Can Anybody tell me why the expectation value of the Index of Coincidence is the following?

For a cipher text string of length $L$, where $n$ is the number of alphabet characters, $k_r = 1/n$, and $k_p = \sum_{i=1}^n p_i^2$

$$ E(IC) = \frac{1}{t}*\frac{L - t}{L - 1}*k_p + \frac{t - 1}{t}*\frac{L}{L - 1}*k_r$$

I forgot to state that $t$ is the period of the poly alphabetic cipher used to encipher the message.

  • $\begingroup$ I don't think it is. In any case - what is $t$? $\endgroup$ Commented Jun 27, 2012 at 19:52
  • $\begingroup$ @StephenHarris - Ops, sorry for not specifying. $t$ is the period of the poly alphabetic cipher used for creating the crypto message. $\endgroup$
    – Matteo
    Commented Jun 28, 2012 at 13:04

1 Answer 1


Expected Index of Coincidence usually refers to a language's expected index of coincidence (1.73 for English, or 0.067 if you're not normalising).

The formula in question is usually used to determine the length of the key ($t$) given the (measured) $IC$ of received cipher-text. $IC$ is the probability that two randomly-selected letters from the cipher text are identical.

Let $X=\{ x_1,x_2,x_3,...x_L\}$ be the cipher-text. If we think the the poly-alphabetic cipher has period $t$, then we would expect each of the following:

\begin{eqnarray} & X_1 =\{x_1, x_{t+1},x_{2t+1},...\}\\ & X_2 =\{x_2, x_{t+2},x_{2t+2},...\}\\ & \vdots \\ &X_t =\{x_t, x_{2t},x_{3t},...\} \end{eqnarray}

to exhibit the same index as the plaintext ($\kappa_p$). So we can reconstruct $IC(X)$ as follows. Pick two letters at random, we want the probability that they match.

The probability that they are in the same $X_i$ is:

$$ \frac{tC(\frac{L}{t},2)}{C(L,2)} = \frac{L*(\frac{L}{t}-1)}{L(L-1)} $$

The probability they are in different $X_i$s is: $$ \frac{C(t,2)*\frac{L}{t}*\frac{L}{t}}{C(L,2)} = \frac{t(t-1)*\frac{L}{t}*\frac{L}{t}}{L(L-1)} $$

  • If they are in the same $X_i$ then they are both enciphered using the same alphabet: so the probability is $\kappa_p$
  • If they are in different $X_i$s then they are enciphered using different alphabets, so we can assume they are randomly distributed: so the probability is $\kappa_r$

So the probability of two random letters matching is approximately

\begin{eqnarray} IC(X) & \approx \frac{L*(\frac{L}{t}-1)}{L(L-1)}\kappa_p + \frac{t(t-1)*\frac{L}{t}*\frac{L}{t}}{L(L-1)}\kappa_r\\ & = \frac{(L-t)}{t(L-1)}\kappa_p + \frac{(t-1)*L}{t(L-1)}\kappa_r \end{eqnarray}

  • $\begingroup$ Ok, so what I found indicated as Expectation Value of the IC, actually is the calculation of the IC?I'm afraid I'm making confusion: I thought that the IC was a random variable, and E(IC) it's expectation value...am I wrong? $\endgroup$
    – Matteo
    Commented Jun 29, 2012 at 7:20
  • $\begingroup$ I forgot to say, thks for help! $\endgroup$
    – Matteo
    Commented Jun 29, 2012 at 7:21
  • $\begingroup$ A random variable is a map from a probability space to a space of 'outcomes'. $IC$ is a probability. The $IC_{expected}$ is the probability that two letters taken from the alphabet match and depends on the statistical characteristics of the language. You compare this with the calculated $IC$ (which is a ratio of number of matches to number of pairings). You then use the above formula to solve for $t$. This gives an estimate for the period used. $\endgroup$ Commented Jun 29, 2012 at 11:56

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