Expected maximum value for a specific distribution, if the trials are dependent I have a Poisson based distribution as follows:
$P(1)=0.1708$;
$P(2)=0.138$;
$P(3)=0.092$; 
...
...
$P(10)=0.000034$; 
I pick numbers between $1$ and $10$ according to this distribution but if a number is previously picked, I re-iterate the routine to find a non-picked number in this interval.
In the beginning, I suppose none of the numbers are picked.
I have to pick $5$ numbers among these $10$ numbers.
How can I find the maximum expected number that is picked? 
Thanks...
 A: According to @JaroslawMatlak's suggestion, I wrote the following Matlab code and found the probabilites of picking maximum numbers as following:
X=(combnk(1:10,5));  % Finds all the 5-combinatons of 10 numbers
Y=sortrows(X,5);  % sorts them according to the last column-the geratest one.
P(1)=0.1708;
P(2)=0.138;
P(3)=0.092;
P(4)=0.0509;
P(5)=0.0234;
P(6)=0.009;
P(7)=0.0029;
P(8)=0.00078;
P(9)=0.00018;
P(10)=0.000034;
Prob=zeros(10,1);

for i=5:10    % Start from 5 because the minimum max-number is 5.
    ind=Y(:,5)==i;   % Find the indices that are equal to i.
    A = Y(ind,:);  % Copy those rows to array A
    for j=1:size(A,1)
        Prob(i)=Prob(i)+P(A(j,1))*P(A(j,2))*P(A(j,3))*P(A(j,4))*P(A(j,5));
    end
end

And I found the probabilities for the numbers to be the maximum one as follows:
P(5)=2,58E-06;
P(6)=2,01E-06; 
P(7)=8,17E-07; 
P(8)=2,36E-07; 
P(9)=5,55E-08; 
P(10)=1,05E-08;

Running the last equation (averaging on x.p(x)) I find 5.81 as the expected value of the maximum number picked. I normalized the probebilities for the numbers >=5 so that their sum is 1. Please comment on the reliability of the result.
A: Because my previous answer was wrong, I will use method suggested by @Henry - bruteforce checking all possible pick schemes:
Note, that 
$$P(a,b,c,d,e)=\frac{P(a)\cdot P(b)\cdot P(c)\cdot P(d)\cdot P(e)}{Q(a)\cdot Q(a,b)\cdot Q(a,b,c)\cdot Q(a,b,c,d)}$$
where $$Q(a,b,...)=1-P(a)-P(b)-...$$
Let $R(x)$ be a function describing the probability, that the greatest drawn ball is $x$. We have then
$R(x)=\sum\limits_{a,b,c,d < x,\\ a\neq b, a\neq c, a\neq d,\\
b\neq c, b\neq d, c\neq d} P(a,b,c,d,x)$
The expected value $E$ is then equal to:
$$E=\sum_{n=5}^{10}nR(n)$$
My python script:
P=[0]*11
R=[0]*11

P[1]=0.1708
P[2]=0.138
P[3]=0.092
P[4]=0.0509
P[5]=0.0234
P[6]=0.009
P[7]=0.0029
P[8]=0.00078
P[9]=0.00018
P[10]=0.000034

# Probabilities don't sum up to 1, so we will normalize them
s=sum(P)
for i in xrange(1,11):
  P[i]=P[i]/s


for i in xrange(1,11):
  for j in xrange(1,11):
    if j != i:
      for k in xrange(1,11):
        if k!=i and k!=j:
          for l in xrange(1,11):
            if l!= i and l!=j and l!=k:
              for m in xrange(1,11):
                if m!= i and m!=j and m!=k and m!=l:
                  maxSelected=max(i,j,k,l,m)
                  r = P[i]*P[j]*P[k]*P[l]*P[m]/((1-P[i])*(1-P[i]-P[j])*(1-P[i]-P[j]-P[k])*(1-P[i]-P[j]-P[k]-P[l]))
                  R[maxSelected]+=r

E=0
for i in xrange(1,11):
  E+=i*R[i]
print 'E = '+E

Out:
E = 5.60641877293
