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Conditional Probabilities And Addition Mod-26
Topic Started: Dec 10 2007, 12:35 AM (234 Views)
jdege Dec 10 2007, 12:35 AM Post #1
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 What I had noticed, when messing about with running keys and auto keys - or rather, had long ago been told of, but not recognized the importance of - is how predictable the sum mod-26 of random plaintext can be.

Consider ordinary English. 12% of English text is 'e'. Add two English texts together, and 1.44% of the output will be 'e' + 'e', or Ii'. Of course, 'd' + 'f' also equals 'I', and 'c' + 'g', etc. But all of these are less likely than 'e' + 'e'. So when you see 'I', it's more likely to be 'e' + 'e' than anything else. I always knew that.

But how much more likely? I'd never run the numbers.

Turns out that assuming normal English distributions for both texts, the probability that an 'I' represents 'e' + 'e' is 31.5% - more than I would have expected. There's a 24% probability that its 'a' + 'i', a 10% probability that it's 'o' + 'u', and an 8% probability that it's 'r' + 'r'.

The most strongly identified ciphertext character is 'X', which is the sum of 'e' and 't' 56% of the time. Next is 'S', which is 'e' + 'o' 43% of the time.

And so on:
Code:
 

X = e + t   56.94%
S = e + o   42.80%
E = a + e   41.04%
R = e + n   39.62%
M = e + i   39.58%
O = a + o   37.40%
B = i + t   35.46%
N = a + n   35.12%
T = a + t   34.76%
H = o + t   33.56%
W = e + s   32.94%
I = e + e   31.55%
When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl.
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PulsarSL Dec 14 2007, 04:13 AM Post #2
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 That's pretty interesting...
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