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Let's say we have a ciphertext of length 1000. We try to decipher it with an Enigma machine with random rotors and initial positions but no plugboard (so only $5 * 4 * 3 * 26^3 \approx 2^{20}$ possibilities). Assuming we find the correct settings for the rotors, then in this configuration, some of the ciphertext letters would become the correct plaintext letters. Indeed, as there were usually 10 pairs of letters connected through the plugboard, this would mean that the resulting plaintext is $6/26\%$ correct and $20/26\%$ junk. This would increase a little the index of coincidence of the resulting plaintext, so by looking for IoC of 0.039 or 0.040, it should be possible to obtain the correct rotors and their positions without having to deal right now with the plugboard.

Is this assumption correct ? I have a script running right now to try this approach, but since it seems it will take $\approx50$ hours to finish, I figured I should better ask first for reviews.

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I don’t think so. Remember that the Enigma process passes through the Stecker twice: once on the input and once on the output. As these are necessarily distinct by the Enigma no-fixed-point property, the chance of causal decryption would be $6\times 5/(26\times 25)$ which leads to a much smaller IOC.

Things might be slightly better if the unplugged letters include common plaintext letters, but I think that there will still be many false positives.

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Indeed it is. See here and here

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  • $\begingroup$ The graph on the first link shows that for 10 plugs (as described in the question) there's a roughly 5% chance of this working with a 1463 length message (which is markedly longer than the 1000 described in the question). $\endgroup$
    – Daniel S
    Oct 29 at 21:38

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