General Report on Tunny


25A Page 153

It is expected that in ΔD/'s will be more numerous T's. Suppose that in the first column of the partial de-chi there are 6/'s and 10 T's: then if the first character of ΔΧ5 is a dot the contribution to ΔD is 6/'s and 10T's; but if the first character of ΔΧ5 is a cross the contribution to ΔD is 10/'s and 6T's therefore the character is more likely to be cross than a dot.

Each character of ΔΧ5 can thus be estimated, though some may be doubtful because the numbers of /'s and T's in a column are too nearly equal. Similarly evidence is obtainable from other pairs of ΔD letters differing only in ΔD5, e.g. it is expected that there will be more 5's than J's, more U's than Q's.

The method does not, of course, require that four Χ's shall be known e.g. if only Χ1, Χ2 are known, ΔΧ4 may be found using the ΔD characteristic: 4 = 1 = 2 is commoner than 4 ≠ 1 = 2.

Nor does it require of a ΔΧi, regarded as known, and used to find ΔDi, that all its characters shall be known; places on Z against unknown ('doubted') ΔΧ characters are simply ignored.

The evidence for a particular character is derived only from places against that character; and, very crudely, the evidence for a dot may be described as 'excess of good letters over bad letters' measured in the first place as so many 'pips'.

Clearly refinements are needed. Even at random ΔD letters will not all be exactly equally numerous, so that it will be necessary to have a criterion to determine whether the bulges are significantly large; and, when they are, to have a method for evaluating the evidence with some precision. (25B). Further the evidence of a wheel-breaking run may conflict with known ΔΧ characteristics, and require adjustment.

A special instance of this is that wheels must have approximately equal numbers of dots and crosses: it might seem reasonable to take as dots and crosses not those characters where 'pippages' are positive and negative, but those whose 'pippages', having regard to sign, are above and below average. The two methods should however agree, just because wheels do contain approximately equal numbers of dots and crosses. Discrepancies are due to


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