The probability law behind every fair price: when events (goals) arrive
independently at a steady average rate, the count of events in a window follows
a Poisson distribution. Enter the average rate λ and a count x to get
P(X = x) plus the four cumulative tails, a range probability, and the full
distribution table — the same arithmetic the
Odds Generation engine runs when it
turns a candidate λ_total into a fair P(over T). It also runs backwards:
hand the solver a target probability and a count — P(X < x), ≤, =, >, or
≥ — and it searches out the λ that produces it, the same higher-or-lower hunt
the engine uses to recover λ_total from a Total quote. Each probability also
shows its fair decimal odds (1 / p).
Seeded from the engine's worked example: λ = 2.6743 expected goals gives
P(X ≥ 3) ≈ 50% — the fair P(over 2.5) its λ_total search converges on. The
inverse solve below runs the same example backwards: P(X ≥ 3) = 0.50 → λ.
↑/↓ steps an input, Shift steps bigger.
Poisson facts
P(X = x) = e−λ · λx / x! · mean = λ · variance = λ · σ = √λ Additivity: X ~ Pois(λₕ) and Y ~ Pois(λₐ) independent ⇒ X + Y ~ Pois(λₕ + λₐ) — why a match's total goals can be priced from λ_total alone.
Inputs
Average rateλ
Events expected per window — e.g. goals per match. Live σ = √λ = 1.6353.
0 < λ ≤ 99
Countx
The number of events to evaluate at.
whole number, 0 … 120
Probabilities at x
Quantity
Meaning
Value
%
Fair dec
P(X = 3)
exactly 3 events
0.21981
21.98%
4.549
P(X < 3)
2 or fewer
0.49994
49.99%
2.000
P(X ≤ 3)
3 or fewer
0.71975
71.97%
1.389
P(X > 3)
4 or more
0.28025
28.03%
3.568
P(X ≥ 3)
3 or more
0.50006
50.01%
2.000
How P(X = x) is computed
0
start at zero events: P(X = 0) = e−λe−2.6743 — the chance the window stays empty
0.06896
1
climb to 1: multiply by λ/1 = 2.6743P(X = 1) = P(X = 0) × λ/1
0.18441
2
climb to 2: multiply by λ/2 = 1.3372P(X = 2) = P(X = 1) × λ/2
0.24658
3
climb to 3: multiply by λ/3 = 0.8914P(X = 3) = P(X = 2) × λ/3
0.21981
✓
direct formula check: e−λ · λx / x!0.068955 × 19.126274 ÷ 6
0.21981
Solve for λ — supply a probability, get the rate back
TailP(X ⊙ x)
Which probability statement the target refers to. P(X ≥ x) is the over-line case: P(over x − 0.5).
Countx
The count in the probability statement.
whole number, 0 … 120
Targetp
The probability the statement should have.
0 < p < 1 — type 0.5 or 50%
Solved λ
the rate at which "3 or more events" happens 50.00% of the time
2.6741
σ = √λ
spread of the count at that rate
1.6353
Check
P(X ≥ 3) at λ = 2.6741
0.50000 = the target ✓
How λ is solved — a higher-or-lower search
·
no formula maps a target probability back to λ — but P(X ≥ 3) only rises as λ grows, so guessing is safeguess a rate, compare P(X ≥ 3) with the 50.00% target, keep the half of the window that must hold the answer
1
window 0.00 – 99.00 → guess the midpoint, λ = 49.5001P(X ≥ 3) at this rate is 100.00% — above the 50.00% target; this tail rises as λ grows, so the answer is lower
▼ keep lower half
2
window 0.00 – 49.50 → guess the midpoint, λ = 24.7501P(X ≥ 3) at this rate is 100.00% — above the 50.00% target; this tail rises as λ grows, so the answer is lower
▼ keep lower half
3
window 0.00 – 24.75 → guess the midpoint, λ = 12.3751P(X ≥ 3) at this rate is 99.96% — above the 50.00% target; this tail rises as λ grows, so the answer is lower
▼ keep lower half
4
window 0.00 – 12.38 → guess the midpoint, λ = 6.1876P(X ≥ 3) at this rate is 94.59% — above the 50.00% target; this tail rises as λ grows, so the answer is lower
▼ keep lower half
5
window 0.00 – 6.19 → guess the midpoint, λ = 3.0938P(X ≥ 3) at this rate is 59.75% — above the 50.00% target; this tail rises as λ grows, so the answer is lower
▼ keep lower half
6
window 0.00 – 3.09 → guess the midpoint, λ = 1.5470P(X ≥ 3) at this rate is 20.30% — below the 50.00% target; this tail rises as λ grows, so the answer is higher
▲ keep upper half
…
and so on — each guess halves the window; 60 halvings pin λ beyond display precision