# What's the chance of an explicit series of integers in a limited random distribution?

Say I collect 40 perfectly random integers between 1 and 400. What's the chance that any integer is repeated consecutively six times in such a random draw?

What I'm looking for is the chance of sequences like [372, 193, 42, 42, 42, 42, 42, 42, 274, 42, 7, ...], [372, 193, 42, 42, 42, 42, 42, 42, 274, 242, 7, ...], or [372, 193, 42, 42, 42, 42, 42, 42, 42, 42, 42, ...] as they all fulfills what I'm looking for. As a counter example [372, 193, 42, 42, 42, 42, 42, 77, 274, 42, 7, ...] does not satisfy my conditions because the six 42's are not consecutively repeated.

The Birthday problem gives that it's an 87% chance that two of the 40 are the same number but I'm failing to go from that to calculating the chance of a certain integer consecutively repeating itself n number of times in the random collection.

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When you say "number", do you mean integer or real number? When you say "random", according to what distribution? –  Zev Chonoles Jun 29 '11 at 23:06
Integer and lets imagine a perfectly random distribution. –  Jonas Elfström Jun 29 '11 at 23:08
Before computing the probability of an event you must carefully define it: the same number repeats itself six times is bad formulated. You mean that some (any) number appears exactly six times? Regardless of that some other number appears more that six times (or six times)? –  leonbloy Jun 29 '11 at 23:16
Any of the integers between 1 to 400 appearing after another six times. Like [301, 23, 42, 42, 42, 42, 42, 42, 255, 120, 42, ...]. This is six times repeated. I will try to clarify this in the question. –  Jonas Elfström Jun 29 '11 at 23:20
Must it be repeating itself exactly six times (of the 40 random numbers)? (i.e., appears exactly 7 times)? or at least 7 times? –  amWhy Jun 29 '11 at 23:39

You can attack this with a Markov chain, with six states (the number of consecutive numbers of the last run), and with an absorbent final state. So the stochastic matrix is given by

 N = 400
P =  [ 1-1.0/N   1.0/N    0.0      0.0      0.0      0.0    ;
1-1.0/N   0.0      1.0/N    0.0      0.0      0.0    ;
1-1.0/N   0.0      0.0      1.0/N    0.0      0.0    ;
1-1.0/N   0.0      0.0      0.0      1.0/N    0.0    ;
1-1.0/N   0.0      0.0      0.0      0.0      1.0/N  ;
0.0       0.0      0.0      0.0      0.0      1.0    ]


And so the desired probability is

octave > ([1 0 0 0 0 0 ] * P^39 )(6)
ans = 3.4097e-012


Because the probabilities are so low, the most naive approximation (just compute the probability that some 6-run have consecutive numbers: the upper bound in Yuval's answer) is quite precise. You might want to try with other numbers to get more interesting/enlightening results.

Added : An asymtotic approach, for large values of tries $L$ and not so low probabilities: Think of the lenghts of the runs as random variables. Approximate them by $m$ iid geometric variables $x_i$ , with $p=1-1/N$, and $m$ chosen so that $m E(x_i) = L$, and compute the probability that all $x_i < k$ (in the original question $N=400$, $L=40$, $k=6$). This approximation (barring errors) gives :

$$\approx 1 - \left( 1 - \frac{1}{N^{k-1}}\right)^{\frac{N-1}{N} L }$$

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By the way, this is (expectedly) pretty close to the trivial upper bound. –  Yuval Filmus Jun 30 '11 at 2:06
Yes, I was precisely commenting that :-) –  leonbloy Jun 30 '11 at 2:08
sorry for a late comment, but somehow this doesn't make sense to me - isn't the matrix only calculating the probabilities of a length of the last run instead of the longest one ? –  Tomas Vana Jun 10 '13 at 17:19
@TomasVana The Markov chain does not compute the length of the longest run, but the probability that it's at least 6. It has an absorbing state, corresponding to "a run of length 6 was found", so, it will be at that state at the end if that event happened at any time. –  leonbloy Jun 10 '13 at 17:33

You can use inclusion-exclusion to get bounds on this. For $A \subset [40]$ of length $6$, let $S_A$ be the event that all numbers in $A$ are the same. The probability $p$ that you want is then bounded by $$\sum_A \Pr[S_A] - \sum_{A \neq B} \Pr[S_A \text{ and } S_B] \leq p \leq \sum_A \Pr[S_A].$$ Since $\Pr[S_A] = 400^{-5}$, it is easily to calculate the upper bound: it is $$\binom{40}{6} 400^{-5} \approx 0.000000375.$$ Calculating the lower bound is more messy so I'll leave it to you. My guess is that it will be pretty close to the upper bound. If it isn't, you can always take more terms in the inclusion-exclusion formula.

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That's under the interpretation that the repetition aren't consecutive. If they are, then the calculation is easier and the probability much lower. –  Yuval Filmus Jun 30 '11 at 1:35
I understand that we are speaking of sixe consecutive numbers, and that hence the upper bound is just $35 * 400^{-5}\approx 3.4180e-012$ –  leonbloy Jun 30 '11 at 1:57
Sorry, the idea was that they should be consecutive repeated. –  Jonas Elfström Jun 30 '11 at 7:27
@leonbloy Could you please explain $35∗400^{−5}$? I find it interesting that the result is very close to the one I reached in my answer. –  Jonas Elfström Jun 30 '11 at 11:46
@Jonas: Imagine the 40 tries numbered from 1 to 40. The probability of a 6-run starting at (say) position 1 is $400^{-5}$. The same if we consider the starting position 2, 3... 35. You have 40-k+1=35 posible positions for the 6-run. The approximation comes from just summing them, and that's only reasonable because, having so low probabities, the events we are "forgetting" (more than one 6-run) are negligible. –  leonbloy Jun 30 '11 at 12:16

Here's my simplistic approach to calculate the probability.

We can look at this like rolling a dice with 400 sides 40 times. It's 1/400 to draw any one number. Gives $1/(400^k)$ chance to draw a certain number $k$ times. We are not interested in a certain number but any of the 400 will do and that gives $400/(400^k)$.

We have $40-6+1$ chances to hit a streak of $6$ consecutive repeated numbers.

$$(L-k+1)N/N^{k}$$

$N=400, L=40, k=6$

$$(40-6+1)400/400^{6} \approx 3,418e{-12}$$

But as leonbloy and Yuval already pointed out $N/N^{k}$ is equal to $N^{1-k}$ and thus it can be written as

$$(L-k+1)N^{1-k}$$

$$35*400^{-5} \approx 3,418e{-12}$$

That's a tiny chance indeed.

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Actually you have $40-5$ chances (possible positions for the k consecutive numbers inside the 40 run). And that's the approximation that arises in Yuval's answer. –  leonbloy Jun 30 '11 at 12:21
Ah, thanks! A classic off-by-one error. –  Jonas Elfström Jun 30 '11 at 13:00