Can someone please explain how this is done, I have the cypher text and 3 options for plain text. The cypher text is encrypted with a one time pad with a key where the letter 'L' was used too often.
Cyphertext: NLBYVBDNSBWBCXEOPPBUDLKEZJOGKLIFYSLKHQKGRWIVFSZWWUCKHZIVWQELHWOIYZPIVKHRYAIZHHQWLIGPQXPQYTREHOGQFZNZEAKJEPMOHNIBZTOWBGYVESZSWYLRHCYZUDVAWLLHMERJTLWVTLOJDEAVYRBOEVGZQEREZUGDGTREWRIMDPZJLBVWTXVBQAGTTYYBZTWCNSOZWPMXVXNAPNWWGATQRLNYESRIFQEAMRCTSKZNILTTQJLOKRSVWVUWHCRJRQZXJXQWFEQIRESCEPGVFBLBPVGQCKEMLMXUHWKGWRGCOUPOPTMSDJJUFYDFBXFMPPJQXEVCJBTOGGUKZQZWQKYMWHSLWHIBRHPTQDUIZDAVUPGPPWSWYZPRYJPLSMIQYIMDGCZL
Plaintext option 1: ITWASABRIGHTCOLDDAYINAPRILANDTHECLOCKSWERESTRIKINGTHIRTEENWINSTONSMITHHISCHINNUZZLEDINTOHISBREASTINANEFFORTTOESCAPETHEVILEWINDSLIPPEDQUICKLYTHROUGHTHEGLASSDOORSOFVICTORYMANSIONSTHOUGHNOTQUICKLYENOUGHTOPREVENTASWIRLOFGRITTYDUSTFROMENTERINGALONGWITHHIMTHEHALLWAYSMELTOFBOILEDCABBAGEANDOLDRAGMATSATONEENDOFITACOLOUREDPOSTERTOOLARGEFORINDOORDISPLAYHADBEENTACKEDTOTHEWALLITDEPICTEDSIMPLYANENORMOUSFACEMORE
Plaintext option 2: CALLMEISHMAELSOMEYEARSAGONEVERMINDHOWLONGPRECISELYHAVINGLITTLEORNOMONEYINMYPURSEANDNOTHINGPARTICULARTOINTERESTMEONSHOREITHOUGHTIWOULDSAILABOUTALITTLEANDSEETHEWATERYPARTOFTHEWORLDITISAWAYIHAVEOFDRIVINGOFFTHESPLEENANDREGULATINGTHECIRCULATIONWHENEVERIFINDMYSELFGROWINGGRIMABOUTTHEMOUTHWHENEVERITISADAMPDRIZZLYNOVEMBERINMYSOULWHENEVERIFINDMYSELFINVOLUNTARILYPAUSINGBEFORECOFFINWAREHOUSESANDBRINGINGUPTHER
Plaintext option 3: ITWASTHEBESTOFTIMESITWASTHEWORSTOFTIMESITWASTHEAGEOFWISDOMITWASTHEAGEOFFOOLISHNESSITWASTHEEPOCHOFBELIEFITWASTHEEPOCHOFINCREDULITYITWASTHESEASONOFLIGHTITWASTHESEASONOFDARKNESSITWASTHESPRINGOFHOPEITWASTHEWINTEROFDESPAIRWEHADEVERYTHINGBEFOREUSWEHADNOTHINGBEFOREUSWEWEREALLGOINGDIRECTTOHEAVENWEWEREALLGOINGDIRECTTHEOTHERWAYINSHORTTHEPERIODWASSOFARLIKETHEPRESENTPERIODTHATSOMEOFITSNOISIESTAUTHORITIESINSIS
The solution to the cypher text is one of the above solutions.
I presume the one-time pad is being used as follows. Each letter A to Z in plaintext and pad is interpreted as number $0$ to $25$, added mod $26$, and then translated back to letters.
If the plaintext is typical of English text, it will have the typical letter distribution (in particular more than 12% E). Those E positions where the one-time pad has L will translate E to P. The cyphertext has 400 letters, of which 21 are P, in positions $17, 18, 67, 84, 87, 106, 182, 210, 217, 282, 289, 311, 313, 329, 330, 363, 374, 376, 377, 383, 387$.
The E's in the three plaintexts are in the following positions: $$\eqalign{\text{option } 1:& 32, 40, 42, 56, 57, 83, 94, 102, 110, 115, 118, 122, 132, 150, 194, 204, 206, 231, 234, 253, 263, 272, 280, 298, 299, 313, 319, 328, 349, 350, 356, 362, 370, 375, 385, 396, 400\cr \text{option }2:& 6, 12, 17, 19, 27, 29, 44, 48, 62, 70, 80, 106, 108, 112, 119, 149, 154, 155, 158, 162, 173, 191, 206, 210, 211, 217, 228, 242, 244, 246, 256, 277, 285, 287, 289, 310, 313, 325, 327, 329, 339, 363, 367, 377, 382, 399\cr \text{option }3:& 8, 10, 18, 27, 38, 47, 50, 66, 69, 80, 90, 91, 99, 102, 111, 112, 123, 137, 139, 158, 160, 172, 182, 194, 202, 207, 212, 219, 223, 225, 234, 238, 242, 254, 258, 262, 264, 266, 278, 284, 287, 290, 292, 294, 306, 311, 315, 329, 331, 347, 350, 353, 355, 359, 371, 382, 394\cr}$$
Option 1 has only one E in the same position where the cyphertext has P, while option 2 has $9$ and option 3 has $4$.
So my vote is for option 2.