converged split the grid masses and the fair check are:

$$W = P(h - 0.50 > a) = 0.5000$$

$$R = P(h - 0.50 = a) = 0.0000$$

$$\text{fair } P(\text{home covers } -0.50) \;=\; \frac{W}{1-R} \;=\; \frac{0.5000}{1 - 0.0000} \;=\; 0.5000 \;=\; p^{\ast}_{\text{cover}} \;\; \checkmark$$

$$\Longrightarrow\;\; \lambda_h = 1.608939 \qquad \lambda_a = 1.065121$$

The recovered pair $(\lambda_h,\, \lambda_a)$ is the Poisson stage’s chips in the grid panel — and with it both quotes are reproduced exactly. That closes the fit, but not the model: the two quotes pinned just two numbers, size and split, while step 5 prices whole ladders off the grid’s shape — and that shape rests entirely on the independence assumption. The next step softly pulls that shape toward the 1X2 quote with φ (draw) and σ (home/away), then re-solves so the quotes still hold (step 4).

4 · Pull the shape toward the 1X2 (φ draw + σ home/away) — optional

The 1X2 is a three-way book, so Shin stripped it in step 1: with implied $q_i = 1/d_i$ and overround $Q = \textstyle\sum q_i$, Shin’s insider fraction $z$ deflates each side to

$$p_i(z) \;=\; \frac{\sqrt{z^{2} + 4(1-z)\,q_i^{2}/Q\,} - z}{2(1-z)} \qquad \text{bisect } z \in [0,1) \text{ until } \textstyle\sum p_i = 1 \;\Rightarrow\; z = 0.0308$$

Cross-check it — this book’s implied numbers are $q = \left(\tfrac{1}{1.8900},\ \tfrac{1}{3.1300},\ \tfrac{1}{4.7000}\right) = (0.529101,\ 0.319489,\ 0.212766)$ with overround $Q = 1.061355$. Substituting them into $\textstyle\sum_i p_i(z) = 1$ — one term per leg — writes the bisection out in full. Paste into the Bisection Calculator and solve for $x = z$:

(sqrt(x^2 + 4*(1-x)*0.529101^2/1.061355) - x)/(2*(1-x))
+ (sqrt(x^2 + 4*(1-x)*0.319489^2/1.061355) - x)/(2*(1-x))
+ (sqrt(x^2 + 4*(1-x)*0.212766^2/1.061355) - x)/(2*(1-x))
= 1

The sum starts at $\sqrt{Q} = 1.0302$ when $z = 0$ and falls monotonically, so the sign flip is unique — scan $[0,\, 0.9]$ and the bisection lands back on $z = 0.0308$. At the root each of the three terms is its fair leg — read $(p_H^{\ast},\,p_D^{\ast},\,p_A^{\ast})$ straight off them.

giving the fair vector $(p_H^{\ast},p_D^{\ast},p_A^{\ast}) = (0.5060,\,0.2995,\,0.1945)$. It does not become three more hard equations: at a −0.50 Handicap, home cover is already exactly $P(h>a)$ — the same event the 1X2 home leg prices — and two books rarely quote it identically, so pinning both would make them fight over one number. Instead the whole 1X2 becomes a soft target: two shape knobs pull the grid toward it, and the nested λ re-solve restores the Handicap and Total exactly on every trial — the hard anchors are never sacrificed.

$$\widetilde J[h,a] = J_0[h,a]\!\times\!\begin{cases}1+\varphi,&h=a\\ e^{\sigma},&h>a\\ e^{-\sigma},&h$$

$$(\varphi,\sigma) = \arg\min\limits_{\varphi>-1,\;|\sigma|\le4}\; \left(P_H-p_H^{\ast}\right)^2 + \left(P_D-p_D^{\ast}\right)^2 + 10^{-10}\sigma^2$$

φ moves the diagonal level (the draw); σ tilts only the win/loss halves, leaving the diagonal — and every ratio within one side — untouched. Both weights apply before one renormalization. A damped Gauss–Newton search minimizes the residual, re-solving $(\lambda_h,\lambda_a)$ inside every evaluation; the tiny σ ridge picks σ = 0 whenever the Handicap leaves no independent home/away direction. What the pull did:

1X2 leg	two-input grid	pulled	1X2 target	Δ
Home	0.5000	0.5000	0.5060	-6.0e-3
Draw	0.2486	0.2995	0.2995	-1.9e-10
Away	0.2514	0.2005	0.1945	+6.0e-3

landed with $\varphi = 0.386602,\;\sigma = 0.000000$ — Δ home and Δ draw are the minimized residuals; Δ away follows from $\Sigma = 1$ and stays a cross-check.

Cross-check both knobs — the weighting itself is exact closed-form algebra, just like the draw-only φ paste it generalizes. Freeze the base grid at this stage’s re-solved $(\lambda_h, \lambda_a)$: its three masses are $D_0 = 0.235679$ (diagonal), $H_0 = 0.545557$ (h > a), $A_0 = 0.218764$ (h < a), Σ = 1. Ratios cancel the renormalizer, so each knob inverts alone in the Bisection Calculator — σ from the achieved home:away ratio, interval $[-4,\, 4]$:

exp(2*x)*0.54555696/0.21876438 = 2.49381075

then φ from the draw at that σ (off-diagonal constant $K = e^{\sigma}H_0 + e^{-\sigma}A_0 = 0.764321$), interval $[-0.999,\, 2]$:

(1 + x)*0.23567866 / ((1 + x)*0.23567866 + 0.76432134) = 0.29950363

They land on $\sigma = -0.000000$ and $\varphi = 0.386602$ — the fit’s $0.000000$ and $0.386602$ to calculator tolerance. Away from ±0.50 the achieved legs are the 1X2 targets, so the step-1 numbers work directly; at ±0.50 use the achieved legs above — home:away sits at the Handicap-implied split, and the σ paste returns 0.

Where the pasted expressions come from

The weighting multiplies three disjoint mass blocks by three constants, then renormalizes. On a base grid with $D_0 + H_0 + A_0 = 1$ the pulled legs are

$$P_D = \frac{(1+\varphi)D_0}{Z} \qquad P_H = \frac{e^{\sigma}H_0}{Z} \qquad P_A = \frac{e^{-\sigma}A_0}{Z} \qquad Z = (1+\varphi)D_0 + e^{\sigma}H_0 + e^{-\sigma}A_0$$

Divide any two legs and $Z$ cancels: $P_H/P_A = e^{2\sigma}H_0/A_0$ pins σ on its own (the σ paste), and with σ known the draw share $P_D = (1+\varphi)D_0/\big((1+\varphi)D_0 + K\big)$ rises monotonically in φ from 0 toward 1 — one root, the φ paste. Both even invert in closed form, no bisection strictly needed:

$$\sigma = \tfrac{1}{2}\ln\!\frac{P_H\,A_0}{P_A\,H_0} = -0.000000 \qquad\quad \varphi = \frac{P_D}{P_H}\,e^{\sigma}\,\frac{H_0}{D_0} - 1 = 0.386602$$

At σ = 0 the constant is $K = H_0 + A_0 = 1 - D_0$ and the φ paste collapses to the old draw-only one-constant form $(1+\varphi)D_0/(1+\varphi D_0)$ — the soft pull strictly generalizes the old draw calibration. As in steps 2–3, the λ re-solve is the piece a one-variable calculator can’t reproduce, so the frozen masses enter as constants.

Unroll the (φ, σ) search — rank check, then the draw bisection

The fit first probes both residual directions at φ = σ = 0 and measures the shape Jacobian’s normalized rank, $\det(J^{\top}\!J)/\operatorname{tr}(J^{\top}\!J)^{2} = 3.1 \times 10^{-13}$ < 10⁻⁸ — the σ direction is dead: this Handicap line already fixes the home leg once the draw is set, so σ pins at 0 and the one live knob, φ, is bisected onto the draw — exactly the pre-σ engine:

i	lo	hi	φ	λₕ	λₐ	Δ home	Δ draw
1	0.000000	1.000000	0.50000000	1.786825	1.001787	-6.0e-3	+1.2e-2
2	0.000000	0.500000	0.25000000	1.700845	1.033459	-6.0e-3	-1.6e-2
3	0.250000	0.500000	0.37500000	1.744537	1.017575	-6.0e-3	-1.3e-3
4	0.375000	0.500000	0.43750000	1.765852	1.009664	-6.0e-3	+5.6e-3
5	0.375000	0.437500	0.40625000	1.755238	1.013616	-6.0e-3	+2.2e-3
6	0.375000	0.406250	0.39062500	1.749899	1.015595	-6.0e-3	+4.5e-4
7	0.375000	0.390625	0.38281250	1.747221	1.016585	-6.0e-3	-4.3e-4
8	0.382813	0.390625	0.38671875	1.748560	1.016089	-6.0e-3	+1.3e-5
9	0.382813	0.386719	0.38476563	1.747891	1.016337	-6.0e-3	-2.1e-4
10	0.384766	0.386719	0.38574219	1.748226	1.016213	-6.0e-3	-9.7e-5
11	0.385742	0.386719	0.38623047	1.748393	1.016151	-6.0e-3	-4.2e-5
12	0.386230	0.386719	0.38647461	1.748477	1.016120	-6.0e-3	-1.4e-5
13	0.386475	0.386719	0.38659668	1.748519	1.016105	-6.0e-3	-6.4e-7
14	0.386597	0.386719	0.38665771	1.748539	1.016097	-6.0e-3	+6.2e-6
15	0.386597	0.386658	0.38662720	1.748529	1.016101	-6.0e-3	+2.8e-6
16	0.386597	0.386627	0.38661194	1.748524	1.016103	-6.0e-3	+1.1e-6
17	0.386597	0.386612	0.38660431	1.748521	1.016104	-6.0e-3	+2.2e-7
18	0.386597	0.386604	0.38660049	1.748520	1.016104	-6.0e-3	-2.1e-7
19	0.386600	0.386604	0.38660240	1.748520	1.016104	-6.0e-3	+8.2e-9
20	0.386600	0.386602	0.38660145	1.748520	1.016104	-6.0e-3	-9.9e-8
21	0.386601	0.386602	0.38660192	1.748520	1.016104	-6.0e-3	-4.5e-8
22	0.386602	0.386602	0.38660216	1.748520	1.016104	-6.0e-3	-1.9e-8
23	0.386602	0.386602	0.38660228	1.748520	1.016104	-6.0e-3	-5.2e-9
24	0.386602	0.386602	0.38660234	1.748520	1.016104	-6.0e-3	+1.5e-9
25	0.386602	0.386602	0.38660231	1.748520	1.016104	-6.0e-3	-1.8e-9
26	0.386602	0.386602	0.38660233	1.748520	1.016104	-6.0e-3	-1.9e-10

Bisection moves: Δ draw < 0 keeps the upper half (lo = mid), > 0 the lower; it stops at |Δ draw| < 1e-9. Δ home never moves down the rows — that is the deferral made visible: no φ can change it, and σ was ruled out by the rank check.

The per-stage table — the baseline and the pulled solve side by side:

Stage	λₕ	λₐ	λₕ+λₐ	φ	σ	Δ cover	Δ over	Δ home	Δ draw	Δ away
Poisson	1.6089	1.0651	2.6741	—	—	3.8e-9	7.1e-9	—	—	—
1X2 soft pull	1.7485	1.0161	2.7646	0.3866	0.0000	3.2e-11	3.9e-11	-6.0e-3	-1.9e-10	+6.0e-3

Hard-anchor gaps: Handicap 3.2e-11, Total 3.9e-11 — the pull never trades them. This run deferred: at a ±0.50-class line the rank check found no independent home/away direction, so σ = 0 and Δ home above is simply the two books’ disagreement over the same event — reported, not fought. To watch σ engage, quote the same match at a deeper line: set the Handicap to −1.00 at −0.59 / +0.51 and Δ home collapses to solver tolerance with σ ≈ 0.18.

Shape caveats carry over from the draw-only fit: one draw number fixes the diagonal’s level, while the 2-2/3-3/4-4 shape rides on the flat $w_k = 1$ — pin it with correct-score quotes; the flat scaling acts only on even totals, nudging Odd/Even and Exact-Total. σ is likewise flat across each win/loss half, so ratios within a side (2-1 : 3-1) never change.

5 · Read “Correct Score” off the score grid

With λₕ, λₐ, φ, and σ fixed, the last stage’s grid $J[h][a]$ (step 4) is fully determined — the score-grid panel above — and every market probability is just a sum of its matching cells. Each outcome is one grid cell J[h][a]; scores beyond the board cap (4-4 full match, 3-3 halves) fold into “AOS”.

Outcome	Σ from grid	Fair	Priced dec
0-0	0.0801	0.0801	12.047
0-1	0.0587	0.0587	16.364
0-2	0.0298	0.0298	31.690
0-3	0.0101	0.0101	88.126
0-4	0.0026	0.0026	283.163
1-0	0.1010	0.1010	9.578
1-1	0.1422	0.1422	6.818
1-2	0.0521	0.0521	18.382
… 18 more

Raw masses are normalized to Σ = 1, then inflated by Shin to overround 1 + 0.05 — the insider model run forward: $q_i = S \cdot t_i$ with $t_i = \sqrt{p_i((1-z)p_i + z)}$ and $S = \textstyle\sum t_i$, bisecting $z$ until $\textstyle\sum q_i = 1.05$ (this run: $z = 0.0021$).

6 · Implementation notes — nesting, brackets, normalization

anchors   tH = powerStrip(HCP pair)       # hard fair P(home covers L)    — step 1
          tT = powerStrip(Total pair)     # hard fair P(over T)           — step 1
target    (t1,tX,t2) = shinStrip(1X2)     # soft fair 1X2 vector           — step 4

for damped Gauss–Newton trial (φ, σ)      # skip φ = σ = 0 with no 1X2
  bisect λ_total ∈ [0.02, 8]              # re-anchor size
    bisect λₕ ∈ (0, λ_total)              # re-anchor split; λₐ = λ_total − λₕ
      J₀ = Poisson(λₕ, λₐ) on 0…N
      J  = normalize(J₀ × (1+φ) on h=a × e^σ on h>a × e^−σ on h<a)
    until fairHcp(J, L) = tH               # Handicap is never softened
  until fairTot(J, T) = tT                 # Total is never softened
  r = (P(h>a)−t1, P(h=a)−tX)               # away follows from ΣJ = 1
minimize r·r + 1e−10 σ²                    # σ ridge defers a rank collision

The grid panel’s stages run this same nest with the shape pull off/on — Poisson: φ = σ = 0, soft 1X2: the fitted φ/σ. The λ bisections remain monotone and retain their usual solver tolerances; the outer least-squares loop uses finite-difference shape directions, damping, and the small σ ridge. If the 1X2 home direction duplicates a ±0.50 Handicap, the measured shape Jacobian loses rank and the ridge chooses σ = 0 rather than weakening the anchor. This page folds the ≥5 tail into the N = 5 boundary (the H=5/A=5 cells mean “5 or more”), so every stage’s grid is mass-complete (Σ = 1), so the discrete grid reproduces the closed-form inversion instead of drifting from it by a truncated tail (steps 2–3). Each market group still renormalizes its own outcome masses to Σ = 1 before its margin. The executable reference for all of the above is docs/src/lib/odds/core.ts + markets.ts; the underlying math is held to the original simulator by the golden-master test in docs/src/lib/odds/__tests__/ (which exercises the default truncated grid — this page opts into the mass-complete capTail variant).