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(a) e = P(Z & B, nA)/P(Z & nB, nA) = P(B, Z & nA)/P(nB, Z & nA)
and X2 have been used cannot be very high. There are two reasons why they
= O(B, Z & nA)
The odds on the highest score of a run (3+4)x/1x2x where the wrong X1
might be greater than 1/ 29x26.
(i) The score for B on (3+4)x may be so enormous that it shews up
(b) 1-m = d'/d = P(Z & nB, nA)/P(Z & nB, A)
even on the random sample of the text given by the wrong
X12 setting. This can be tested by counting (3+4)x.
(ii) A may be wrong, but a good slide of correct settings. This
is a more troublesome possibility.
This is the ratio of chances of getting the observed scores with
the highest one wrong, starting from wrong and right X12 settings
We shall generally be considering cases where Z contains a pretty good
score, so that we shall not often be dealing with Z such that P(Z, nA)>P(Z,A).
Let us then also consider It unlikely that P(Z & nB, nA)> P(Z & nB, A).
On the other hand we may get P(Z & nB, nA)
possibility of B's being wrong but a good slide of the correct settings.
These considerations shew that e and m are small compared with 1,
therefore we can take it that
O(A & B, Z) approx =P(A). O(B,Z & A) = po.
In that case po is an overestimate.
If e is not negligible, it is because X1 or X2 is only a good slide.
if m is not negligible it is (probaby) because of slides in X3 and X4
improving P(Z & nB, A) at the expense of P(Z & nB, nA). In fact if there
are strong rivals on tte second run m is not negligible and po is an