Odds Generation — soccer · goal · regular · DIBP

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 — and step 4 extends the default'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 (0-0, 1-1, 2-2, …) with a fixed decay ω you set — never fitted. 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.

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.

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 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).

l₁ 1.7832l₂ 1.0407λ₃ 0.0000θ 0.0872ω 0.35μ_total 2.8239anchors Δp 2.1e-11 / 7.6e-111X2 Δ -6.0e-3 / +7.1e-10 / +6.0e-3draw 0.2486 → 0.2995grid N=5 · Poisson → DIBP

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

Poisson

l₁ 1.6089l₂ 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

DIBP

l₁ 1.7832l₂ 1.0407λ₃ 0.0000θ 0.0872draw 0.2995

	A=0	A=1	A=2	A=3	A=4	A=5+
H=0	11.10	5.64	2.93	1.02	0.26	0.07
H=1	9.66	12.05	5.23	1.82	0.47	0.12
H=2	8.62	8.97	5.36	1.62	0.42	0.11
H=3	5.12	5.33	2.77	1.21	0.25	0.06
H=4	2.28	2.38	1.24	0.43	0.20	0.03
H=5+	1.14	1.18	0.61	0.21	0.06	0.04

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	-0.61	+0.53
-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	+0.49	-0.57
2/2.5	+0.73	-0.81
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)

1-01017

2-01132

2-11118

3-01987

3-11851

3-23457

4-041276

4-139172

4-273190

4-3187289

0-0

8.7

1-1

8.0

2-2

18

3-3

75

4-4

347

AOS

26

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

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 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.

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 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

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

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 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.

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

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

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

$\mathrm{BP}$ 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 diagonal (the draw cells 0-0, 1-1, 2-2, …); ω sets how fast its mass decays, 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	0.5000	0.5000	0.5060	-6.0e-3
Draw	0.2486	0.2995	0.2995	+7.1e-10
Away	0.2514	0.2005	0.1945	+6.0e-3

landed with $\lambda_3 = 0.000000,\;\theta = 0.087190$ at fixed $\omega = 0.3500$ — 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 $H_0 = 0.547759$ (h > a), $D_0 = 0.232594$ (diagonal), $A_0 = 0.219647$ (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 — $\theta = 1 - P_H/H_0 = 0.087190$, $\;1 - P_A/A_0 = 0.087190$, $\;(P_D - D_0)/(1 - D_0) = 0.087190$ — all landing on the fit’s $\theta = 0.087190$. 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 lower ω piles the mass onto 0-0, a higher ω spreads it up the diagonal to 1-1, 2-2, … 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, $\det(J^{\top}\!J)/\operatorname{tr}(J^{\top}\!J)^{2} = 2.3 \times 10^{-14}$ < 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
1	0.000000	0.900000	0.45000000	3.747603	0.613501	-6.0e-3	+1.9e-1
2	0.000000	0.450000	0.22500000	2.171726	0.980638	-6.0e-3	+7.7e-2
3	0.000000	0.225000	0.11250000	1.842218	1.032039	-6.0e-3	+1.4e-2
4	0.000000	0.112500	0.05625000	1.716638	1.050168	-6.0e-3	-1.8e-2
5	0.056250	0.112500	0.08437500	1.776899	1.041564	-6.0e-3	-1.6e-3
6	0.084375	0.112500	0.09843750	1.808883	1.036926	-6.0e-3	+6.5e-3
7	0.084375	0.098438	0.09140625	1.792728	1.039275	-6.0e-3	+2.4e-3
8	0.084375	0.091406	0.08789063	1.784773	1.040426	-6.0e-3	+4.0e-4
9	0.084375	0.087891	0.08613281	1.780826	1.040997	-6.0e-3	-6.1e-4
10	0.086133	0.087891	0.08701172	1.782797	1.040712	-6.0e-3	-1.0e-4
11	0.087012	0.087891	0.08745117	1.783784	1.040569	-6.0e-3	+1.5e-4
12	0.087012	0.087451	0.08723145	1.783291	1.040641	-6.0e-3	+2.4e-5
13	0.087012	0.087231	0.08712158	1.783044	1.040676	-6.0e-3	-3.9e-5
14	0.087122	0.087231	0.08717651	1.783167	1.040659	-6.0e-3	-7.6e-6
15	0.087177	0.087231	0.08720398	1.783229	1.040650	-6.0e-3	+8.2e-6
16	0.087177	0.087204	0.08719025	1.783198	1.040654	-6.0e-3	+2.6e-7
17	0.087177	0.087190	0.08718338	1.783183	1.040656	-6.0e-3	-3.7e-6
18	0.087183	0.087190	0.08718681	1.783190	1.040655	-6.0e-3	-1.7e-6
19	0.087187	0.087190	0.08718853	1.783194	1.040655	-6.0e-3	-7.3e-7
20	0.087189	0.087190	0.08718939	1.783196	1.040654	-6.0e-3	-2.3e-7
21	0.087189	0.087190	0.08718982	1.783197	1.040654	-6.0e-3	+1.4e-8
22	0.087189	0.087190	0.08718960	1.783197	1.040654	-6.0e-3	-1.1e-7
23	0.087190	0.087190	0.08718971	1.783197	1.040654	-6.0e-3	-4.7e-8
24	0.087190	0.087190	0.08718976	1.783197	1.040654	-6.0e-3	-1.7e-8
25	0.087190	0.087190	0.08718979	1.783197	1.040654	-6.0e-3	-1.2e-9
26	0.087190	0.087190	0.08718980	1.783197	1.040654	-6.0e-3	+6.5e-9
27	0.087190	0.087190	0.08718980	1.783197	1.040654	-6.0e-3	+2.6e-9
28	0.087190	0.087190	0.08718979	1.783197	1.040654	-6.0e-3	+7.1e-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: 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	1.6089	1.0651	—	—	3.8e-9	7.1e-9	—	—	—
DIBP	1.7832	1.0407	0.0000	0.0872	2.1e-11	7.6e-11	-6.0e-3	+7.1e-10	+6.0e-3

Hard-anchor gaps: Handicap 2.1e-11, Total 7.6e-11 — the pull never trades them; $\omega = 0.3500$ is held fixed throughout. This run deferred: with both knobs draw-directional and the ±0.50 line already pinning $P(h>a)$ — 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 deeper Handicap (e.g. −1.00 at −0.59 / +0.51) so the anchored event becomes $P(\text{win by }2{+})$ — distinct from $P(h>a)$ — and the home residual frees up: the rank climbs and the fit switches to Gauss–Newton, engaging λ₃ to bring Δ home down too.

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 (even totals), Odd/Even and Exact-Total at even totals move with θ and ω while odd totals do not; and since λ₃ mostly reshapes the low cells through the Total trade, its footprint on the wider ladders is small.

5 · Read “Correct Score” off the score grid

With l₁, l₂, λ₃, and θ 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	0.1110	0.1110	8.722
0-1	0.0564	0.0564	17.011
0-2	0.0293	0.0293	32.172
0-3	0.0102	0.0102	87.481
0-4	0.0026	0.0026	276.367
1-0	0.0966	0.0966	10.001
1-1	0.1205	0.1205	8.042
1-2	0.0523	0.0523	18.309
… 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
fixed     ω  (geometric-diagonal decay)   # user input, never fitted       — step 4
          w[k] ∝ (1−ω)ω^k on k=0…N (Σ=1)  # D(ω) profile

for damped Gauss–Newton trial (λ₃, θ)     # skip λ₃ = θ = 0 with no 1X2
  bisect λ_total = l₁+l₂ ∈ [0.02, 8]      # re-anchor size (mean total = l₁+l₂+2λ₃)
    bisect l₁ ∈ (0, λ_total)             # re-anchor split; l₂ = λ_total − l₁
      B  = BP(l₁, l₂, λ₃) on 0…N          # common-shock bivariate Poisson
      J  = (1−θ)·B + θ·D(ω)               # additive diagonal inflation — Σ stays 1
    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 λ₃²                   # λ₃ ridge defers a rank collision

The grid panel’s stages run this same nest with the shape pull off/on — Poisson: λ₃ = θ = 0, DIBP: the fitted (λ₃, θ). The λ bisections remain monotone and retain their usual solver tolerances — the diagonal injection is fully visible to fairHcp/fairTot, which evaluate on the mixed grid, so both anchors stay exact through the mixture. The outer least-squares loop uses finite-difference shape directions, damping, and the small λ₃ ridge. Because both knobs are draw-directional and neither owns the home leg, the measured shape Jacobian is quasi-rank-deficient more often than the φ/σ candidate’s; when it collapses the ridge pins λ₃ = 0 and θ is bisected onto the draw 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) — the DIBP grid stays mass-complete because both mixture components do — 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 is docs/src/lib/odds/core.ts + markets-dibp.ts; the shared strip/solve/settlement 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).