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OCR for page 163
Forensic Analysis Weighing Bullet Lead Evidence
I
Birthday Problem Analogy
The committee has found a perceived similarity between determining the false match probability for bullet matches and a familiar problem in probability, the Birthday Problem: Given n people (bullets) in a room (collection), what is the probability that at least two of them share the same birthday (analytically indistinguishable composition)? Ignoring leap-year birthdays (February 29), the solution is obtained by calculating the probability of the complementary event (“P{A}” denotes the probability of the event A):
P{no 2 people have the same birthday} = P{each of n persons has a different birthday} = P{person 1 has any of 365 birthdays} · P{person 2 has any of the
Then P{at least 2 people have the same birthday} When n = 6, 23, 55, p(n) = 0.04, 0.51, 0.99, respectively (Ref. 1).
That calculation seems to suggest that the false match probability is extremely high when the case contains 23 or more bullets, but the compositional analysis of bullet lead (CABL) matching problem differs in three important ways.
First, CABL attempts to match not just any two bullets (which is what the birthday problem calculates), but one specific crime scene bullet and one or more of n other potential suspect bullets where n could be as small as 1 or 2 or as large as 40 or 50 (which is similar to determining the probability that a specific person shares a birthday with another person in the group). Hence, bullet matching by CABL is a completely different calculation from the birthday problem.
OCR for page 164
Forensic Analysis Weighing Bullet Lead Evidence
Second, as stated in Chapter 4, a match indicates that the two bullets probably came from the same source or from compositionally indistinguishable volumes of lead, of which thousands exist from different periods, for different types of bullets, for different manufacturers, and so on, not just 365. Even if interest lay in the probability of a match between any two bullets, the above calculation with N = 5,000 and n + 1 = 6 or 23 or 55 bullets yields much smaller probabilities of 0.003, 0.0494, and 0.2578, respectively.
Third, if bullets manufactured at the same time tend to appear in the same box, and such boxes tend to be distributed in geographically nondispersed locations, the n potential suspect bullets are not independent, as the n persons’ birthdays in the birthday problem are assumed to be.
We conclude that this analogy with the birthday problem does not apply.
REFERENCE
1. Chung, K.-L. Elementary Probability Theory with Stochastic Processes, 2nd Ed., Springer-Verlag: New York, NY, 1975; p 63.
Representative terms from entire chapter:
suspect bullets
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When n = 6, 23, 55, p(n) = 0.04, 0.51, 0.99, respectively (Ref. 