my GF's middle name

[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.

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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...


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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.

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


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1 in 5? [img]/images/graemlins/wink.gif[/img]

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