Arthur Conan Doyle, "The Sign of the Four" (1890):
"How
came he, then?" I reiterated. "The door is locked, the window is
inaccessible. Was it through the chimney?"
"The
grate is much too small," he answered. "I had already considered that
possibility."
"How
then?" I persisted.
"You
will not apply my precept," he said, shaking his head. "How often
have I said to you that when you have eliminated the impossible whatever
remains, however improbable, must be the truth? We know that he did not come
through the door, the window, or the chimney. We also know that he could not
have been concealed in the room, as there is no concealment possible. Whence,
then, did he come?"
"He
came through the hole in the roof!" I cried.
Let the four
hypotheses—the came in through the door, window, chimney and hole in the roof—be
represented as (respectively): Hd, Hw, Hc and Hh. Let the a priori
probabilities of the first three hypotheses be much more likely than the forth—say,
P(Hd) = P(Hw) = P(Hc) = 33/100
P(Hh) = 1/100
Now suppose the
first three hypotheses are, as Holmes says, “impossible”—i.e., subsequent
evidence is incompatible with the first three hypotheses (but compatible with
the last hypothesis):
P(Hd & E) = P(Hw & E) = P(Hc
& E) = 0
P(Hh & E) ≠ 0
Part (a):
Show that
P(E|Hd) = P(E|Hw) = P(E|Hc) = 0
Part (b):
Show that,
on the evidence, the last (a priori) “improbable” hypothesis “must be the truth”—i.e.,
P(Hh|E) = 1
Solution:
Part (a):
By the General Conjunction Rule of probability:
0 = P(Hd & E) = P(Hw
& E) = P(Hc & E)
= P(E|Hd)P(Hd)
= P(E|Hw)P(Hw) = P(E|Hc)P(Hc)
= P(E|Hd)(.33) = P(E|Hw)(.33) = P(E|Hc)(.33)
= P(E|Hd) = P(E|Hw) = P(E|Hc),
as claimed.
Part (b):
By Bayes’ Rule:
P(Hh|E)
=
P(E|Hh) x P(Hh)
P(E|Hh) x P(Hh) + P(E|Hd) x P(Hd) + P(E|Hw) x
P(Hw) + P(E|Hw) x P(Hw)
But by Part (a), the latter three summands are zero, so
P(Hh|E) = P(E|Hh) x P(Hh) / P(E|Hh) x P(Hh) = 1. It "must be the truth," as Holmes
claims.
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