Two numbers $r$ and $s$ are drawn one at a time, without replacement from the set $\{1,2,...,n\}$ Find $P(r\leq p/s\leq p)$ where $p$ belongs to the set

  • $\begingroup$ $p/s \le p$ seems unnecessary. Check the source to make sure you stated the problem correctly. $\endgroup$ – quasi Sep 18 '17 at 18:09
  • $\begingroup$ I agree it seems unnecessary but that's how the question is. The question can still be solved $\endgroup$ – Ava Sep 18 '17 at 18:12
  • $\begingroup$ Is the answer supposed to be expressed in terms of the unknown $p$? $\endgroup$ – quasi Sep 18 '17 at 18:13
  • $\begingroup$ I don't have the answer sorry. $\endgroup$ – Ava Sep 18 '17 at 18:14
  • $\begingroup$ But I think that since p can be anything it should in terms of that $\endgroup$ – Ava Sep 18 '17 at 18:15

Fix an integer $n>1$.

If $r,s$ are drawn randomly, one at a time, without replacement, from $\{1,...,n\}$, then for fixed $p \in \{1,...,n\}$, the probability that $r \le {\large{\frac{p}{s}}} \le p\;$can be expressed as $$ \frac { {\displaystyle{ \left( \sum_{r=1}^p \left\lfloor{\frac{p}{r}}\right\rfloor \right) -\lfloor{\sqrt{p}}\rfloor }} } { \phantom{\large{X^X}}{n(n-1)}\phantom{\large{X^X}} } $$

  • $\begingroup$ How!I don't understand how you got P^1/2 $\endgroup$ – Ava Sep 18 '17 at 19:08
  • $\begingroup$ You need to subtract off the counts for the pairs $(r,r)$ where $r^2 \le p$. $\endgroup$ – quasi Sep 18 '17 at 19:14
  • $\begingroup$ Test the formula for small values of $n$ and $p$. $\endgroup$ – quasi Sep 18 '17 at 19:16

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