my GF's middle name

[MicroBob]

hello all....not sure if there's anything that anyone can do to help me with this HUGELY important problem.

my GF is extremely embarassed by her middle name and won't tell me what it is.
she promises to tell me if i actually do get it correct and i have procured a few clues out of her that narrows it down somewhat.
i was hoping to put together some sort of print-out that lists every single possibility and just bombard her with all of the guesses until i get it. but i don't know how to write a computer program and i still have to large a range of possibilities for this to be realistic.


anyway...the clues...
she is of latin american decent....she says that she and her mom are the only ones that she has ever heard of with this name.

8 letters...
begins with L
ends with A
2nd letter is a vowel
letters definitely not in the name include Z, Y, Q, W


multiplying it out (5 possibilities for 2nd letter, 22 possibilities for 3rd through 7th letters) you get 25,769,160.

that would be a rather huge printout.

obviously possibilities like Laaaaaaa and Lalalala and Layoyoya and Lerrrhha are out.
i'm assuming it's a fairly reasonably pronounced word and not a bunch of nonsense letters.

perhaps i can find out if any letter is repeated back to back....i suspect the name does not use the same letter 3 times in a row sense that would be pretty weird.


at first she miscounted and thought there wre only 7 letters (which also cut down the possibilites significantly) but she has now informed me that it's actually 8 letters. i told her that that fact alone took it from around 1 million to 25 million and she was rather proud of herself.

guesses that did not make the cut when i thought it was 7 letters included Levitra (a TV commercial was on while we were talking about this) and LaBamba.


anybody have any ideas on how i can go about this??
apologies if you think i am wasting everyone's time with this goofy post.

[SnakeRat]

Laetitia


Good luck in your quest.

[pzhon]

[ QUOTE ]
she says that she and her mom are the only ones that she has ever heard of with this name.

[/ QUOTE ]
Why don't you ask her mother? <font color="white"> Umm, it would be cheating? </font>
Say you want to get the spelling right for a personalized coffee mug.

[M.B.E.]

Would be helpful to know whether the name contains eight distinct letters, and if not how many distinct letters it does contain. Also would be nice to know precisely how many of the 8 letters are consonants.

With that information you might be able to narrow it down to a few hundred thousand possibilities.

[SuitedSixes]

[ QUOTE ]
Say you want to get the spelling right for a personalized coffee mug.

[/ QUOTE ]

Now that's funny, right there.

[deacsoft]

It may be time to tell her to grow up.

[daryn]

i agree.. this is pretty stupid /images/graemlins/smile.gif

[MicroBob]

[ QUOTE ]
It may be time to tell her to grow up.

[/ QUOTE ]

not sure that would be the healthiest way to help our relationship grow. /images/graemlins/smile.gif

actually, she would have every justification in the world to demand that i grow up...certainly far more then the reverse. yet she never does this.

i would certainly rather have her, interesting little quirks and all, then have to deal with many of the judgemental wives and GF's out there who are so disapproving of this poker thing regardless of how many times you show them that you are a winning player in the long-run.
no offense to the many who have significant others who have a difficult time understanding....i know how it is.
i'm just glad that she doesn't happen to be one.

she doesn't question my ways and nor do i hers.




okay....other ideas...i like the whole 'mug' thing...except for the fact that her madre no habla englais.
our conversations don't go too much past 'hola. como estas? buenos noches' etc etc.
and she likely wouldn't tell me anyway.


i might try to narrow it down to see if it's 8 'different' letters....number of consonants, etc etc.
but even when if i do this...i really can't get any further then typing out all the possibilities myself.

anyway, thanks for the suggestions and my apologies again to those who think i'm wasting anyone/everyone's time.

[nopants]

Lilianna.
Lucretia?

Hey, post #1 for me. I came to a poker forum to make bold statements about Latin middle names.

In the probability forum.

What are the chances of that?

Steve

[MicroBob]

lilianna seems unlikely....especially given my latest clues (which i divulge later)
i think i tried something close enough to lucretia.


indeed, what a strange way to make a first post.
there are definitely better threads around here i assure you. welcome to the forums.



couple more clues...here's the whole summary again.

former clues -
8 letters
begins with L
ends with A
2nd letter is a vowel
Q, W, Y, Z are not in the name

new clues -
one of the consonants is used twice
one of the vowels is used multiple times (could not establish if it was twice or thrice)
the repeated letters do NOT appear consecutively


this makes calculating the exact number of possiblities rather tricky.
anyone care to take a shot at determining the precise figure?
also...anyone care to take a shot at determining the chances that the 2nd letter is an 'A'??

remember, one of the vowels is used multiple times and the last letter is definitely an A and the 2nd letter is definitely a vowel...


anyway, getting too tricky for me.

[SnakeRat]

I dont think this is a math problem unless she will allow you guess Xmillion potential names. Did you try Laetitia?

[RocketManJames]

Now, why would she be so embarrassed...?

Here's a hint.

The first 5 letters of her name are likely to be LABIA or some "embarrassing" string.

This reminds me of twin sisters who had the names... ReGina and VaGina. Yes, she obviously went by her middle name, and I heard that she legally changed her name when she was of age.

-RMJ

[tech]

[ QUOTE ]
she says that she and her mom are the only ones that she has ever heard of with this name

[/ QUOTE ]

This would be my avenue for finding the solution. A little strategic research into some public records should net you the answer rather quickly.

[Duke]

Ask her how many points the total word is worth if played in scrabble, assuming no special squares or blanks. That'll help to narrow it down as well.

Hell, ask her if it's a playable scrabble word.

~D

[MicroBob]

these are all fairly amusing ideas....but i'm actually more interested in the math involved then anything else.


as far as public records....i'm not sure how to go about doing that.
she was not born in the US and i believe her driver's license and INS papers, etc all just list a middle-initial (as if it would be appropriate to root through all her crap anyway).

[elwoodblues]

[ QUOTE ]
as far as public records....i'm not sure how to go about doing that.

[/ QUOTE ]

If you're really interested in doing this, PM me.

[Ghazban]

Write a computer program that will just list every single possibility (that shouldn't be too hard), then print them all out and hand her the list. You're guaranteed to have guessed it in there somewhere and she'll have to admit you did even though you don't know which guess is correct. At this point, you can argue logically that, as you already know, she shouldn't have any problem with telling you.

[george w of poker]

la cucaracha

you know its illegal to date underage girls right?

[MicroBob]

i would love to write such a program...but i'm a computer-idiot and would have virtually no chance of being able to write ANY program.

is this something that could be done on excel?? is so, i still wouldn't have much chance of knowing how to do it as excel confuses the hell out of me.
i really should get around to learning more about computers.

[gamblore99]

Its a shame that this is my 100th post. oh well.

hurray me /images/graemlins/grin.gif

[deacsoft]

The EZ MACROS software could do it in a few minutes.

[Ghazban]

I haven't done any programming in a while but back in college I know I had to write a ForTran program that did some sort of similar enumeration and it didn't take too long. Alternatively, you could go through her purse when she's out of the room and look at her driver's licence (please tell me she's at least old enough to drive..... /images/graemlins/tongue.gif)

[submariner]

It seems to me that since she said she had never heard of anyone but her and her mother having this name, that it is probably a name made up by the family, possibly a feminized form of a male family name, something like that. See if you can find out names of relatives, and if one could be made into something that would fit.

[koolmoe]

[ QUOTE ]
anyone care to take a shot at determining the precise figure?


[/ QUOTE ]

Don't have the precise figure, but an upper bound is 1,390,260 ways

[ QUOTE ]

also...anyone care to take a shot at determining the chances that the 2nd letter is an 'A'??


[/ QUOTE ]

1 in 5? /images/graemlins/wink.gif

[ZZZ]

This is slightly more fun than working. I found a list of female names on the web, and wrote a script to search them according to your hints. Here's what it came up with:

LAKEISHA
LASHONDA
LUCRETIA
LAURETTA
LATRICIA
LAKESHIA
LASHANDA
LORRETTA
LUCRECIA
LOUVENIA
LARHONDA
LASHUNDA
LEONARDA
LATASHIA
LASANDRA
LATRISHA
LAKIESHA
LATARSHA
LUDIVINA
LAURINDA
LISANDRA
LAKENDRA
LAKEESHA
LILLIANA
LAMONICA
LACRESHA

ZZZ

[sthief09]

if you know how to use Access, you can paste append in a list of spanish words (since names often come from words), and all combinations of "words" that fit your qualifications. then, if you write a query to link them, it'll give you the spanish words that can be made given your criteria. of course this wouldn't work if her name isn't a word in spanish.

[EnderFFX]

[ QUOTE ]
hello all....not sure if there's anything that anyone can do to help me with this HUGELY important problem.

my GF is extremely embarassed by her middle name and won't tell me what it is.
she promises to tell me if i actually do get it correct and i have procured a few clues out of her that narrows it down somewhat.
i was hoping to put together some sort of print-out that lists every single possibility and just bombard her with all of the guesses until i get it. but i don't know how to write a computer program and i still have to large a range of possibilities for this to be realistic.


anyway...the clues...
she is of latin american decent....she says that she and her mom are the only ones that she has ever heard of with this name.

8 letters...
begins with L
ends with A
2nd letter is a vowel
letters definitely not in the name include Z, Y, Q, W


multiplying it out (5 possibilities for 2nd letter, 22 possibilities for 3rd through 7th letters) you get 25,769,160.

that would be a rather huge printout.

obviously possibilities like Laaaaaaa and Lalalala and Layoyoya and Lerrrhha are out.
i'm assuming it's a fairly reasonably pronounced word and not a bunch of nonsense letters.

perhaps i can find out if any letter is repeated back to back....i suspect the name does not use the same letter 3 times in a row sense that would be pretty weird.


at first she miscounted and thought there wre only 7 letters (which also cut down the possibilites significantly) but she has now informed me that it's actually 8 letters. i told her that that fact alone took it from around 1 million to 25 million and she was rather proud of herself.

guesses that did not make the cut when i thought it was 7 letters included Levitra (a TV commercial was on while we were talking about this) and LaBamba.


anybody have any ideas on how i can go about this??
apologies if you think i am wasting everyone's time with this goofy post.

[/ QUOTE ]

I'd start looking for spanish names/words/curses/female body parts that follow that criteria.

[TeeVeeDude]

Ask her if it rhymes with a female body part. Then go through her purse and look at her driver's license.

(If this post doesn't make sense, then you are not a Seinfeld Fan.)

[MicroBob]

[ QUOTE ]
also...anyone care to take a shot at determining the chances that the 2nd letter is an 'A'??



--------------------------------------------------------------------------------



1 in 5?

[/ QUOTE ]


is it really 1 in 5??

the last letter is an A.
one of the vowels is repeated (and not conseuctively).
we also know that at least one of the letters in spots 3, 4, 5, 6, or 7 is a consonant (although we also know that if there aren't more consonants it would be a REALLY silly name).



i don't know how to use any of the programs that are referred to here. i don't even know if they are already on my computer or if i have to download them or buy them somewhere or what.
some people don't seem to believe me when i say that i would have no idea how to write a computer program.
i guess i could take some sort of class to learn but i don't really feel the incentive to do that.


a couple of people have mentioned her driver's license....it only lists here middle initial on the license.

[MicroBob]

[ QUOTE ]
LAKEISHA
LASHONDA
LUCRETIA
LAURETTA
LATRICIA
LAKESHIA
LASHANDA
LORRETTA
LUCRECIA
LOUVENIA
LARHONDA
LASHUNDA
LEONARDA
LATASHIA
LASANDRA
LATRISHA
LAKIESHA
LATARSHA
LUDIVINA
LAURINDA
LISANDRA
LAKENDRA
LAKEESHA
LILLIANA
LAMONICA
LACRESHA



[/ QUOTE ]


you didn't read the 2nd list of clues.....
one of the vowels is repeated (potentially more than once).
one of the consonants is repeated.
there are no letters used consecutively.

amazingly, just scanning your list, it appears that NONE of them qualify.
as far as i could tell, any names that used the same consonant twice used it back to back.


this makes the caluclation of total possiblities pretty difficult doesn't it??

how does one caluclate the figure for this?

2nd letter - 5 possibilities
3rd letter - 21 - the 4 'banned' letters and the previous letter can't qualify...but we don't know what the 2nd letter is....so do we really have only 21 possiblities here?
and so on...
and somewhere in there we have to use the same consonant twice and the same vowel twice.


i received a PM from a guy who has a latin-american friend with a name that fit the first post's criteria.
after my 2nd post he saw that it still could be a possiblity. i'll try that one out when i get a chance.




if someone wants to steer me more in the right direction of writing such a program with EZ-whatever (if it really is so easy) and can successfully explain to me how to do it then i will certainly look into it. but i doubt that i will be able to do it very easily and i really don't want to bang my head against the monitor for a couple of hours trying to figure out how to write some silly GF-middle-name program.

[BradleyT]

Just open some of her mail and find out.

[aloiz]

I think that the number of possibilities is something like 5*21^4*20. The fact that you have a repeating consonant and vowel doesn't tell you much given that the first letter is a consonant and the last is a vowel. The last term being a twenty is because you know it doesn't end in aa, plus the fact that it can't be the same as the third to last letter.

Writing a program would be straight forward, but pointless as you're going to spit out about 1,000,000 results. I think the easiest way would be to have a bunch of nested for loops, the innermost being the second to last letter, the outermost being the second letter.

You’re going to need a ton more info before you can whittle it down to a reasonable number of names.

Oh, also you are right about the second letter having a different probability than 1 in 5 of being an A, but without knowing the total number of vowels, or the relative probabilities that the name contains x number of vowels you can't come up with an exact answer (I think). However if the name had four total vowels, then the probability that the second letter was an A given that you had one vowel that occurred twice would be 1/6. If the name only had three vowels it'd be 1/3. Pretty sure this is right but if someone else could verify...

aloiz

Edit: Now that I'm thinking about it, 5*21^4*20 (should be an upper bound given the info) is not the correct answer. The correct answer is much more complicated, and since I've always sucked at permutations so maybe someone else can come up with an exact number.

[MicroBob]

some people don't have their middle name on their mail or driver's license or virtually anything else... etc etc.

[MicroBob]

[ QUOTE ]
I think that the number of possibilities is something like 5*21^4*20. The fact that you have a repeating consonant and vowel doesn't tell you much given that the first letter is a consonant and the last is a vowel.

[/ QUOTE ]

i am still curious if it is possible to account for this possiblity and come up with an exact number of possiblities.
i suspect that there isn't a way to come with an EXACT number of names that fit the qualifications specified without writing them all out.
this fact alone i find to be most interesting because the stipulations that we are putting on the name aren't THAT complicated.

i seem to remember that there are other math problems that can't be easily solved by mere calculations and you actually need to write out all the possiblities to come up with a precise figure.
this isn't terribly practical for this problem....but i think it might qualify.


[ QUOTE ]
Writing a program would be straight forward, but pointless as you're going to spit out about 1,000,000 results.

[/ QUOTE ]

i was hoping to take the list and read the whole thing at her as quickly as possible.
then i could come back with a new thread titled 'the answer to my EX-GF's middle name!!'



[ QUOTE ]
Oh, also you are right about the second letter having a different probability than 1 in 5 of being an A, but without knowing the total number of vowels, or the relative probabilities that the name contains x number of vowels you can't come up with an exact answer (I think). However if the name had four total vowels, then the probability that the second letter was an A given that you had one vowel that occurred twice would be 1/6. If the name only had three vowels it'd be 1/3. Pretty sure this is right but if someone else could verify...


[/ QUOTE ]

interesting, and i think you may be right.
i think you can take all of the combinations of vowels for spaces 3 through 7 and come up with some figures.
spot 7 can't be an A...nothing can be repeated....and at least of the letters has to be a non-vowel.

for purposes of practicality...what if we assumed that there were at least two consonants represented between letters 3 and 7??
we already know there is at least one more because a consonant is repeated but that could still be the L.
but to ONLY have one more L surrounded by a bunch of vowels probably won't make anything we can use.

also lets assume that there is at least one more vowel between 3 and 7 (in other words, it's not ALL consonants...that would be silly).

so for the 5 spots that are letters 3 through 7 we SHOULD have between 1 and 3 vowels (not 0 through 4 as we started when we only knew that one consonant was repeated).

that should narrow it down to a point where you can actually determine the number of possibilities of vowel combinations by hand i think.

LAxAxxxA
LAxxAxxA
LAxxxAxA
LAExxxxA
LAxExxxA
LAxxExxA

and so on for each possiblity that there is only one vowel in the 3-through-7 spots.
then you would have to do it again for two vowels, and so forth...obviously making sure you didn't have any consecutive letters and making certain that you repeated a vowel at least once.


after that, you could even take the number of combinations of consonants for all of the possibilities of total number of consonants and multiply it together.

first you do it with all the combos that include L as the first represented consonant perhaps...
LxxLxBxA
LxxLxxBA
LxxxLBxA
LxxxLxBA
LxxxxLBA

that's 10 possibilties...there are also 10 possibilities if B is the first letter and L is the latter.

so there are a total of 20 possibilities in there (disregarding vowels) if L and B are the only consonants in spaces 3 through 7.
we have 17 letters that we can substitue for the L (26 minus 5 vowels minus the 4 'banned' consonants).
but if the 2nd letter is also an L there are only 6 possibilties.
20 X 16 = 320 + 6 = 326.

after you work through the vowels combos and consonants combos you should jsut be able to multiply the two, correct??

it would be some work....but a bit more realistic then i previously would have thought (which was writing out each one of the possibilities by longhand).


i really suck at permutations and combos too...and anyone who is decent at them can see i am starting from very much a beginner's point here and am going through some basic stuff....but i still find it interesting nonetheless.

[MicroBob]

somebody pm'ed me that they have a latina friend named Liudmila but the GF tells me that's not it....and says she wishes that was it because Liudmila is much better than what she's stuck with (she's really embarassed by this thing but i'm positive she is overreacting and it's not THAT bad).

anyway....
there are a couple more additions to the clues list -

here's the whole list....

8 letters
begins with L
ends with A
2nd letter is a vowel
letters DEFINITELY not in the name include Q, W, X, Y, Z
one consonant is used twice (and only twice)
one of the vowels is used twice (and only twice)
none of the letters appear consecutively


please note, the addition of Z to the banned letter list.
also note, there is only one vowel that is used exactly twice (previously i wasn't certain whether it could be used 3x in the name or not).


she's pretty sure i will never get it....which naturally encourages me to forge onward. maybe i really can somehow print out the whole list of possibilities (1-million or so) and bombard her with all of the guesses until i get it.

[koolmoe]

No, I was joking. It's actually better than 1 in 5.

Let's take a simple case to see this. Assume you have three letters. The first is a vowel, the second could be any letter, and the third is an A. You know that the word contains a repeated vowel and that the vowel is repeated only once.

You have three cases to consider.

1) Assume that the second letter is a consonant (probability 21/26). In that case, the probability that the first letter is an A is 1.

2) Assume that the second letter is an A (1/26). In this case, the probability that the first letter is an A is 0.

3) Assume that the second letter is a vowel besides A and that it is equally likely that the fist letter repeats the second letter vs. repeating the A. In this case, the probability that the first letter is an A is 1/2.

The total probability is the sum of each conditional probability multiplied by the probability that the condition occurs:

P[first = A] =
P[first = A | second = consonant] P[second = consonant] +
P[first = A | second = A] P[second = A] +
P[first = A | second = other vowel] P[second = other vowel]

= (1)(21/26) + (0)(1/26) + (1/2)(4/26) = 23/26

Notice that each condition is mutually exclusive of the others and they describe the entire space of possibilities.

You can do this for your problem, but there are a lot more conditions to consider...

[aloiz]

Actually in all likelyhood the probability that the second letter is an A is less than 1 in 5. In our case we don't know how many total vowels are in the name, so I don't think we can come up with an exact answer. However since there are 5 letters bettween the second vowel and last letter 'A' it looks like there are either 3, 4, or 5 total vowels. If there are 3 then the second letter would be an 'A' 1/3, however I think only three vowels is pretty unlikely. If there are 4 then it would be 1/6, and for 5 1/10.

aloiz

[BigBiceps]

[ QUOTE ]
anyway....
there are a couple more additions to the clues list -

here's the whole list....

8 letters
begins with L
ends with A
2nd letter is a vowel
letters DEFINITELY not in the name include Q, W, X, Y, Z
one consonant is used twice (and only twice)
one of the vowels is used twice (and only twice)
none of the letters appear consecutively



[/ QUOTE ]

Your new rules ruined my guess.

I thought for sure it was Lavagina, which is spanish for the vagina.

Maybe it is Lebabosa which confirms to your new rules, which I guess means the slug.

[koolmoe]

You can compute the probability exactly. It is about 74.6% likely if you assume that all permutations that fit the criteria listed (even the ones with 5 consonants between the 2nd and 8th letters) are allowed.

If you discount all the the names with only two vowels, then the probability would be less than 74%, but it would still be more than 20%.

But I don't think that's what the assignment...er...his girlfriend is asking for.

[koolmoe]

[ QUOTE ]
Don't have the precise figure, but an upper bound is 1,390,260 ways


[/ QUOTE ]

I misread your note and thought that the repeated consonant could not be an L. Assuming that it can, there are about 18.5 million possibilities according to my calculations.

[ QUOTE ]

[ QUOTE ]

also...anyone care to take a shot at determining the chances that the 2nd letter is an 'A'??


[/ QUOTE ]

1 in 5? /images/graemlins/wink.gif

[/ QUOTE ]

It's about 75% assuming that all permutations that fit your criteria are allowed.

[aloiz]

Could you show me how you calculated that?

thanks,
aloiz

[RollaJ]

[ QUOTE ]
Lavagina, which is spanish for the vagina

[/ QUOTE ]
Thanks for clearing that up /images/graemlins/laugh.gif

[koolmoe]

Sure. Enumerate the cases:

Case 1: 2nd letter is an A, L is the repeated consonant (11,162,880 possibilities)
Case 2: 2nd letter is an A, L is not the repeated consonant (2,643,840 possibilities)
Case 3: 2nd letter is not A, L is the repeated consonant, A or the 2nd letter is the repeated vowel (3,760,128 possibilities)
Case 4: 2nd letter is not A, L is not the repeated consonant, A or the 2nd letter is the repeated vowel (783,360 possibilities)
Case 5: 2nd letter is not A, L is the repeated consonant, neither A nor the 2nd letter is the repeated vowel (117,504 possibilities)
Case 6: 2nd letter is not A, L is not the repeated consonant, neither A nor the 2nd letter is the repeated vowel (34,560 possibilities)

The first two cases account for about 13.8 million of the 18.5 million possibilities, or about 75%. Notice that a large percentage of the posssibilities in case 1 contain 5 consonants in positions 3-7. If you disallow this, the probability will drop dramatically. I contend that it will always be &gt; 1/5, though I can't verify it at the moment.

The tricky part is figuring out the number of ways you can distinctly arrange the letters in positions 3-7 for each case.

[aloiz]

Alright I wrote a little program to spit out all of the possibilities. I used all of the hints, plus I made a small assumption that the only repeating letters were the two that she mentioned. Taking this into account there are 3,113,046 total possibilities. This would really be cut down significantly if you could find out another letter or two, or find out how many vowels there were.

aloiz

[MicroBob]

would it be possible to get my hands on that little program?? he-he.


that's interesting that you came up with 3-mil+.
higher than i expected (or was previously guesstimated).

i'll get to work on trying to acquire more information....maybe i can get the specific number of vowels.


and fwiw....i don't think she's bothered because it is a dirty word in spanish or english or anything. i think she just really doesn't like the name...that's all.

[aloiz]

Ran some more tests. If you can get the total number of vowels you limit the possibilities to the following:
3 vowels: 1,315,412
4 vowels: 380,804
5 vowels: 37,316

If you can get the number of vowels and one more consonant you'd have
3 vowels: ~ 60,000
4 vowels: ~ 14,000
5 vowels: ~ 800

Yea, I'd be happy to send you the program, just PM me and let me know.

aloiz

[Moozh]

Molva!

[SuitedSixes]

I hate to bring this up and make it more difficult than necessary, but the Spanish alphabet has more letters than the English one. You probably need clarification what she considers letters. The biggest problem I see is that in Spanish 'll' is considered one letter.

[BruceZ]

[ QUOTE ]
I hate to bring this up and make it more difficult than necessary, but the Spanish alphabet has more letters than the English one. You probably need clarification what she considers letters. The biggest problem I see is that in Spanish 'll' is considered one letter.

[/ QUOTE ]

So are 'rr' and 'ch'. The other extra letter is n with a ~ over it, called (phonetically) "enye" and pronounced as "ny". 'll' is pronounced like a 'y' or a 'j', or somewhere in between, depending on the dialect.

[mostsmooth]

lavagina
oops, looks like somebody guessed this already

[MicroBob]

but these numbers dont add up to the 3.1 mil figure you mentioned earlier.
i'll do my best to narrow it down...but since she really doesnt want me to know i may have gotten all the clues out of her that i can.


i think she is counting 1 letter as 1 letter (so to speak).
when you type ll you have to hit l 2x. just because it would be pronounced like a y i don't see why it would count as 1 letter per-se.
and i'm sure she would count this as the same letter repeating (which she says does not occur in the name).

anyway....lets just assume that we are all on the same page regarding what a single letter is. lets also assume that 'n' and 'n with a squiggle thing' are the same letter as i am hoping she would give me credit for either.


however....there CAN be a huge difference to be sure.

a few mths ago i was talking with a friend of mine on the phone who knows spanish and asked him to teach me to say 'Happy New Year' so i could wish that upon my GF.

later when i tried it and she started laughing i learned that
anos and anos (one with the enye and one without) are very different words.

one means 'year' and the other means 'anus'.
i had accidentally wished my GF a 'Happy New Anus!!'

okay...i'm ready for the variety of witty comebacks i'm sure will ensue.

[aloiz]

A large portion of the 3.1 million came from names with only two vowels (the second and last letters both being 'A'). Since there's no way that there's only two vowels we could easily eliminate over half the 3.1 million, but I just want to give you an exact number based only on what we know for sure.

Also I think the point about the extra letters is at least something you should ask her about.

aloiz

[MicroBob]

you mean you don't think Lasdskha or Larcrfsa are possibilities??
what about Lajkjpta??

hmmm.....maybe she really DOES have cause to be so embarassed by this name.

[magic7s]

lynx -dump http://www.20000-names.com/female_l_names.htm |egrep -i "L.{6}A:"
LAETITIA: see Letitia.
LAURETTA: Diminutive of Laura, "victory; triumph." Variant, Loretta,
LUCRETIA: Feminine form of Roman clan name meaning "wealth."

[magic7s]

LEOCADIA

[mikewvp]

What is wrong with you? Look at her friggin driver's license.

I read later that this is for your own math related entertainment, in which case it is ok, just not my thing.

[MicroBob]

you also missed the part where i mentioned (several times) that it's not on her DL or any other papers.

[MicroBob]

new clues i have procured to be added to the old ones:

2nd letter is NOT an A or an I
L, B, V are now on the list of 'banned' letters in the 3-through-7 spots.
the 7th letter is a consonant.

she also confirmed that there is no double-letter as a single-letter situation such as ll. it is an 8-letter name in the sense that us silly gringos would recognize it as 8 letters.


this should narrow it down considerably i think.
couldn't get a figure on exact number of vowels though.


all clues combined

8 letters
begins with L
ends with A
2nd letter is E, O or U
7th letter is a consonant
letters DEFINITELY not in the 3-through-7 spots are:
L, B, V, Q, W, X, Y, Z
one (and only one) of the consonants is used twice (and only twice)
one (and only one) of the vowels is used twice (and only twice)
at no point does the same letter appear consecutively.


aloiz....if you want to run these new variables into the program feel free.
i'll probably be sending you a PM in the near future.

maybe we are closing in on having a reasonable number to possibly print-out.

[aloiz]

Those helped a ton. Down to 74,880.

aloiz

[koolmoe]

[ QUOTE ]
Those helped a ton. Down to 74,880.

aloiz

[/ QUOTE ]

If you just assume that A is the repeated vowel and that the 7th letter is the repeated consonant, you have 147,420 possibilities. This is a lower bound on the total number of names.
choices for 2nd letter: 3
chocices for 7th letter: 13
choices for letters 3-6: 1, 1, 15, 14
ways to arrange letters 3-6: 3*3*2*1 (the first 3 is due to the fact that the repeated consonant cannot appear in the 6th position)

Total number of names fitting this assumption: 3*13*1*1*15*14*(3*3*2*1) = 147,420

You can enumerate 6 cases to compute the total number of names. In addition to the case above, I have:

Case 2: RV is A, RC not 7th - 39,312
Case 3: RV is 2nd, RC is 7th - 114,660
Case 4: RV is 2nd, RC not 7th - 32,760
Case 5: RV not A nor 2nd, RC is 7th - 8,190
Case 6: RV not A nor 2nd, RC not 7th - 2,808

The total number I calculated for the new conditions is 345,150.

[elwoodblues]

A public records search of either her name or her mom's name might help. It might not be on any of her info, but might show up on mom's.

[Benholio]

Once you get it down to a 'short' list of 100k names or so (via some sort of script or program), have it throw those names at google, and then weed out the names that come up with 0 matches, or fewer than X matches, or just have it show you the top X% most commonly used names. One way or another, Google knows everything. I know this isn't a 'common' name, but for example, 'leocadia' comes up with ~10k matches.

Ben

[aloiz]

hmm, my brain is really working right now, but when I get home from work I'll double check.

aloiz

[elwoodblues]

Or just use her first &amp; last name in a search of public records (real property records, credit beaureau info, etc.) Chances are that one of them has the full name of either her or her mother.

[MicroBob]

chances are that it is not listed on any of these.

she was not born in this country and is not a resident of this country.

unless i can get my hands on some birth certificates from peru i don't think i'm going to track it down.


she just uses the middle initial.

[MicroBob]

this is an intersting idea...

i was just thinking of going through the list and weeding out the more extreme unlikely possibilities...

anything that has a 'k' or 'j' or 'h' in an unrealistic arrangement

lerahjra
lejkraja
luhjukja
lomomhja

but now that i look at them they don't seem THAT impossible. but are still pretty darn unrealistic.


so i'll cross out the names that look just TOO unlikely to be it and come at her with the remaining ones. if i have to go back to the list of 'mucked' names i can do that later.


thanks for looking into this guys.
i hope you're getting some sort of warped enjoyment out of this little puzzle.


btw - more clues are unlikely for the next week or so as she has a lot of shifts at work.
for now, i think we're stuck with what we've got.

[BigBiceps]

well .... what is the name?

[Ghazban]

Yeah, we're all dying to know here. After all this discussion, you must've gotten it by now, right? Maybe you found out and are now also ashamed.... /images/graemlins/tongue.gif

[kevyk]

You can almost certainly eliminate any names containing a K as well. Just about the only spanish words with a K in them have been adapted from English (e.g. el Kilowatt, etc).

[caretaker1]

Levitraa /images/graemlins/shocked.gif Sorry, had to tack on an A

[PapaSan]

you are pussywhipped..She is latina you can't let her do things like this to you its all a game. Grab her hair and smack her you are the man make her call you Papi Chulo and forget her middle name.

[ODB72]

Luanna

[sunshn4life]

leopolda?