Comment Re:Welcome to the club (Score 1) 112
The odds are 1 in ({number of different symbols per character} to the power of {length of ciphertext}) - 1 of it being the plaintext.
If the OTP is truely random, then a ciphertext can be decoded to any possible permutation of the same length with equal probability.
Take for example the ciphertext TIGZIFZOMASDRVBTJFVTS:
With an OTP of ABCDEFGHIJKLMNOPQRSTU it decodes to THEWEATHERISFINETODAY,
while an OTP of LPOIIXMGZUQDYDBGGCHNA yields ITSRAININGCATSANDDOGS.
Without knowing the original OTP you have no way of knowing which of these decodes (or all other legible) is the correct one.
Of course many decodes will result in gibberish as well (e.g. OTP 123456789012345678901).
In this case the probability that any given message is the correct one (without being able to decide that, of course) is 1 in 7,31e+43.
Its likelier to win the lottery and be struck by lightning on the same day.
To try it out for your self look here: http://www.braingle.com/braint...