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I'm trying to find the probability that two randomly-selected letters from "average" text in a language will be the same.

For example, if my hypothetical language contains four letters which each occur on average with the following frequency:

A = 60%
B = 25%
C = 10%
D = 5%

What is the probability that selecting any two letters from a representational text will be the same?

My intuition for solving this is first to find the chance that they're different, so the sum over the probabilities that a letter is chosen and then some other letter is chosen next, over each letter in the alphabet:

(0.6  * (1 - 0.6)  + 
 0.25 * (1 - 0.25) + 
 0.1  * (1 - 0.1)  + 
 0.05 * (1 - 0.05)) = 0.565

Then the chance that they are the same: 

1 - 0.565 = 0.435

Is this reasoning sound? It seems like a very basic probability problem, but I always seem to be thinking about these things in the wrong way and would appreciate a sanity check (and any pointers to materials which would help me be more confident about this kind of thing in the future!)

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  • $\begingroup$ A derivative question: What frequency distribution minimizes this probability that two randomly-selected letters from "average" text in a language will be the same? Is it all letters equally common distribution? $\endgroup$
    – curious_cat
    Commented Mar 21, 2013 at 6:49
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    $\begingroup$ @curious_cat yes a discrete uniform minimizes the probability. (This is related to the birthday problem - if you assume anything but uniform births, the probability goes up.) $\endgroup$
    – Glen_b
    Commented Mar 21, 2013 at 7:26
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    $\begingroup$ @curious_cat Based on Glen_b's answer, you may interpret the chance that the two letters are the same as the squared distance to the origin of the "probability vector" $(p_1, p_2, \ldots, p_n)$. The restriction that $p_1+p_2+\cdots+p_n=1$ confines this vector to a plane, yielding a geometric interpretation of your question: what point on this plane is nearest the origin? At any such point, the vector to the origin must be perpendicular to the plane. Obviously such a point is unique and just as obviously $(1,1,\ldots,1)$ is perpendicular: thus all the $p_i$ must equal one another, QED. $\endgroup$
    – whuber ♦
    Commented Mar 21, 2013 at 20:52
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    $\begingroup$ What you are calculating is basically Friedman’s index of coincidence. See for example en.wikipedia.org/wiki/Index_of_coincidence. $\endgroup$
    – schotti
    Commented Dec 19, 2023 at 1:40

2 Answers 2

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You got the right answer but you kind of did it the hard way.

If $L_1$ is the first letter and $L_2$ is the second letter, then

the probability is $\sum_x P(L_1=x) \times P(L_2=x|L_1=x)$.

If you assume independence (which would be the case if you were selecting from the whole population at random for each letter, for example), that's

$\sum_x P(L_1=x) \times P(L_2=x)$.

But of course then the two probability distributions will be the same. Let's call the probability that a random letter has the value $x$ "$p_x$".

That is, the overall probability = $\sum_x p_x^2$

In your case, that's 0.6*0.6 + 0.25*0.25 + 0.1*0.1 + 0.05*0.05 which comes to 0.435.

--

generalization to $k$ letters:

To find the probability that $k$ letters selected from the population are all identical:

If $L_1$ is the first letter, $L_2$ is the second letter, ..., and $L_k$ is the $k^\mathrm{th}$ then

the probability is $\sum_x P(L_1=x) \times P(L_2=x|L_1=x)\times P(L_3=x|L_1=x,L2=x)\times ... \times P(L_k=x|L_1=x,L2=x, ...,L_k=x)$.

If you assume independence (which would be the case if you were selecting from the whole population at random for each letter, for example), that's

$\sum_x P(L_1=x) \times P(L_2=x)\times ... \times P(L_k=x)$.

But of course then all the probability distributions will be the same. Let's call the probability that a random letter has the value $x$ "$p_x$".

That is, the overall probability = $\sum_x p_x^k$

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    $\begingroup$ Thanks, this is exactly the sort of answer I was hoping for. It seems that if I wanted to extrapolate to 3 letters, I would just sum over $p_x^3$ then? $\endgroup$
    – Arne Roomann-Kurrik
    Commented Mar 21, 2013 at 17:19
  • $\begingroup$ Updated answer to cover this case. Short answer: yes. $\endgroup$
    – Glen_b
    Commented Mar 21, 2013 at 20:16
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I think you can compute it by $A^2+B^2+C^2+D^2=0.6^2+0.25^2+0.1^2+0.05^2=0.435$ And $A^2$ means you will get $2A$s, $B^2$ means you will get $2B$s and so on.

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