Retrieval check answer. P(flagged) = 0.95×0.02 + 0.10×0.98 = 0.019 + 0.098 = 0.117. P(caused it | flagged) = 0.019 / 0.117 ≈ 0.162 — about 16%. A bot with 95%/10% hit/false-alarm rates, applied to a rare (2%) event, still flags mostly-innocent deploys most of the time. Same mechanism as Lesson 7’s disease test — different noun.
Main answer: LR ≈ 2.848.
P(data | p = 0.75) = 0.75^6 × 0.25^2 = 0.177979 × 0.0625 ≈ 0.011124
P(data | p = 0.5) = 0.5^6 × 0.5^2 = 0.5^8 ≈ 0.003906
LR = 0.011124 / 0.003906 ≈ 2.848 (exactly 729/256)
With C(8,6) = 28 folded into both sides instead:
P(6 heads in 8 | p=0.75) = 28 × 0.011124 ≈ 0.311462
P(6 heads in 8 | p=0.5) = 28 × 0.003906 ≈ 0.109375
LR = 0.311462 / 0.109375 ≈ 2.848 — identical.
Same ratio either way. C(8,6) multiplies both the numerator and the denominator by the same
factor, 28, so it divides straight out. That’s the whole mechanism: in a ratio between two
hypotheses on the same data, any factor that doesn’t depend on the hypothesis (p) cancels.
C(n,k) never depends on p — it’s a fact about counting orderings, not about the coin — so it’s
always safe to drop from a likelihood ratio, even though it’s essential to the absolute
probability of “k heads in some order.”
Now the colleague’s claim — where it’s right, and where it overreaches.
Check A (same composition, different order): TRUE, they’re equal. H H T H H H T H and
H T H H H H T H both have 6 heads and 2 tails. Under p = 0.75, any specific sequence with that
exact composition has probability 0.75^6 × 0.25^2 ≈ 0.011124 — order genuinely doesn’t matter once
the composition (how many heads) is fixed. This part of the intuition is correct, and it’s why
C(n,k) exists in the first place: it’s literally counting how many equally-probable orderings share
one composition, so you can add them up into “probability of exactly k heads, any order.”
Check B (different composition): FALSE, they’re wildly unequal. A specific 6-heads sequence and a
specific 2-heads sequence, both under p = 0.75:
P(specific 6H,2T sequence | p=0.75) = 0.75^6 × 0.25^2 ≈ 0.011124
P(specific 2H,6T sequence | p=0.75) = 0.75^2 × 0.25^6 ≈ 0.000137
ratio ≈ 81× (exactly (0.75/0.25)^4 = 3^4 = 81)
Under a biased coin, a sequence with more heads is dramatically more probable than a same-length
sequence with fewer heads — composition matters enormously, only order-within-a-composition doesn’t.
The colleague’s mistake is exactly the gap between Check A and Check B: “order doesn’t matter
among sequences with the same head-count” (true) got overgeneralized to “every sequence is equally
probable” (false, unless p = 0.5, where every sequence — any composition — genuinely is equally
likely, since 0.5^k × 0.5^(n-k) = 0.5^n regardless of k). At p = 0.5 the colleague’s blanket
claim would be correct. At p = 0.75, it silently smuggles in the fair-coin case as if it were
general.
Why this matters for the likelihood ratio specifically. The colleague was reaching for a true
fact (C(n,k) cancels in the ratio) via a false justification (“sequences are equally probable”). The
right justification is narrower and more useful: C(n,k) cancels because it’s a p-independent
constant, full stop — not because the data itself carries no information about which composition is
more or less likely. The composition (6 heads vs. 2 heads) carries plenty of information about p;
it’s only the ordering-count that’s inert in a ratio.
The general shape. For any two hypotheses p₁, p₂ and observed k successes in n trials, the
likelihood ratio is:
LR = [C(n,k) p₁^k (1−p₁)^(n−k)] / [C(n,k) p₂^k (1−p₂)^(n−k)] = (p₁/p₂)^k × [(1−p₁)/(1−p₂)]^(n−k)
C(n,k) is gone from the right-hand side entirely — it was never carrying hypothesis-relevant
information, only orderings-per-composition information, and a ratio between hypotheses doesn’t need
that.
Where this goes: the likelihood ratio you just computed (≈2.848) is exactly the ingredient
Lesson 9’s odds form eats directly — posterior odds = prior odds × LR. Next lesson turns back to
the beta-binomial posterior itself and asks a sharper question about how much evidence it actually
takes to narrow one down.