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, plus, in live, the same current score input (default 0-0) — but step 4 swaps the default page'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.

Live. The same remaining-goals frame as the default page: the rest-of-match Handicap reads the grid unchanged, the absolute Total line drops by the goals already scored, and final-result markets settle on current + remaining. The difference is step 4's live reach — the default page's λ₃ shock can only lift the centred draw, so its 1X2 fit greys out while a side leads; the φ/σ tilt weights the final-result regions themselves, which shift with the deficit, so the pull keeps fitting at any score. What stays reachable is still set by the anchors: when the anchored cover event is a union of tilt blocks (prematch, the ±0.5-class line; live, line and deficit aligned) the rank check defers to the φ-only draw solve, exactly as before. At 0-0 the pipeline is byte-identical to prematch.

Where the default page'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 — proposed (2026-07-16, pending team approval) to replace the default after the 0-0 measurements — which extends the 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.

Score (live) 0-0 = prematch

HomeAway

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 16 lines in quarter steps around your line; 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). The score is whole goals (0-0 = prematch); in live enter the quotes as the book states them — Handicap rest-of-match, Total absolute (its line must sit above the current total), 1X2 on the final result. The φ/σ pull follows the final-result regions, so it keeps fitting at any score — no level-score gate (the default page's λ₃ shock has one; step 4 compares).

λₕ 1.757096λₐ 1.012926λ_total 2.770022φ 0.411700σ 0.000000anchors Δp 4.3e-11 / 3.6e-111X2 Δ -9.2e-3 / -3.8e-10 / +9.2e-3draw 0.248599 → 0.302298grid N=5 · Poisson → 1X2 soft pull

Score grid

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

Poisson

λₕ 1.608939λₐ 1.065121draw 0.248599

	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 0.500000 draw Σ 0.248599 away Σ 0.251401

1X2 soft pull

λₕ 1.757096λₐ 1.012926φ 0.411700σ 0.000000draw 0.302298

	A=0	A=1	A=2	A=3	A=4	A=5+
H=0	8.07	5.79	2.93	0.99	0.25	0.06
H=1	10.04	14.36	5.15	1.74	0.44	0.11
H=2	8.82	8.93	6.39	1.53	0.39	0.09
H=3	5.17	5.23	2.65	1.26	0.23	0.05
H=4	2.27	2.30	1.16	0.39	0.14	0.02
H=5+	1.11	1.12	0.57	0.19	0.05	0.02

home Σ 0.500000 draw Σ 0.302298 away Σ 0.197702

shade ∝ probability · each stage re-solves (λₕ, λₐ) against the same fair targets · the φ/σ blocks ride the final-result regions, so in live they shift with the score

Markets

HandicapMargin (Ladder)

HDP	-2/2.5	-2.0	-1.5/2	-1.5	-1/1.5	-1.0	-0.5/1	-0.5	-0/0.5	0.0	+0/0.5	+0.5	+0.5/1	+1.0	+1/1.5	+1.5
H	-0.21	-0.23	-0.33	-0.44	-0.51	-0.61	-0.82	+0.96	+0.66	+0.37	+0.28	+0.22	+0.14	+0.07	+0.06	+0.06
A	+0.13	+0.15	+0.25	+0.36	+0.43	+0.53	+0.74	+0.96	-0.74	-0.45	-0.36	-0.30	-0.22	-0.15	-0.14	-0.14

Over/UnderMargin (Ladder)

Goal	0.5/1	1.0	1/1.5	1.5	1.5/2	2.0	2/2.5	2.5	2.5/3	3.0	3/3.5	3.5	3.5/4	4.0	4/4.5	4.5
Over	+0.08	+0.09	+0.19	+0.29	+0.35	+0.45	+0.70	+0.96	-0.84	-0.64	-0.54	-0.47	-0.36	-0.25	-0.23	-0.22
Under	-0.16	-0.17	-0.27	-0.37	-0.43	-0.53	-0.78	+0.96	+0.76	+0.56	+0.46	+0.39	+0.28	+0.17	+0.15	+0.14

1X2

Margin (Book)

Outcome	Odds
Home	1.95
Draw	3.06
Away	4.68

Home is the −0.5 Handicap price (+0.96 Malay — the favourite, same event, same odds); the Draw and Away share the rest of the book to 1 + margin.

Correct Score

Margin (Shin)

1-09.600-116

2-0100-232

2-1101-218

3-0180-389

3-1181-353

3-2352-359

4-0410-4288

4-1401-4182

4-2772-4203

4-32013-4311

0-0

11

1-1

6.70

2-2

15

3-3

71

4-4

441

AOS

27

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.

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

Handicap -0.50 → P(home covers -0.50)

	Home	Away
Malay quote	+0.96	+0.96
Decimal $d$	1.960000	1.960000
Fair $p = q^{1/x}$	0.500000	0.500000
Fair decimal $1/p$	2.000000	2.000000

$0.510204^{1/x} + 0.510204^{1/x} = 1 \;\Rightarrow\; x = 0.970854$

Total 2.50 → P(over 2.50)

	Over	Under
Malay quote	+0.95	+0.95
Decimal $d$	1.950000	1.950000
Fair $p = q^{1/x}$	0.500000	0.500000
Fair decimal $1/p$	2.000000	2.000000

$0.512821^{1/x} + 0.512821^{1/x} = 1 \;\Rightarrow\; x = 0.963474$

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

	Home	Draw	Away
Decimal quote $d$	1.890000	3.130000	4.900000
Fair $p$ (Shin)	0.509241	0.302298	0.188461
Fair decimal $1/p$	1.963708	3.307996	5.306125

$\textstyle\sum_i p_i(z) = 1 \;\Rightarrow\; z = 0.026465$

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

Eight decimals on purpose — step 1’s table shows this same number at 6 dp, a rounding. The λ-inverse is stiff (a wobble in the target’s 4th decimal moves the recovered λ by several units in its 4th decimal), so the cross-checks below and step 3’s target carry full precision.

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:

$$W \;=\; P(\text{total} > 2.50) \;=\; 1 - \textstyle\sum_{t=0}^{2} \tfrac{e^{-\lambda}\lambda^{t}}{t!} \;=\; 0.500000$$

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

$$\text{fair } P(\text{over } 2.50) \;=\; \frac{W}{1-R} \;=\; \frac{0.500000}{1 - 0.000000} \;=\; 0.500000 \;=\; p^{\ast}_{\text{over}} \;\; \checkmark$$

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

(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 ($\lambda_a = \lambda_{\text{total}} - \lambda_h = 1.065171$):

$$0.275697 + 0.175465 + 0.042809 + 0.005557 + 0.000447 + 0.000024 \;=\; 0.500000 \;=\; p^{\ast}_{\text{cover}} \;\; \checkmark$$

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:

$$W = P(h - 0.50 > a) = 0.500000$$

$$R = P(h - 0.50 = a) = 0.000000$$

$$\text{fair } P(\text{home covers } -0.50) \;=\; \frac{W}{1-R} \;=\; \frac{0.500000}{1 - 0.000000} \;=\; 0.500000 \;=\; p^{\ast}_{\text{cover}} \;\; \checkmark$$

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

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 the market boards above price 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

$$p_i(z) \;=\; \frac{\sqrt{z^{2} + 4(1-z)\,q_i^{2}/Q\,} - z}{2(1-z)} \qquad \text{bisect } z \in [0,1) \text{ until } \textstyle\sum p_i = 1 \;\Rightarrow\; z = 0.026465$$

Cross-check it — this book’s implied numbers are $q = \left(\tfrac{1}{1.890000},\ \tfrac{1}{3.130000},\ \tfrac{1}{4.900000}\right) = (0.529101,\ 0.319489,\ 0.204082)$ with overround $Q = 1.052671$. 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)*0.529101^2/1.052671) - x)/(2*(1-x))
+ (sqrt(x^2 + 4*(1-x)*0.319489^2/1.052671) - x)/(2*(1-x))
+ (sqrt(x^2 + 4*(1-x)*0.204082^2/1.052671) - x)/(2*(1-x))
= 1

The sum starts at $\sqrt{Q} = 1.025998$ when $z = 0$ and falls monotonically, so the sign flip is unique — scan $[0,\, 0.9]$ and the bisection lands back on $z = 0.026465$. 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 $(p_H^{\ast},p_D^{\ast},p_A^{\ast}) = (0.509241,\,0.302298,\,0.188461)$. 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	0.500000	0.500000	0.509241	-9.2e-3
Draw	0.248599	0.302298	0.302298	-3.8e-10
Away	0.251401	0.197702	0.188461	+9.2e-3

landed with $\varphi = 0.411700,\;\sigma = 0.000000$ — 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 $D_0 = 0.234841$ (diagonal), $H_0 = 0.548342$ (h > a), $A_0 = 0.216817$ (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)*0.54834209/0.21681680 = 2.52905713

then φ from the draw at that σ (off-diagonal constant $K = e^{\sigma}H_0 + e^{-\sigma}A_0 = 0.765159$), interval $[-0.999,\, 2]$:

(1 + x)*0.23484111 / ((1 + x)*0.23484111 + 0.76515889) = 0.30229786

They land on $\sigma = -0.000000$ and $\varphi = 0.411700$ — the fit’s $0.000000$ and $0.411700$ 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:

$$\sigma = \tfrac{1}{2}\ln\!\frac{P_H\,A_0}{P_A\,H_0} = -0.000000 \qquad\quad \varphi = \frac{P_D}{P_H}\,e^{\sigma}\,\frac{H_0}{D_0} - 1 = 0.411700$$

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	-9.2e-3	+9.4e-3
2	0.000000	0.500000	0.25000000	1.700845	1.033459	-9.2e-3	-1.9e-2
3	0.250000	0.500000	0.37500000	1.744537	1.017575	-9.2e-3	-4.1e-3
4	0.375000	0.500000	0.43750000	1.765852	1.009664	-9.2e-3	+2.8e-3
5	0.375000	0.437500	0.40625000	1.755238	1.013616	-9.2e-3	-6.0e-4
6	0.406250	0.437500	0.42187500	1.760556	1.011639	-9.2e-3	+1.1e-3
7	0.406250	0.421875	0.41406250	1.757900	1.012627	-9.2e-3	+2.6e-4
8	0.406250	0.414063	0.41015625	1.756570	1.013121	-9.2e-3	-1.7e-4
9	0.410156	0.414063	0.41210938	1.757235	1.012874	-9.2e-3	+4.5e-5
10	0.410156	0.412109	0.41113281	1.756902	1.012998	-9.2e-3	-6.3e-5
11	0.411133	0.412109	0.41162109	1.757069	1.012936	-9.2e-3	-8.7e-6
12	0.411621	0.412109	0.41186523	1.757152	1.012905	-9.2e-3	+1.8e-5
13	0.411621	0.411865	0.41174316	1.757110	1.012920	-9.2e-3	+4.7e-6
14	0.411621	0.411743	0.41168213	1.757089	1.012928	-9.2e-3	-2.0e-6
15	0.411682	0.411743	0.41171265	1.757100	1.012924	-9.2e-3	+1.4e-6
16	0.411682	0.411713	0.41169739	1.757095	1.012926	-9.2e-3	-3.3e-7
17	0.411697	0.411713	0.41170502	1.757097	1.012925	-9.2e-3	+5.1e-7
18	0.411697	0.411705	0.41170120	1.757096	1.012926	-9.2e-3	+9.2e-8
19	0.411697	0.411701	0.41169930	1.757095	1.012926	-9.2e-3	-1.2e-7
20	0.411699	0.411701	0.41170025	1.757096	1.012926	-9.2e-3	-1.3e-8
21	0.411700	0.411701	0.41170073	1.757096	1.012926	-9.2e-3	+3.9e-8
22	0.411700	0.411701	0.41170049	1.757096	1.012926	-9.2e-3	+1.3e-8
23	0.411700	0.411700	0.41170037	1.757096	1.012926	-9.2e-3	-3.8e-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	1.608939	1.065121	2.674060	—	—	3.8e-9	7.1e-9	—	—	—
1X2 soft pull	1.757096	1.012926	2.770022	0.411700	0.000000	4.3e-11	3.6e-11	-9.2e-3	-3.8e-10	+9.2e-3

Hard-anchor gaps: Handicap 4.3e-11, Total 3.6e-11 — the pull never trades them. This run deferred: the rank check found no independent home/away direction — the ±0.50-class line already pins the home leg once the draw is set— and σ = 0, so Δ home above is simply the two books’ disagreement over the same event — reported, not fought. To watch σ engage, quote a line whose cover event cuts across the 1X2 regions: 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.