Dividing amount unequally in tournament winners? Hi I am trying to write the algorithm to calculate the winning amount of the winners. My problem is as below,
1) I have open ended tournament in which participants can be in ratio of 2 (i.e 2,4,8,16 and so on)
2) I am going to declare 50% or 25% of them as winners for each respective position 
Ex.- If i have 32 participants then in case of 50% winners my winners are as below,
 Position    Number Of winners
   1st            1
   2nd            1
   3rd            2
   4th            4 
   5th            8

and if i have 25% of winners then winners will be till 4th position
3) Now i want to  distribute the winning amount in them unequally as winner of 1st position will get more and then 2nd is little less than 1st and 3rd is little less than 2nd and so on 
4) The multiple number of winners on position 3rd on words will get the same amount 
Ex. If amount to distribute is $800 then in 8 winners then the rough distribution for this will be like as below,
  Position     Number of Winners       Amount for each Winner
     1                 1                      300
     2                 1                      150
     3                 2                       75(Each)
     4                 4                       50 (Each)
                             -------------------------------
                              Total           800

(this is rough calculation we need formula or directions which will help us or guide to how to get the winning amount)
We are trying various methods but we are weak in math so unable to find any formula any help would be great.we are not expecting the exact formula (if have then great) but any direction   would be great. Thanks in advance. 
Update
I tried this formula
Divide 800 in 8
(X + 4) + (X + 3) + (X + 2) + (X + 1) + (X - 1) + (X - 2) + (X - 3) + (X - 4)
In above case X = 800/8 =100
it will give 
104 + 103 + 102 + 101 + 99 + 98 + 97 + 96 = 800.
But now the problem is I have 2 players on 3rd position and 4 players on 4th position, and players on same position must have same prize.
 A: As tony noted, there are more than one solution to this problem. Let me restate the problem the way I understand it:
You have $2^n$ players. You would like to split the prize money M, between $2^n/2=2^{n-1}$ (in the $50\%$ case) or $2^n/4=2^{n-2}$ (in the $25\%$ case) winners.
With the ecxeption of the 1st position, the number of people on the k-th position is  $2^{k-2}$. For the $50\%$ case, the highest $k$ you want to use is $k=n$, for $25\%$ case $k=n-3$. Lets focus on the $50\%$ case from now on
What you are trying to do is find winnings $m_i$ for each position such that $M=m_1+m_2+m_32^{3-2}+m_42^{4-2}+\ldots+m_n2^{n-2}=m_1+m_2+2(m_3+m_42^{4-3}+\ldots+m_n2^{n-3})=m_1+m_2+2(m_3+2(m_4+\ldots+m_n2^{n-4}))$. 
This gives us an algorithm for splitting the prize money M (800 in your example):


*

*Choose prize money for the first position, $m_1<M$ and subtract it from $M$, $M=M-m_1$. In your example, you chose $m_1=300$, $M=500$.

*Chose prize money for the the second position $m_2<M$ and subtract it from $M$, $M=M-m_2$. In your example, you chose $m_1=150$, $M=350$.

*Divide M by two, $M=M/2$, choose prize for k-th position, $m_k<M$ and subtract it from M, $M=M-m_k$. 
In your example, for $k=3$,you get $M=350/2=175, m_3=75 \text{ and } M = 175-75 = 100$

*Repeat 3. until $k=n$ or until you run out of money. 


As you can see, you can choose any amount for any position as long as that amount is smaller than the amount of money you have left. You can pick many different strategies for splitting the prize money, for example, people in each position will together get half or quarter the money of the position above them. 
