Probability that a sequence repeats itself

Given an infinite sequence $a_n$ of uniformly random integers $0$ to $9$, what is the probability there exist an integer $m$ such that the sequence $a_1$ to $a_m$ is equal to that from $a_{m+1}$ to $a_{2m}$?

What if we restrict to two symbols, or $k$ symbols?

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You probably want the squence $a_1$ to $a_m$ and $a_{m+1}$ to $a_{2m}$ so the lengths match and you don't insist that the first and last digits are the same. Also, is "rnaomd" supposed to be a random version of "random"? –  Ross Millikan Oct 23 '12 at 18:31
Rnaomd was a wonderfully serendipitous typo! –  Brian M. Scott Oct 23 '12 at 18:42
At first blush, 1/9 suggests itself: 1/10+1/100+1/1000+... but this isn't right because the probabilities are not independent. So it appears the answer is between 1/10 and 1/9. –  Fixee Oct 23 '12 at 19:16
@Fixee yup, I think so too, well at least we can conclude that eventually it will never repeat again... –  Jean-Sébastien Oct 23 '12 at 19:27
@Fixee: That calculation assumes that the events are disjoint, not independent. It would predict a probability of $1$ in the binary case. (See also my answer.) –  joriki Oct 24 '12 at 9:23

I can't give an exact answer, but an interesting connection and some numerical results.

Some dabbling with short examples shows that the events for different integers $m$ are nearly but not exactly independent. For instance, some sequences that don't repeat at $m=3$ repeat at $m=2$ but not at $m=1$, whereas for sequences that do repeat at $m=3$ repetition at $m=2$ implies repetition at $m=1$. However, the effect from the dependence appears quite small, so

$$\prod_{m=1}^\infty(1-k^{-m})$$

should be a good approximation to the probability that there is no repetition. Incidentally, if $k$ is a prime power, this is the probability that a large random square matrix over $\mathbb F_k$ is invertible; see Probability that a random binary matrix is invertible. In that answer, I computed the product for $k=2$ to be approximately $0.288788$, which would make the probability for a binary sequence to exhibit repetition approximately $0.711212$.

Here are numerical results for $k=2$ for repetitions up to $m=17$. Each row contains the number of strings of length of $2m$ that repeat, the number of strings that don't repeat, the proportion $p_m$ of strings that repeat, and a value $\alpha_m=-\log_2(1-(1-p_m)/(1-p_{m-1}))$ that would be $m$ if repetition at $m$ were independent of previous repetitions (here's the code):

$$\begin{array}{rrrlr} m&\text{repeating}&\text{non-repeating}&\text{proportion}&\alpha_m\\\hline 1 & 2 & 2 & 0.5 & 1.00000\\ 2 & 10 & 6 & 0.625 & 2.00000\\ 3 & 44 & 20 & 0.6875 & 2.58496\\ 4 & 182 & 74 & 0.7109375 & 3.73697\\ 5 & 738 & 286 & 0.720703125 & 4.88753\\ 6 & 2972 & 1124 & 0.7255859375 & 5.83794\\ 7 & 11924 & 4460 & 0.7277832031 & 6.96450\\ 8 & 47768 & 17768 & 0.7288818359 & 7.95290\\ 9 & 191214 & 70930 & 0.7294235229 & 8.96725\\ 10 & 765136 & 283440 & 0.7296905518 & 9.98483\\ 11 & 3061104 & 1133200 & 0.7298240662 & 10.98340\\ 12 & 12245530 & 4531686 & 0.7298904657 & 11.99044\\ 13 & 48984342 & 18124522 & 0.7299235761 & 12.99397\\ 14 & 195941804 & 72493652 & 0.7299401015 & 13.99640\\ 15 & 783776080 & 289965744 & 0.7299483567 & 14.99761\\ 16 & 3135122038 & 1159845258 & 0.7299524820 & 15.99838\\ 17 & 12540523572 & 4639345612 & 0.7299545438 & 16.99901\\ \end{array}$$

The agreement with $0.711212$ is OK but not marvelous. As suggested by the $m=3$ example, the dependence slightly increases the probability of repetition because a repetition implies correlations between the conditions of repetition at lower values of $m$; but this effect is strongest at $m=3$ and becomes negligible at higher values of $m$, where $\alpha_m\approx m$ shows that almost exactly the expected proportion of sequences repeat for the first time. Thus we can get an accurate estimate of the limit probability from $p\approx p_{17}+(p_{17}-p_{16})\approx0.7299566056$, which should be accurate to eight or nine digits.

I checked OEIS for the sequences of both the repeating and the non-repeating counts; no hit.

P.S.: Note that as t.b. pointed out in a comment under the answer linked to above, by the pentagonal number theorem the above product is given by

\begin{align} \prod_{m=1}^\infty(1-k^{-m})&=\sum_{j=-\infty}^\infty(-1)^jk^{-j(3j-1)/2)}\\&=1-k^{-1}-k^{-2}+k^{-5}+k^{-7}-k^{-12}-k^{-15}+k^{-22}+k^{-26}-\dotso\;, \end{align}

which leads to an interesting pattern of digits in its representation in base $k$:

\begin{align} \prod_{m=1}^\infty(1-2^{-m})&=0.01001001111011100000010000111111110111110000000000100\ldots_2\;, \\ \prod_{m=1}^\infty(1-10^{-m})&=0.89001009999899900000010000999999998999990000000000100\ldots_{10}\;. \end{align}

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(I'm not an expert on this (I'm a newbie), but I suddenly hit something while reading the question, it may be wrong)

Assuming $_m$ is a constant:

It will depend on how big $_m$ is. If the $_m = 1$, then the probability would be $\frac1{10}$. If $_m = 2$, then $\frac1{100}$. (ie if you had $19$ as the first two, there are $100$ possibilties of the second two, so the probability of $19$ coming up is $\frac1{100}$.

Therefore, the probablity is $$\frac1{10^m}$$

If $_m$ can be any number:

Then..... I don't know how to work that out. (Unless I suddenly have a brainwave...)

(This is my first answer, I like mathjax.)

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It depends on how you see the question, in your example for $m=2$, say you got $11$ instead. Then you already had a repeating sequence. Your idea is the right one I think, you want to compute the probability for each $m$, and then sum over all $m's$ to get the probability that it ever repeats. –  Jean-Sébastien Oct 23 '12 at 19:03
@Jean-Sébastien, yeah I think you're right. However, if $_m$ can change, I see all sorts of compliacted stuff, and I don't think I can work that out. –  ACarter Oct 23 '12 at 19:08