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Math Help - Random Question -= What are the chances

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    Random Question -= What are the chances

    This may sound random but I was writing my girlfriends name this afternoon and noticed that out initials were in perfect alphabetical order. She has the initials JK and I have LM. So that's 'JKLM'. Can I somehow workout what the chances or odds are for this happening with two people? Is this a combinatorics problem, 1 - 26 letters?
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    Re: Random Question -= What are the chances

    Quote Originally Posted by floorplay View Post
    This may sound random but I was writing my girlfriends name this afternoon and noticed that out initials were in perfect alphabetical order. She has the initials JK and I have LM. So that's 'JKLM'. Can I somehow workout what the chances or odds are for this happening with two people? Is this a combinatorics problem, 1 - 26 letters?
    The most charitable I can be is to say that this is an ill posed statement.
    It is not clear if you mean four distinct letters or can a letter be repeated?

    It is not clear if the letters must be consecutive as in your example or could it be AEJP?
    There are only 23 strings of consecutive four letters.

    There are combination 26 chose 4, \binom{26}{4}, ways to have four distinct letters in alphabetical order.

    There are \binom{29}{3} ways to have any four letters(repeats allowed) in alphabetical order.

    Here is a separate issue: are we to assume a person has only two initials?
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    Re: Random Question -= What are the chances

    The odds can be calculated as follows - assuming that the chance of an initial being any particular letter is 1/26:

    1. Your girl friend's first initial could have been any letter from A through W (not X. Y or Z or there would have been no way for the next three letters to follow in order). The odd of this is 22/26.

    2. Then the next three letters must follow in order, the odds of which are (1/26)^3.

    The combined odds is theresfore 22/(26^4). However, it could have been that your first initial (not hers) comes first, so the odds that a two people have initials that can be arranged alphabetically is two times this, or 44/(26^4) = 0.0000963.

    As noted earlier this assumes that any letter is as likely to be used as an initial as any other. But in reality I suspect the odds are a bit higher, as some letters are more common than others, and I suspect that the letters J, K, L and M are more common than, say W, X, Y and Z.
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    Re: Random Question -= What are the chances

    @ Plato What I was asking was consecutive letters 1 after another in alphabetical order. Just like the initials in my OP, 'JKLM'. (ABCEDFGHI'JKLM'NOPQRSTUVWXYZ) The chance of that happening with two people. No middle names.

    @ ebaines Thanks for explaining that. Your right there would be other fine tuning variables that would give slightly different outcomes. But that result you post (0.0000963) is that the percentage chance of two people having the their initials in consecutive alphabetical order?!

    Maybe I should buy a lotto ticket.
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    MHF Contributor ebaines's Avatar
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    Re: Random Question -= What are the chances

    Yes - that's the chance of two randomly-selected people having all 4 initials in consecutive rising alphabetical order. But be aware that sometimes these "fluke chance" events are flukes only because of how we define them. There are lots of ways that 4 letters could be considered special which may seem even more "special" than what you are asking. For example, if you and your GF both had the same initials - such as JJJJ, or JKJK, or if your initials were in reverse alphabetical order such as MLKJ, or if they spelled out a word such as "LOVE" - all those combinations are be considered to be failures for this analysis yet may seem more special than merely being in order.
    Last edited by ebaines; June 5th 2013 at 07:19 AM.
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    Re: Random Question -= What are the chances

    Hello, floorplay!

    Assuming all letters are equally likely, there are: . 26^2\!\cdot\!26^2 possible outcomes.

    There are 23 sets of four consecutive letters, from ABCD to WXYZ.
    However, you could have the first two initials and she could have the last two
    . . or vice versa.

    Hence, there are: 2\times 23 \,=\,46 successful outcomes.


    Therefore, the probability is:. \frac{46}{26^4} \;=\;0.000100662
    . . about one in ten thousand.


    That seems to be quite a rare event.
    But consider the number of couples in, say, California.
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