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Simple Combinatorics: What are odds of picking same number?

  1. Feb 22, 2009 #1
    This isn't a hwk question, it's just something I've been trying to show my dad. I'm probably wrong. Okay my question is, suppose you pick ONE letter out of the alphabet.

    The odds of me picking your letter are 1/26, if I only have one guess. However, if I have two guesses, aren't the odds a LITTLE bit more than 2/26? The reason why I think so is if you do the calculation this way:

    Find out what the prob. is that you DONT pick the letter. Then it is (25/26)*(24/26). Since there are 25 ways to pick the wrong first letter, and then 24 ways to pick the wrong letter a second time. Now you just do 1 minus the above to get the probability it is that you pick the CORRECT letter. Is my reasoning right?

    Also, obviously, if this is true, you can then extend it to three guesses, and four, etc. by:
    25*24*23*22*.../26^(n) where 'n' is how many tries you took. Thanks!
  2. jcsd
  3. Feb 22, 2009 #2


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    Dearly Missed

    Well, 1-(25/26)*(24/26) represent the chance of picking the right letter AT LEAST once!

    But, obviously, if you pick the right letter the first time, then you do NOT, in the described situation, make yet another guess; thus, for your situation, it would be incorrect to retain the probability of getting two correct guesses!

    To do this properly, we can decompose our P(getting the right letter on max two guesses) as follows:

    P(getting the right letter on max two guesses) =P(getting it right the first time)+P(chance of getting it right the second time, having failed the first)

    This equals:
  4. Feb 22, 2009 #3
    Thanks arild! Um.. There's just one thing. I'm guessing the letters simelaneously. So there's no feedback on wether my first guess was right. I write down my two guess , and then put it on a table, and he reads it. What are the odds he tells me one of my letters are right. That's the question..

    Does this change the 2/26 ?
  5. Feb 22, 2009 #4


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    Well, but you DO know that you choose two DIFFERENT letters, don't you?
    To put this in another way:
    Thus, you can pick one combination out of 26*25 different combinations.

    Label the two letters picked as "A" and "B". Now, the number of combinations in which
    "A" is the right letter (but not B) is evidently 25. Equally, there are 25 possiblities that "B" is right and "A" not. Thus, you have 2*25 favourable combinations to pick between.

    Thus, the probability is (2*25)/(26*25)=2/26.
  6. Feb 22, 2009 #5
    It doesn't matter if you guess at the same time or one after
    the other, so long as you pick different letters.

    Probability of getting it right on the first one:

    1/26. Easy.

    Probability of missing it on the first, then getting it
    right on the second:

    (25/26)*(1/25) = 1/26. Not too bad.

    Since these two events are mutually exclusive, we have
    that P(A AND B) = 0. And since

    P(A OR B) = P(A) + P(B) - P(A OR B)
    P(A OR B) = P(A) + P(B)
    P(A OR B) = 1/26 + 1/26 = 2/26.

    Looks like 2/26 to me.
  7. Feb 22, 2009 #6


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    Right. And if you can't control duplicates -- say, you're rolling a 26-sided die labeled with the letters, and C might come up twice -- the chance of guessing it is 2/26 - 1/26^2 = 51/676.
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