# Random Intervals with One Dominant

### Problem

### 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 problemo 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 $X_{max}$}

Thus

Numerator= $2 n (2 n-2)!$

Denominator(total permutations)= $(2n)!$$\displaystyle p=\frac{1}{2 n-1}$

By mirroring, same $n (2n-2)!$

Note: thanks to Michael Wiener for correcting for the cound 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 $X_{max}.$

### Solution 4

Locate the point on the left(min). It has $\displaystyle \frac{1}{2}$ chance to be odd with $\displaystyle \frac{1}{2n-1}$ probability that the next index point is max. Similarly, $\displaystyle \frac{1}{2}$ chance to be even and with $\displaystyle \frac{1}{2n-1}$ chance previous index point to be max.$

### A Variant

Same setup, but this time we don't see the numbers drawn. We are told that they happen to be in increasing order. What is the probability that the first number drawn was a 1?

### Solution 1 to the Variant

$\displaystyle\begin{align}&\frac{\text{# of ways to choose four other numbers when 1 has already been chosen}}{\text{# of ways to choose 5 numbers}}\\ &=\frac{{98\choose 4}}{{99\choose 5}}=\frac{5}{99}. \end{align}$

### Solution 2 to the Variant

Let $A$ be the event that a $1$ is drawn. Let $B$ be the event that the numbers came out in increasing order. I asked for $P(A|B).$ But $A$ and $B$ are independent. So, $P(A|B)=P(A).$ Five numbers are drawn; the probability that one of them is $1$ is $\displaystyle \frac{5}{99}.$

### 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.

### Geometric Probability

- Geometric Probabilities
- Are Most Triangles Obtuse?
- Barycentric Coordinates and Geometric Probability
- Bertrand's Paradox
- Birds On a Wire (Problem and Interactive Simulation)
- Buffon's Noodle Simulation
- Averaging Raindrops - an exercise in geometric probability
- Rectangle on a Chessboard: an Introduction
- Marking And Breaking Sticks
- Random Points on a Segment
- Semicircle Coverage
- Hemisphere Coverage
- Overlapping Random Intervals
- Random Intervals with One Dominant
- Points on a Square Grid
- Flat Probabilities on a Sphere

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