Originally posted by FISHY1118
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Ive just looked at Druitt again. I was too harsh on him. So I’ve amended it to the score that I was originally going to give him.
Kelly > 2 - 2 - 4 - 2 - 2 - 1 - 0 = 13
Bury > 2 - 2 - 4 - 0 - 2 - 1 - 0 = 11
Cutbush > 2 - 2 - 3 - 2 - 0 - 1 - 0 = 10
Hyams > 2 - 2 - 3 - 2 - 0 - 0 - 0 = 9
Kosminski 2 - 2 - 1 - 2 - 1 - 0 - 0 = 8
Grainger > 2 - 1 - 3 - 0 - 1 - 1 - 0 = 8
Chapman > 2 - 2 - 1 - 0 - 1 - 0 - 1 = 7
Tumblety > 1 - 1 - 0 - 0 - 2 - 2 - 1 = 7
Thompson > 2 - 2 - 0 - 0 - 0 - 1 - 1 = 6
Barnado > 2 - 2 - 0 - 1 - 0 - 1 - 0 = 6
Cohen > 2 - 2 - 1 - 1 - 0 - 0 - 0 = 6
Levy > 2 - 2 - 0 - 1 - 0 - 0 - 1 = 6
Druitt > 2 - 2 - 0 -1 - 1 - 0 = 6
Barnett > 2 - 2 - 0 - 0 - 0 - 1 - 0 = 5
Stephen > 2 - 1 - 0 - 2 - 0 - 0 - 0 = 5
Stephenson > 2 - 1 - 0 - 0 - 0 - 0 - 1 = 4
Bachert > 2 - 2 - 0 - 0 - 0 - 0 - 0 = 4
Cross > 2 - 2 - 0 - 0 - 0 - 0 - 0 = 4
Hardiman > 2 - 2 - 0 - 0 - 0 - 0 - 0 = 4
Hutchinson > 2 - 2 - 0 - 0 - 0 - 0 = 4
Mann > 2 - 2 - 0 - 0 - 0 - 0 - 0 = 4
Gull > 1 - 1 - 0 - 0 - 0 - 0 - 1 = 3
Maybrick > 2 - 1 - 0 - 0 - 0 - 0 - 0 = 3
Sickert > 2 - 1 - 0 - 0 - 0 - 0 - 0 = 3
If it could be shown that it was reasonably possible that they were in England…
Deeming 2 - 1 - 4 - 0 - 0 - 0 - 0 = 7
Feigenbaum 2 - 1 - 4 - 0 - 0 - 0 - 0 = 7
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