Determining the third number - given 5 and 13

Basic information

        Balls drawn         Richard's success
# 1     03 05 14 22 30 44   two matches, one off by one
# 2     06 12 15 16 31 44   one match
# 7     09 17 32 36 42 44   two matches, three off by four
# 8     02 05 21 22 25 32   one match

Select three from 05 13 28 46 48
Both 5 and 13 selected

Lottery 7

The two known numbers are four off 9 and 17 respectively. The final near miss from lottery 7 must be chosen from:

28 46 48

Richard also had two exact matches, chosen from:

32 36 42 44

Lottery 8 already has one match (5) so 32 cannot be selected. This gives:

5 13 [28 46 48] [36 42 44]x2

Lemma - 44 is not selected

The exact matches from lottery 7 must therefore be 36 and 42. They cannot be used for near misses, ruling out 46. This gives:

5 13 [28 48] 36 42

Lottery 1 requires a second match (other than 5), but none of these numbers will provide Richard's match in Lottery 2.

Lemma Fails!

Hence 44 is known

This gives:

5 13 [28 46] [36 42] 44

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