Editing One-time pad (section)

=== Attempt at cryptanalysis ===
To continue the example from above, suppose Eve intercepts Alice's ciphertext: <code>EQNVZ</code>. If Eve tried every possible key, she would find that the key <code>XMCKL</code> would produce the plaintext <code>hello</code>, but she would also find that the key <code>TQURI</code> would produce the plaintext <code>later</code>, an equally plausible message:
     4 (E)  16 (Q)  13 (N)  21 (V)  25 (Z) ciphertext
 −  19 (T)  16 (Q)  20 (U)  17 (R)   8 (I) possible key
 = −15       0      −7       4      17     ciphertext-key
 =  11 (l)   0 (a)  19 (t)   4 (e)  17 (r) ciphertext-key (mod 26)
In fact, it is possible to "decrypt" out of the ciphertext any message whatsoever with the same number of characters, simply by using a different key, and there is no information in the ciphertext that will allow Eve to choose among the various possible readings of the ciphertext.<ref name="schneierotp">{{cite web |last=Schneier |first=Bruce |title=One-Time Pads |url=http://www.schneier.com/crypto-gram-0210.html#7 |url-status=live |archive-url=https://web.archive.org/web/20050403200231/http://www.schneier.com/crypto-gram-0210.html#7 |archive-date=2005-04-03}}</ref>

If the key is not truly random, it is possible to use statistical analysis to determine which of the plausible keys is the "least" random and therefore more likely to be the correct one. If a key is reused, it will noticeably be the only key that produces sensible plaintexts from both ciphertexts (the chances of some random ''incorrect'' key also producing two sensible plaintexts are very slim).