I am indebted to Frode Weierud of the Crypto Simulation Group at CERN for supplying me with the data on this page.
It is advantageus to be able to predict how many times the bombe is likely to stop for a particular menu. Different menus are of different strengths in that they yield different numbers of stops. Clearly, it is desirable to have the least number of stops. In general, the more loops in the menu the fewer the number of stops.
When a menu is constructed from a crib there are often a number of webs, usually one main web and one or two other smaller webs consisting of two to five letters.
At Bletchley Park the following rough rule of thumb was used to predict the number of stops per wheel order for menus running on the three wheeled bombes.
number of 
number of loops
in main web 
number of stops
per drum order 
any number

4

1

7

3

5

8

3

2

9

3

1

10

2

10

11

2

4

12

2

1

13

1

7

14

1

2

15

0

7.5

16

0

1.5

Turing developed the following formula for calculating the number of stops.
number of stops = 26^{(4c)} * HM
where c is the number of loops.
The HM factor was tabulated as:
number of
letters in web 
HM factor

2

0.92 
3

0.79 
4

0.62 
5

0.44 
6

0.29 
7

0.17 
8

0.087 
9

0.041 
10

0.016 
11

0.006 
12

0.0018 
13

0.00045 
14

0.000095 
15

0.000016 
16

0.0000023 
The HM factor represents the effect of the diagonal board. Without the diagonal board the bombe is unable to exploit letters which are not connected to the main web, that is, the web to which the test register is connected.
Turing may have made a few errors in calculating this table, but the errors are small and all values are correct for up to at least 10 letters. The very small values may have some rounding errors.
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