Decrypting a one-time pad with a bias around letter 'L' with solution options. 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.
 A: Making the same (standard) encoding assumptions: encryption is addition modulo 26, while letters A-Z correspond to 0-25.
One easily (I did it in Python; less tedious) computes the pads that must have been used if plain1, plain2 and plain3 would have been the true plaintext:
(not as digits as but as letters):
pad1 = FSFYDBCWKVPIAJTLMPDMQLVNRYOTHSBBWHXIXYOCASQCOKPOJOJDZIPRSDIDUEVULHDACDAJGYBRUUWXMXCMIKWCRLZDQKGYMRAZRWFEQYTVTJQZZEKDUCDNTODKJVTGZNJVRNBSUBAJTXAVZFPCMHIYDMISKDKWQQLROLDNBIGQOLDREYBYJJSWXIFCLVLQSWTFZSRILEFYSOBGWXQPEMZVJWODNCQWZSIHQGNVMMNSZLCIEXTRASMMIXSHGKSKLZUYPQNYYCUWVPFSCCQHQEMYECDHUYUBJJGXKKLYYITHEIFYDREODGVXLQXELQFDMKPUBGZIKBSIKBHOSYLWRVUMSQWVMGLTWFIHTOUIKDTTFSMPWZLNSWCMXOGHNBPEUWBUGYOYTIKZUOIH
pad2=LLQNJXVVLPWXRFQCLRXUMTKYLWKLGUWXLPEWLFWTLHRRDKHSLWVKMRVPLILSWSARLLDUIGJJLOKKNQYSLVDCCEIILNCEQVYOLONILMCWLLVKPUWXLGWPNPUNLLLYQRSJLOEORLVSLLKTSLRYLSDKPLBGLAWCRNFOLRPBBEALLPNWCXDNLOATVXZNLDNPTCRNLXPLYQLVLORJGOWKLLIKVKKJLHCLGHLDLSGUCKAGLFEHEDPXLGMJNHCLLBYLYTARLQOFTGJWLKIPXXPILLXBNSEILIKOBOHGLEYXUSEJLAIRQOLHLTTOTQDNLCEFRLRGLNHYXKBRLYBLPRSQLJPDBYHPLFFJXKHELJDLCPAOLGLOCMQGLYVNHTGYLPECGVXRLGOUKZCILCSONVVU
pad3 = FSFYDIWJRXEIOSLGDLJMKPKMGCKKWUQMKNSCVMSYYAIDMLVWQQOFLRQSIEWSLWWPRVPCRWCMKMXRPADSTQYWUXXXRPNPTMZCAYJOWWFBLTMWOGEXKFMPNBQICBVPCNDYJUFDULCTSTHHUQEVOAOPMSGQHEICRNJKEDSMCZOEIKTZOBJLARQTWLHUUTIQFSONBWYAXYGISPAUAZKIIKJTDINSYRSPGXPVNUPFXKECEMZMVNIBWGSNFYFAJBYIJNNHFRAELYVFAMZMYRCOSYNAAAQJLBZRFGHOTRKMLGEBAGJMUQHYFNEJVNLAWMIBHJLMSGWRKEMFLAFZPQSGJJBABGDZRGVDJGJVSPOYDSEKJTMAJDBQLRWHPHNXCIKEQVXYYPWEEVAXQEUVTKRT
with 18, 77 and 16 'L''s respectively. So text 2 is the original plaintext, clearly, and pad2 the key. (The expected number of L's is 400 (size of pad) / 26 = 15.3, so the first and last values are quite normal. 
