Submitted by mrmonkey on Fri, 10/20/2017 - 15:54
Forums:
iii) You will not be able to separate the players if $$r>n-r+1,$$ $$2r>n+1,$$ $$r>\frac{1}{2} (n+1)$$ ? On the solutions it has $r>\frac{1}{2}(n-1)$ as the case instead so I was a bit confused, is this an error in the solutions or have I missed something?
Thanks a lot
You're correct. The non-zero
You're correct. The non-zero probability case is only valid for $r \leqslant \frac{1}{2}(n+1)$. I'm sure the mistake will be corrected soon, thanks for pointing it out.
[Incidentally, the blue text in part (ii) is also "wrong". You can write it in a nicer (subjective) way as $\displaystyle (n-r+1) {{n}\choose {n-r}}^{-1}$, which you might like to think about the alternative argument it suggests.
Thanks!
Will sort it today. Thanks for pointing it out!
Sorted!
Solutions have been updated, and I have tweaked the "blue comment" as well. Thanks both!