When given a set of distinct numbers, the probability of choosing a number greater than the median is 0.5 correct? Is there a condition that the probability would not equal 0.5?

  • $\begingroup$ The probability need not be $0.5$ : If we have the numbers $1,3,5$, the median is $3$. The probability to choose a number greater than $3$ is $\frac{1}{3}$. Your statement gets correct, if we know that we have an even number of numbers. $\endgroup$ – Peter Nov 30 '18 at 22:49
  • $\begingroup$ @Peter. Or if you have an even number but several are equal to the median... $\endgroup$ – fleablood Nov 30 '18 at 22:54
  • $\begingroup$ @fleablood Can this happen if we have distinct numbers ? $\endgroup$ – Peter Nov 30 '18 at 22:56
  • $\begingroup$ If you have distinct numbers, no, and ... oh, I missed that that was a requirment. If there are $2n$ distinct elements the prob is $\frac 12$. and if there are $2n + 1$ distinct elements the prob is $\frac n{2n+1}$. If the elements are not distinct then ... it's anyones game. $\endgroup$ – fleablood Nov 30 '18 at 23:12

The median might have a non-zero probability of being chosen; indeed, this is so whenever the sample contains an odd number of values. In that case, exceeding and being less than the median can't both have probability 0.5, and with an odd-sized sample neither would.

  • $\begingroup$ It's also true for even numbers and some numbers are equal to the median. Say. ${1,5,5,5,7,8}$ Median is $5$. Pro of less is 1/6. Prob of more is 1/3. Prob of exactly is 1/2. $\endgroup$ – fleablood Nov 30 '18 at 22:57
  • $\begingroup$ @fleablood On the other hand, no value is equal to the median in $\{1,\,2,\,3,\,4\}$. $\endgroup$ – J.G. Nov 30 '18 at 22:58
  • $\begingroup$ Well, yes, but that's not a counterexample. If any value equals the median the prob is less than 1/2 and just what it is could be anything. If no value equals the median then the prob is 1/2. For no value to equal the median it must be that there are even terms but it is not sufficient that there are even terms. $\endgroup$ – fleablood Nov 30 '18 at 23:09
  • $\begingroup$ Okay, .... I missed the requirement that the elements are distinct.... so in that case the prob is $\frac {\lfloor \frac n2\rfloor}n$ which is $\frac 12$ iff and only if $n$ is even. You're right. $\endgroup$ – fleablood Nov 30 '18 at 23:15
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    $\begingroup$ @fleablood My point was only that some even-sized sets contain their median and some don't. And in my answer, I focused on odd-sized sets because they all contain their median. $\endgroup$ – J.G. Nov 30 '18 at 23:30

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