Algorithm for modified knapsack problem We have n-players in a game. We have a population of players we can choose from. Each player score is a normally distributed random variable and each player has a cost to add to the team. We are limited to n-players and the total cost must be less that or equal to m.
The total score for a team is calculated as the sum of all of the individual player's score on that team. Our goal is to maximize the team score while staying under our allowed player weight and number of players.
Now, suppose that in this game we drop the score of the player that has the lowest score on the team and do not consider it in the total team score. 
Does a deterministic algorithm exist that will allow us to reach the optimal solution?
 A: Each player in a population of $N$ players is denoted by the triple $(\mu_k, \sigma_k, c_k)$ which represents the mean score, standard deviation of score and cost respectively. The score of a player is a normally distributed random variable,
$$X_k \sim N(\mu_k, \sigma_k^2)$$
A team can have $n<N$ players and the total score of a team is
$$S = \sum_{i=1}^n X_i - X_{(1)}$$
where $X_{(1)} = \min(X_1, X_2, \cdots X_n)$. The goal is to produce an algorithm which efficiently computes the $n$ players which should be chosen to maximize $E(S)$ subject to the constraint $\sum_{i=1}^n c_i \leq M$. 

Calculating Expected Score
Let $\phi(x)$ denote the standard normal density function:
$$\phi(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}$$
and let $\Phi(x)$ denote the standard normal distribution function:
$$\Phi(x) = \int_{-\infty}^x\phi(t)dt$$
The expected score, by linearity of expectation, is given by:
$$E(S) = \sum_{i=1}^n \mu_i - E(X_{(1)})$$
So that the problem boils down to finding the expected value of the minimum order statistic. We can start by finding the distribution function of $X_{(1)}$. This is a standard calculation and details can be found in many places. 
$$F_1(x) = 1 - \prod_{i=1}^n\left[1-\Phi\left(\frac{x-\mu_i}{\sigma_i}\right)\right]$$
The density function can be found by taking the derivative. Using the product rule of derivatives, we obtain:
$$f_1(x) = \sum_{i=1}^n\left\{\frac{1}{\sigma_i}\phi\left(\frac{x-\mu_i}{\sigma_i}\right)\prod_{j\neq i}\left[1-\Phi\left(\frac{x-\mu_j}{\sigma_j}\right)\right]\right\}$$
Now we find the expected value of $X_{(1)}$.
\begin{align*}
E(X_{(1)}) &= \int_{-\infty}^\infty x \cdot \sum_{i=1}^n\left\{\frac{1}{\sigma_i}\phi\left(\frac{x-\mu_i}{\sigma_i}\right)\prod_{j\neq i}\left[1-\Phi\left(\frac{x-\mu_j}{\sigma_j}\right)\right]\right\} dx \\
&= \sum_{i=1}^n\int_{-\infty}^\infty g_i(x)dx
\end{align*}
where $$g_i(x) = \frac{x}{\sigma_i}\phi\left(\frac{x-\mu_i}{\sigma_i}\right)\prod_{j\neq i}\left[1-\Phi\left(\frac{x-\mu_j}{\sigma_j}\right)\right]$$
I find it very unlikely that an analytic solution to this integral exists, but these integrals can be approximated numerically to high precision in a deterministic way. Here is some R code which calculates $E(S)$. 
g_i <- function(x, i, mu, sigma){
  res <- rep(NA, length(x))
  for(k in 1:length(x)){
    res[k] <- x[k]/sigma[i]*prod(1-pnorm((x[k]-mu[-i])/sigma[-i])*dnorm((x[k]-mu[i])/sigma[i])
  }
  return(res)
}

expected_score <- function(mu, sigma){
  n <- length(mu)
  res <- sum(mu)
  for(i in 1:n){
    I_i <- integrate(g_i, lower=-Inf, upper=Inf, i=i, mu=mu, sigma=sigma)$value
    res <- res - I_i
  }
  return(res)
}

Note that this implementation takes approx a half second on my machine to compute $E(S)$ for $n=30$. You may want to look into the numerical integration and see if it can be sped up. 

Optimization with Dynamic Programming
Now that we have a method of computing $E(S)$, we can try to solve this optimization problem with Dynamic Programming. Note that your restrictions make this a challenging problem. I will try to give a general path forward, but you will need to fill in many of the details for implementation.
Let $E(i, j, m)$ denote 

the max expected score for a team of $j\leq n$ players selected from the first $i \leq N$ available players with a budget of $m\leq M$.

So the total max expected score is given by $E(N, n, M)$. We define $$E(i,j,m) = 0$$ whenever


*

*$j=0$ (no players on team)

*$i<j$ ("forfeit condition", not enough available players to build a team)

*$m<\min(c_1, c_2, \cdots c_i)$ (Budget not sufficient for any players)


You should also explicitly account for the situation where a single players cost is above the budget.
$$E(i,j,m) = E(i-1,j,m), \ c_i > M$$
Now we can construct the general recursive relation.
\begin{align*}
E(i,j,m) = \max\{&E(i-1, j, m), \\
&E(i-1, j-1, m-c_i) + \mu_i + [L(i-1, j-1, m-c_i) - L_\star] \}
\end{align*}
The first and second lines above represent the cases without selecting player $i$ and with selecting player $i$ respectively. In the second case, we are adding in the players mean score $(\mu_i)$, but we must also account for this players contribution to $E(X_{(1)})$. Thus $L(i,j,m)$ represents the loss ($E(X_{(1)})$) given the same interpretation of $i$, $j$ and $m$. The $L_\star$ refers to the expected loss given the optimal set of players corresponding to $E(i-1, j-1, m-c_i)$ in addition to player $i$. 
When implementing this, you will need to keep track of the optimal player set corresponding to each $E(i,j,m)$ and use it to efficiently compute the appropriate $E(X_{(1)})$. This is a very difficult DP problem, and computation of the integrals numerically will make it costlier than the standard knapsack problem, but a partial DP algorithm exists and will be far more efficient than a brute force method. Good luck!
