Odds Generation — soccer · goal · regular · DIBP

A comparison candidate to the default page, proposed (2026-07-16, pending team approval — ADR-029 addenda) to replace it: on two disjoint live-Saba samples (59 + 124 real prematch Correct Score boards) this engine is unbiased at 0-0 where the λ₃ shock over-prices it by ≈ e^λ₃, fits the 1X2 draw exactly, and has the best per-cell CS error of the draw-lifted engines; ω = 1 held on both samples. 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) — and step 4 extends the default page's common shock λ₃ into the diagonal-inflated bivariate Poisson (Karlis–Ntzoufras 2005):

J = (1 − θ) · BP(l₁, l₂, λ₃) + θ · D(ω)

The independent-Poisson grid gains a common-shock coupling λ₃ and a slug of diagonal probability θ·D(ω), where D(ω) is a geometric distribution over the draw cells starting at 1-1 (1-1, 2-2, 3-3, …) with a fixed decay ω you set — never fitted (ω < 1 decays from 1-1, ω = 1 is uniform, ω > 1 leans toward the high diagonal). 0-0 deliberately takes no injection: measured against Saba's own Correct Score boards (59 real prematch matches, 2026-07-16), books price 0-0 at the independent-Poisson level and keep their draw surplus in the mid/high diagonal — a 0-0 share only ever over-prices that cell (ADR-029 addendum). The two fitted knobs are (λ₃, θ), and after every trial the engine re-solves (λₕ, λₐ) so the Handicap and Total still hold exactly.

Live. The same remaining-goals frame as the default page: Handicap rest-of-match, the absolute Total line drops by the goals already scored, final-result markets settle on current + remaining. The difference is step 4's live reach — D(ω) follows the final-score diagonal: live it injects on the band h − a = d (each final draw k-k keeps its ω^(k−1) weight; final 0-0 stays exempt), so the θ pull keeps fitting while a side leads, where the default page's centred-diagonal λ₃ shock greys its 1X2 out. λ₃ itself still reaches the draw only through the Total trade, which concentrates the centred cells — off level it loses most of its grip and the fit usually defers to the θ-only solve. At 0-0 the pipeline is byte-identical to prematch.

The mechanism has sharp limits, and the walkthrough narrates them. Both knobs are draw-directional and one-sided: θ injects diagonal mass head-on, while λ₃ reaches the draw only sideways — the common shock cancels in the goal margin (X − Y = W₁ − W₂), so λ₃ moves the 1X2 only by trading against the Total (a higher λ₃ forces a lower-scoring, more draw-heavy match). Neither is a structural home:away lever the way the φ/σ candidate's σ is: at the common ±0.50-class Handicap the anchor already pins P(h > a) — the 1X2 home event — so the two knobs collapse onto one direction and the fit defers to a θ-only draw solve, a more fundamental rank collapse than the φ/σ candidate's. At a deeper line the home leg frees up and Gauss–Newton engages λ₃ as well. And because the whole diagonal is one 1X2 outcome, ω is unidentifiable from the 1X2: move it and the fitted 1X2 holds while Correct Score and Exact Total visibly reshape — the diagonal's shape is imposed by ω rather than inherited from the baseline grid.

See also the sibling φ/σ-tilt candidate; the default page keeps the common shock but drops the diagonal inflation.

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 · DIBP 1X2 pull

Home1DrawXAway2ωdiag decay

↑/↓ 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 (λ₃, θ) pull keeps fitting at any score — θ's injection follows the final-score diagonal — with no level-score gate (the default page's λ₃-only shock has one; step 4 compares).

l₁ 1.643231l₂ 0.933443λ₃ 0.000000θ 0.074934ω 1.00μ_total 2.576674anchors Δp 4.5e-11 / 5.6e-111X2 Δ -9.2e-3 / +9.3e-11 / +9.2e-3draw 0.248599 → 0.302298grid N=5 · Poisson → DIBP

Score grid

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

Poisson

l₁ 1.608939l₂ 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

DIBP

l₁ 1.643231l₂ 0.933443λ₃ 0.000000θ 0.074934draw 0.302298

	A=0	A=1	A=2	A=3	A=4	A=5+
H=0	7.03	6.56	3.06	0.95	0.22	0.05
H=1	11.56	12.29	5.03	1.57	0.37	0.08
H=2	9.50	8.86	5.64	1.29	0.30	0.07
H=3	5.20	4.85	2.27	2.20	0.16	0.04
H=4	2.14	1.99	0.93	0.29	1.57	0.01
H=5+	0.95	0.89	0.41	0.13	0.03	1.51

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 · θ injects on the final-result draw band, so in live it shifts 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	-0.20	-0.22	-0.32	-0.42	-0.49	-0.59	-0.81	+0.96	+0.66	+0.37	+0.28	+0.22	+0.14	+0.07	+0.06	+0.06
A	+0.12	+0.14	+0.24	+0.34	+0.41	+0.51	+0.73	+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	+0.07	+0.08	+0.19	+0.31	+0.37	+0.47	+0.72	+0.96	-0.84	-0.64	-0.55	-0.48	-0.38	-0.28	-0.26	-0.24
Under	-0.15	-0.16	-0.27	-0.39	-0.45	-0.55	-0.80	+0.96	+0.76	+0.56	+0.47	+0.40	+0.30	+0.20	+0.18	+0.16

1X2

Margin (Book)

Outcome	Odds
Home	1.960
Draw	3.064
Away	4.685

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)

1-08.415

2-01031

2-11119

3-01893

3-12059

3-24170

4-044316

4-147213

4-295250

4-3258395

0-0

14

1-1

7.9

2-2

17

3-3

42

4-4

59

AOS

23

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

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

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 λ₃ and θ 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 λ₃/θ diagonal inflation perturbs it, so that stage re-solves size and split jointly against the same hard targets — both recovered (l₁, l₂) 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.

This margin picture is λ₃-invariant within the BP part. Step 4’s common shock adds the same $W_3$ to both scores, so it cancels in the difference — $X - Y = (W_1 + W_3) - (W_2 + W_3) = W_1 - W_2$ — a Skellam in $(l_1, l_2)$ alone. The θ injection is the exception: its diagonal slug does sit in margin space (on the draw band), which is exactly why step 4 re-solves $(l_1, l_2)$ on the mixed grid — the anchors see the injection and trade against it, so they hold through the mixture.

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 pulls that shape toward the 1X2 quote with the diagonal-inflated bivariate Poisson — θ (direct diagonal mass) and λ₃ (common shock, felt through the Total trade) — then re-solves so the quotes still hold (step 4).

4 · Pull the shape toward the 1X2 (λ₃ + θ diagonal inflation) — 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 is a soft target, and this page pursues it with a diagonal-inflated bivariate Poisson (Karlis–Ntzoufras 2005): mix the independent-Poisson grid with a common-shock coupling $\lambda_3$ and a diagonal-inflation weight $\theta$, then re-solve $(\lambda_h, \lambda_a)$ so the Handicap and Total still hold exactly on every trial.

$$J[h,a] \;=\; (1-\theta)\,\mathrm{BP}(l_1,l_2,\lambda_3)[h,a] \;+\; \theta\,D(\omega)[h,a]$$

$$D(\omega)[h,a] = \begin{cases} w_h, & h=a \\ 0, & h\neq a \end{cases}, \qquad w_0 = 0, \quad w_k \;=\; \frac{\omega^{k-1}}{\sum_{j=1}^{N}\omega^{j-1}}, \quad k = 1\ldots N$$

$$(\lambda_3,\theta) \;=\; \arg\min\limits_{\lambda_3\in[0,3],\;\theta\in[0,0.9]}\; \left(P_H-p_H^{\ast}\right)^2 + \left(P_D-p_D^{\ast}\right)^2 + 10^{-10}\lambda_3^2$$

$\mathrm{BP}$ is the common-shock construction $X = W_1 + W_3,\; Y = W_2 + W_3$ with independent $W_1 \sim \mathrm{Poisson}(l_1),\; W_2 \sim \mathrm{Poisson}(l_2),\; W_3 \sim \mathrm{Poisson}(\lambda_3)$; at $\lambda_3 = 0$ it is exactly the independent-Poisson grid of steps 2–3. $D(\omega)$ is a geometric distribution laid along the final-score diagonal starting at 1-1 (the draw cells 1-1, 2-2, 3-3, … — final 0-0 deliberately takes none: measured against Saba’s own Correct Score boards, books price 0-0 at the independent-Poisson level and keep their draw surplus in the mid/high diagonal, so any 0-0 share only over-prices that cell); ω sets how the mass tilts (ω < 1 decays from 1-1, ω = 1 is uniform, ω > 1 leans toward the high diagonal), and $\textstyle\sum_k w_k = 1$ so $D(\omega)$ is a proper distribution and the mixture stays mass-complete.

Where each knob’s authority comes from. θ injects draw mass directly — a slug of diagonal probability, straight onto the equal-score cells. λ₃ has no direct line to the 1X2: $X - Y = W_1 - W_2$, so the shock $W_3$ cancels in the goal margin — inside the BP part every handicap event and the raw win/draw/loss split depend on $(l_1, l_2)$ alone. λ₃ reaches the draw only indirectly, through the Total anchor: the expected total is $\mathbb{E}[X+Y] = l_1 + l_2 + 2\lambda_3$, so raising λ₃ forces $l_1 + l_2$ down to hold the Total — a lower-scoring match, which crowds probability toward the low, draw-heavy cells. So both knobs are draw-directional: θ pushes the draw up head-on, λ₃ pushes it up sideways.

This is the sharp contrast with the $\varphi/\sigma$ candidate’s pull (the φ/σ page). There σ is a genuine home:away lever — it tilts the win and loss halves against each other and owns the home direction structurally, dead only at the ±0.50 line, where the Handicap already pins $P(h>a)$. Here neither knob is a home:away lever — both move the draw — so the home leg is reachable only when the anchored Handicap event differs from $P(h>a)$. At a ±0.50-class line the two coincide, the home residual is frozen, and the two draw-directional knobs collapse onto one direction — the fit defers (θ only, λ₃ = 0). At a deeper line the home leg frees up and the two knobs regain independent traction (θ touches only the diagonal, λ₃ reshapes the low-score region through the Total trade), so Gauss–Newton fits both legs with λ₃ engaging — but as a byproduct of the differing anchor, not a dedicated tilt like σ. The displayed rankRatio quantifies which regime this quote is in. And where φ rescales the diagonal it inherits (multiplicative, $\times(1+\varphi)$, keeping each draw cell’s share of the diagonal), θ overwrites the diagonal toward ω’s geometric profile (additive) — visible below in the Correct-Score draw cells and the even-total Exact-Total buckets.

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	+9.3e-11
Away	0.251401	0.197702	0.188461	+9.2e-3

landed with $\lambda_3 = 0.000000,\;\theta = 0.074934$ at fixed $\omega = 1.000000$ — Δ home and Δ draw are the minimized residuals; Δ away follows from $\Sigma = 1$ and stays a cross-check.

Cross-check θ — the mixture is additive, so the 1X2 legs come out in closed form with no bisection at all. Freeze the BP part at this stage’s re-solved $(l_1, l_2, \lambda_3)$: its three masses are $H_0 = 0.540502$ (h > a), $D_0 = 0.245781$ (diagonal), $A_0 = 0.213717$ (h < a), Σ = 1. Because $D(\omega)$ sits entirely on the diagonal (its home/away mass is 0, its draw mass is 1), the achieved legs are a straight-line map in θ:

$$P_H = (1-\theta)H_0 \qquad P_D = (1-\theta)D_0 + \theta \qquad P_A = (1-\theta)A_0$$

so θ is recoverable from any leg — $\theta = 1 - P_H/H_0 = 0.074934$, $\;1 - P_A/A_0 = 0.074934$, $\;(P_D - D_0)/(1 - D_0) = 0.074934$ — all landing on the fit’s $\theta = 0.074934$. ω never appears: the whole diagonal is one 1X2 outcome, so it is unidentifiable from the 1X2 (see below). What a one-variable calculator can’t reproduce is the $(l_1, l_2)$ re-solve and λ₃ — those enter $H_0/D_0/A_0$ as constants, the same division of labour as steps 2–3.

Where the additive leg-map comes from — and the ω unidentifiability

The mixture layers a diagonal-only distribution onto the BP grid, then renormalizes nothing (both components already sum to 1). Splitting $J = (1-\theta)\mathrm{BP} + \theta D(\omega)$ into its three 1X2 blocks, the off-diagonal blocks get only the $(1-\theta)$ factor while the diagonal picks up the full $\theta$ (since $\textstyle\sum_k w_k = 1$):

$$P_H = (1-\theta)H_0 \qquad P_D = (1-\theta)D_0 + \theta\!\!\sum_{k}w_k = (1-\theta)D_0 + \theta \qquad P_A = (1-\theta)A_0$$

Contrast the φ/σ candidate, whose $\varphi/\sigma$ weighting is multiplicative and needs a renormalizer $Z$ — $P_D = (1+\varphi)D_0/Z$, so φ rescales the diagonal cells it inherits, keeping their internal shape (2-2 : 3-3 fixed). The DIBP map instead replaces the diagonal’s shape: the surviving $(1-\theta)D_0$ keeps the BP profile, but the injected $\theta w_k$ imposes ω’s geometric decay — so which draw cells grow is set by ω, not inherited.

ω is unidentifiable from the 1X2. The total injected diagonal mass is $\theta\sum_k w_k = \theta$ whatever ω is — so the achieved draw $P_D = (1-\theta)D_0 + \theta$ (and hence home, away) is blind to ω. Move the ω field above and watch: the fitted 1X2 holds to solver tolerance while the Correct-Score draw cells and even-total Exact-Total buckets visibly reshape — a low ω piles the mass onto 1-1, ω = 1 spreads it evenly, ω > 1 leans it up the diagonal toward 3-3, 4-4, … (0-0 never takes injection). That is the deliberate pedagogy of this candidate: the 1X2 pins the diagonal’s level, ω is a free hand on its shape.

Unroll the (λ₃, θ) search — rank check, then the draw bisection

The fit first probes both residual directions at λ₃ = θ = 0 and measures the shape Jacobian’s normalized rank, $\det(J^{\top}\!J)/\operatorname{tr}(J^{\top}\!J)^{2} = 1.9 \times 10^{-14}$ < 10⁻⁸ — the two knobs collapse onto one direction (both raise the draw, and this Handicap already fixes the home leg). λ₃ pins at 0 — the ridge prefers the pure diagonal-inflation explanation — and the one live knob, θ, is bisected onto the draw:

i	lo	hi	θ	l₁	l₂	Δ home	Δ draw
2	0.000000	0.450000	0.22500000	1.725941	0.607856	-9.2e-3	+1.0e-1
3	0.000000	0.225000	0.11250000	1.661924	0.860680	-9.2e-3	+2.6e-2
4	0.000000	0.112500	0.05625000	1.634331	0.967856	-9.2e-3	-1.3e-2
5	0.056250	0.112500	0.08437500	1.647825	0.915619	-9.2e-3	+6.6e-3
6	0.056250	0.084375	0.07031250	1.641006	0.942060	-9.2e-3	-3.3e-3
7	0.070313	0.084375	0.07734375	1.644397	0.928922	-9.2e-3	+1.7e-3
8	0.070313	0.077344	0.07382812	1.642697	0.935511	-9.2e-3	-7.8e-4
9	0.073828	0.077344	0.07558594	1.643546	0.932222	-9.2e-3	+4.6e-4
10	0.073828	0.075586	0.07470703	1.643121	0.933868	-9.2e-3	-1.6e-4
11	0.074707	0.075586	0.07514648	1.643333	0.933045	-9.2e-3	+1.5e-4
12	0.074707	0.075146	0.07492676	1.643227	0.933456	-9.2e-3	-4.9e-6
13	0.074927	0.075146	0.07503662	1.643280	0.933251	-9.2e-3	+7.2e-5
14	0.074927	0.075037	0.07498169	1.643254	0.933353	-9.2e-3	+3.4e-5
15	0.074927	0.074982	0.07495422	1.643241	0.933405	-9.2e-3	+1.4e-5
16	0.074927	0.074954	0.07494049	1.643234	0.933431	-9.2e-3	+4.8e-6
17	0.074927	0.074940	0.07493362	1.643231	0.933443	-9.2e-3	-7.2e-8
18	0.074934	0.074940	0.07493706	1.643232	0.933437	-9.2e-3	+2.3e-6
19	0.074934	0.074937	0.07493534	1.643231	0.933440	-9.2e-3	+1.1e-6
20	0.074934	0.074935	0.07493448	1.643231	0.933442	-9.2e-3	+5.3e-7
21	0.074934	0.074934	0.07493405	1.643231	0.933443	-9.2e-3	+2.3e-7
22	0.074934	0.074934	0.07493384	1.643231	0.933443	-9.2e-3	+7.9e-8
23	0.074934	0.074934	0.07493373	1.643231	0.933443	-9.2e-3	+3.7e-9
24	0.074934	0.074934	0.07493368	1.643231	0.933443	-9.2e-3	-3.4e-8
25	0.074934	0.074934	0.07493370	1.643231	0.933443	-9.2e-3	-1.5e-8
26	0.074934	0.074934	0.07493372	1.643231	0.933443	-9.2e-3	-5.8e-9
27	0.074934	0.074934	0.07493372	1.643231	0.933443	-9.2e-3	-1.0e-9
28	0.074934	0.074934	0.07493373	1.643231	0.933443	-9.2e-3	+1.3e-9
29	0.074934	0.074934	0.07493373	1.643231	0.933443	-9.2e-3	+9.3e-11

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: neither knob can change it once the Handicap pins P(h>a), and λ₃ was ruled out by the rank check.

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

Stage	l₁	l₂	λ₃	θ	Δ cover	Δ over	Δ home	Δ draw	Δ away
Poisson	1.608939	1.065121	—	—	3.8e-9	7.1e-9	—	—	—
DIBP	1.643231	0.933443	0.000000	0.074934	4.5e-11	5.6e-11	-9.2e-3	+9.3e-11	+9.2e-3

Hard-anchor gaps: Handicap 4.5e-11, Total 5.6e-11 — the pull never trades them; $\omega = 1.000000$ is held fixed throughout. This run deferred: with both knobs draw-directional and the anchored cover event already pinning $P(h>a)$ (the ±0.50 line) — the same event the 1X2 home leg prices — the two collapse onto one direction, so λ₃ = 0 and Δ home above is simply the two books’ disagreement over that event, reported not fought. Quote a Handicap whose anchored event cuts across the 1X2 regions (e.g. −1.00 at −0.59 / +0.51, anchoring P(win by 2+)) and the home residual frees up: the rank climbs and the fit switches to Gauss–Newton.

Shape caveats: the 1X2 fixes the diagonal’s level (its total mass), but its shape — which draw cells carry it — is imposed by ω, not read from a quote. Pin the correct-score profile with correct-score quotes if you need it. Because $D(\omega)$ only touches the diagonal from 1-1 up (even totals ≥ 2 — never the 0-goal bucket), Odd/Even and Exact-Total at those even totals move with θ and ω while odd totals and 0-0 do not; and since λ₃ mostly reshapes the low cells through the Total trade, its footprint on the wider ladders is small.