The total of the player placing 2 was? 
8 players compete in a tournament,every one plays everyone else just
  once.The winner of the game gets 1,the loser 0 or each gets
  $\frac{1}{2}$ if draw.The final result is that every one gets a
  different score and the player placing second gets the same score as
  the total of the four bottom players.
"The total of the player placing 2 was?"

I can do it if the case where a draw can happen is neglected.But I'm stuck with this one...Help!
 A: Let $s_1<s_2\dots < s_8$ be the $8$ scores of players $1,2,3\dots 8$.
Notice $s_7=s_1+s_2+s_3+s_4\geq\binom{4}{2}=6$ (because there are $6$ games between players $1,2,3,4$)
$s_7=6$ is possible, it happens when player $i$ beats player $j$ if $i>j$.
The maximum score for a person is $7$, notice $s_7=6.5$ is impossible, because $s_8$ would have to be $7$ and these scores are mutually exclusive, $s_7=7$ is also clearly impossible.
A: Let $G$ be a directed graph with vertex set $[8]=\{1,\ldots,8\}$ representing the players and an edge $\langle k,\ell\rangle$ if and only if $k$ beat $\ell$. For each $k\in[8]$ the score, $s_k$, of $k$ is 
$$s_k=\operatorname{out-deg}k+\frac12(7-\operatorname{in-deg}k-\operatorname{out-deg}k)=\frac72+\frac12(\operatorname{out-deg}k-\operatorname{in-deg}k)\;.$$
Let $d_k=\operatorname{out-deg}k-\operatorname{in-deg}k$, so that $s_k=\frac12(7+d_k)$; $d_k$ is simply the number of games won by $k$ minus the number lost. Clearly each $d_k$ is an integer in the range $[-7,7]$, $d_1+\ldots+d_8=0$, and $d_1<\ldots<d_8$. Moreover, $s_k$ is an integer, so $d_k$ must be odd. There are only $8$ odd integers in the range $[-7,7]$, so $\{d_1,\ldots,d_8\}=\{-7,-5,-3,-1,1,3,5,7\}$. In particular, $d_7=5$, and $s_7=\frac12(7+5)=6$.
You can do without the graph theory by letting $w_k$ be the number of games won by $k$, $\ell_k$ the number lost, and $d_k=w_k-\ell_k$. Then
$$s_k=w_k+\frac12(7-\ell_k-w_k)=\frac72+\frac12(w_k-\ell_k)=\frac12(7+d_k)\;.$$
Clearly 
$$28=\binom82=s_1+\ldots+s_8=8\cdot\frac72+\frac12(d_1+\ldots+d_8)=28+\frac12(d_1+\ldots+d_8)\;,$$
so $d_1+\ldots+d_8=0$, and the argument concludes as before.
A: A very simple answer is possible if you assume the question is well posed and has a unique answer. As you imply, if there are no draws and everybody gets a different score, the second ranked player gets a score of $6$, which equals the sum of the bottom four players at $0,1,2,3$.  As this result meets the requirement, if there is a unique result it is $6$.  We have not proven that this is the only result possible.
