# Random Intervals with One Dominant

### Solution 1

Our sample space consists of $(2n)!$ permutations of the given point indices. An interval $J_j$ is dominant in $2(2n-2)!$ cases, as the $2n-2$ indices can be arbitrarily permuted between the two extremes. One of the intervals is dominant in $n\cdot 2(2n-2)!=2n(2n-2)!$ cases. Thus the probability of there being a dominant interval is $\displaystyle \frac{2n(2n-2)!}{(2n)!}=\frac{1}{2n-1}.$

### Solution 2

The problem is reduced to finding adjacent $X_{max}$and $X_{min}$(since only the interval $[X_{max}, X_{min}]$ covers all other intervals) in permutations of $X_1,\ldots,X_{2n}$.

Note that, given that $2 j$ is even, to be compatible with the previous formulation of the problem, we consider intervals $[X_1,X_2]$ and$[X_3,X_4]$,but not $[X_2,X_3]$.

• $X_{max}$ followed by $X_{min}$

There are $(2n-2)!$ sequences of each

$n \text{ lines}\left\{ \begin{array}{l} (X_{max},X_{min},X_3,X_4,\ldots,X_{2n})\\ (X_1,X_2,X_{max},X_{min},\ldots,X_{2n})\\ ...\\ (X_1,X_2,\ldots,\ldots,X_{max},X_{min})\\ \end{array} \right.$

for a total of $n (2n-2)!$

• $X_{min}$ followed by $X_{max}$

• By mirroring, same $n (2n-2)!$

• Thus

Numerator= $2 n (2 n-2)!$
Denominator(total permutations)= $(2n)!$

$\displaystyle p=\frac{1}{2 n-1}$

Note: thanks to Michael Wiener for correcting for the count of the number of lines.

### Solution 3

Easier path is that $X_{min}$ is equally likely to be paired with any of the other $2n-1$ values. Only $1$ is $X_{max}.$

### Acknowledgment

The problem has been proposed by N. N. Taleb as a follow-up to an earlier question about overlapping intervals. Solution 2 is by N. N. Taleb; Solution 3 is by Michael Weiner; Solution 4 is by Zhuo Xi. Michael Weiner came up with a variant of the problem. Solution 1 to the Variant problem is by Josh Jordan; Solution 2 is by Michael Weiner.

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