Best Position In Line For Marble Draw In a game you have $N$ players where player $N_i$ will play on turn $i$. On each turn the current player draws without replacement from a bag of marbles and will either win or lose depending on if they draw a winning marble. On the first turn this bag contains $N$ marbles of which $W$ are winning marbles.
I am looking for both an intuitive and mathematical explanation for which player it is best to be for a game given values of $N$ and $W$. My intuition tells me that if $N$ isn't significantly (say 10x) higher than $W$ and $W$ is not 1 then it's better to wait out at least $W$ turns. I'm not sure how to approach the question mathematically using combinatronics. I asked a friend of mine what he thought and he shares this intuition but is not sure. I also feel that this question is similar to the marriage problem.
 A: Let $N_i=1$ if $i^{th}$ player draws the winning ball and $0$ otherwise.
$P(N_1=1)= \frac{W}{N}$ ; $P(N_1=0) = \frac{N-W}{N}$
$\begin{align}P(N_2=1) &= P(N_2 = 1 | N_1 =1)P(N_1=1) +  P(N_2 = 1 | N_1  0)P(N_1=0) \\&= \frac{W-1}{N-1} \times \frac{W}{N} + \frac{W}{N-1} \times \frac{N-W}{N} \\&= \frac{W}{N}\end{align}$
$\begin{align}
P(N_3=1) &= P(N_3=1|N_1=1,N_2=0)P(N_1=1,N_2=0)
\\&+P(N_3=1|N_1=1,N_2=1)P(N_1=1,N_2=1) 
\\&+P(N_3=1|N_1=0,N_2=0)P(N_1=0,N_2=0) 
\\&+ P(N_3=1|N_1=0,N_2=1)P(N_1=0,N_2=1) 
\\&=\frac{1}{N(N-1)(N-2)}\times [W(N-1-(W-1))(W-1) 
\\&+ W(W-1)(W-2)
\\&+(N-W)(N-1-W)(W)
\\&+(N-W)W(W-1)]
\\&=\frac{W}{N}
\end{align}$
$\dots$
A: If there is $1$  winning marble ($W=1$), then all players are equally likely to win, since the winning marble is equally likely to be on any of the $N$ draws.
If there is more than $1$ winning marble ($W>1$), then it is best to go first.  
To see this, start by comparing just the first player's chance of winning to the second player's.  (Call these probabilities $P_1$ and $P_2$ respectively.)
We have $P_1=\frac{W}{N}$ and $P_2=\frac{N-W}{N}\cdot\frac{W}{N-1}$.  Thus $P_2=P_1\cdot\frac{N-W}{N-1}$.  But for $W\ge 2$ this extra factor is less than $1$, and so $P_1>P_2$.
By the same reasoning, reducing $N$ by $1$ and keeping $W$ the same, $P_2>P_3$ and so on.  
Of course after a while the probability drops to $0$ for the later players winning since if all the losing marbles are drawn by the first $N-W$ players, the next player must win.

Edit: As pointed out by rightskewed, as stated the game doesn't end upon the selection of the first winning ball.  Therefore if every player gets a ball, it doesn't matter what order a player picks in.  All players have a probability of $\frac{W}{N}$ of winning.
I think the problem is more interesting if the game ends upon the draw of the first winning ball.
Thanks @rightskewed for pointing out by mistake.
