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I am stuck on the following problem: I have a set $A$ of the first $n$ natural numbers. I define a new set $B$ picking randomly $m$ numbers from $A$. What is the probability to have at least $k$ couples of consecutive numbers in $B$? For example, I put $n=100$, $m=5$, $k=2$. So I have: $$A=\{j=1,2,3,...,n\}$$ What is the probability that in the set $$B=\{b_1,b_2,b_3,b_4,b_5\}$$ I have $2$ couples of consecutive numbers? For instance, the set $$B=\{5,6,34,12,13\}$$ contains two valid couples. The set: $$B=\{90,20,12,13,66\}$$ doesn't qualify as a success. Thanks

Joffan
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Riccardo.Alestra
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  • Do you want the probability of exactly $k$ couples in $B$, or at least $k$? How do you deal with a situation like $B={12,13,14, 20, 25}$? Does that count as two couples? – kccu May 27 '16 at 14:20
  • @kccu: at least $k$ couples. In your example: ${B=12,13,14,20,25}$ is good because it contains $3$ couples – Riccardo.Alestra May 27 '16 at 14:25
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    @Riccardo.Alestra: I get the reason for ${12,13,14,20,25}$ counting as two couples, but why three? – Jack D'Aurizio May 27 '16 at 14:31
  • @Riccardo.Alestra By $n$ couples, do you mean there are $n$ consecutive numbers? – DylanSp May 27 '16 at 14:38
  • @DylanSp: $k$ couples means $k$ couples of consecutive numbers ${j,j+1}$ – Riccardo.Alestra May 27 '16 at 14:49
  • I predict no (complete) answers to this question. See here http://math.stackexchange.com/questions/915400/how-many-subsets-does-the-set-1-2-dots-n-have-that-contain-no-two-con for a far simpler special case. – almagest May 27 '16 at 15:09

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I don't see an easy way. For your example, there are ${100 \choose 5} = 75287520$ possibilities for $B$. Of those, ${96 \choose 5}=61124064$ have no neighboring pairs. To count the number with neighboring pairs, you have $2$ choices pair at the end, then ${95 \choose 3}=138415$ ways to choose three non-neighbors out of what is rest, plus $97$ ways to choose a central pair and ${94 \choose 3}=134044$. The ways to choose $B$ without two pairs is then $61124064+2*138415+97*134044=74403162$ The chance of two pairs is then $1-74403162/75287520\approx 0.01175$

Ross Millikan
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