# Captain Walter Fried's Fish Notes Annex to #F 71, Screed on Delta D Counts and Colossus runs

### Page 12

Tony Sale's
Codes and Ciphers

| previous page | index page |

Appendix p2

(a) e = P(Z & B, nA)/P(Z & nB, nA) = P(B, Z & nA)/P(nB, Z & nA)
= O(B, Z & nA)

The odds on the highest score of a run (3+4)x/1x2x where the wrong X1
and X2 have been used cannot be very high. There are two reasons why they
might be greater than 1/ 29x26.

(i) The score for B on (3+4)x may be so enormous that it shews up
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.

(b) 1-m = d'/d = P(Z & nB, nA)/P(Z & nB, A)

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.

If e is not negligible, it is because X1 or X2 is only a good slide.
In that case po is an overestimate.

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
understatement.