How many players needed for the game to have the highest probability of finishing the fastest? Welcome to the fictional game of "color-tag"; the not-so-fast-paced cousin of paintball Where marking your opponent is all that counts!
If $A$ marks $B$ with his/her color, then $B$ will be permanently marked with $A$'s color,
but at the same time all other people marked with $B$'s color
will instantly be permanently marked with $A$'s color as well.
The first one to have marked everyone wins, and the game is finished!
$m<x$ where $m = 16$, and the number of players, $x$, is put in a room and assigned an unique color.
With every hour that progress, all the players have a probability of $1 - \frac{m}{x}$ of successfully marking another player.
There is no limit as to how many times a player marks another player.
In the event that $A$ marks another player, the player, that is being marked, $B$, is chosen at random; Although players cannot mark themselves.
Now, I would like to know what the optimal number of players for the game having the highest probability of being the fastest.
That is, how many players are needed for the game to have the highest probability of finishing the fastest?
Sounds simple enough, right?
It turns out that it sounds simpler than it is; At least in my ears.
I've tried to warp my head around it, but I fail to find a viable approach.
I started out by calculating $(1 - \frac{16}{32} * \frac{1}{32})^{32}$, thinking that $(1 - \frac{m}{x} * \frac{1}{x})^x$
was the way to go about calculating the fastest possible game with respect to the number of players.
I soon realized that there was no way that was going to give me anything near correct results, so
I came up with something along the lines of $\sum_{n=0}^{x-1} ((1 - \frac{m}{x}) * \frac{x-n}{x})^n$, which I hoped would get me somewhere;
But all it did was feed me an absurdly small number and make me realize that I've never been faced with a problem like this, and that I have no idea of how to solve it.
Right now, I'm not even considering that I must find the cases with the highest probabilities of success, then cross check to see which is he fastest;
Which I do think is paramount in this problem, alas I know of no way to approach this.
Any input (especially tag additions) is greatly appreciated!

EDIT 1:
Just to make sure there is no confusion:
If a player $A$ and another player $B$ both get to mark an opponent (opponents being all the other players) within a hour, then there is a chance, and it is allowed that $A$ marks $B$, and $B$ marks $A$.
Marking another player happens instantaneously and simultaneously.
 A: In short, you want to minimize the players in the game to expect the game to finish as quickly as possible. We could interpret the original question literally - 'Now, I would like to know what the optimal number of players for the game having the highest probability of being the fastest' - then we are simply looking for the number of players to finish the game in as few steps as possible. The shortest time possible a game could last only goes up as more players are added to the game, as it takes more steps for a player's color to spread to all the players in the game.
If on the other we're not trying to be the grammar police then we are trying to minimize the expected time it takes to finish the game. Representing the problem mathematically is very difficult: you have a dynamic Markov process in which the state of every player's color influences all future states of the game. I have not been able to find any probability distributions that fit the game; the closest I have been able to find is the Dirichlet-multinomial.
The answer is still that you want to minimize the number of players in the game to minimize the expected time to finish. With x players in the game, any given player needs to spread his color to x-1 separate players. For a player to win the game, every single player needs to be hit. As the number of players in the game increases, the chances that a specific player is hit in a given round decreases as x goes up. With two players in the game, the probability a player has of being hit is 1 - $\frac{m}{x}$. As x increases, this probability converges to $\frac{e-1}{e}$ or 0.63 for each player (granted I am suspicious of 'm' missing from the result. Also, resembles hat check problem). 
And for some more math:
The probability a player does not hit in a given hour is $\frac{m}{x}$.
Let dt,c be the number of players at time t with a particular color c. The probability none of the players with color c are hit by other players in a given round is represented by $ (\frac{m}{x} + (1 - \frac{m}{x}) * \frac{x-1-d_{t,c}}{x-1} )^{x-d_{t,c}} $. 
dt,c is an unknown discrete probability distribution for each color.
I say unknown because attempting to represent the distribution as a probability could result in the universe imploding.
I believe the best way to prove the minimum time needed to finish needed to finish the game would be using mathematical induction. To do so you'd likely need to simplify the problem first, by demonstrating that a particular outcome is less than/greater than another outcome. And on that, good luck.
A: Based on comments to bjorn's answer, there seems to be some confusion about the question. The way I understood it was that each person has an assigned color at any particular time. If player B shoots player A, then A changes her color to B's color. If player C then shoots player B, then players A and B both change to C's color. We want to know how long it takes for everyone to have the same color.
Let $G$ be the graph whose vertices are players and where the vertices $i$ and $j$ are joined by an edge if $i$ has shot $j$. Then two players have the same color exactly if they are in the same connected component of $G$.
Let's first look at the easier question where taggings happen completely at random: Every after every $\frac{\mbox{1 hour}}{(1-m/x)x}$, some player tags another player at random. So, after $h$ hours, the expected number of edges is $h(x-m)$, and any set of $h(x-m)$ edges is as likely as any other. 
According to a theorem of Erdos and Renyi, if $h(x-m)$ is much more than $(1/2) x \log x$, the graph is extremely likely to be connected and, if $h(x-m)$ is much less than $(1/2) x \log x$, the graph is extremely unlikely to be connected. More precisely: Fix a constant $c$. Consider graphs with $x$ vertices and $(1/2) x \log x + cx$ edges, as $x \to \infty$. The probability that such a graph is connected is $\exp(- \exp(- 2c))$: extremely close to $0$ for $c$ negative and extremely close to $1$ for $c$ positive.
So I would expect connectivity after about $(1/2) \log x$ hours. The constant $m$ only produces second order effects.
Your situation is a bit different, because one shooter cannot tag two people in the same hour. I would predict the final answer is the same. You might try adapting my answer here to your setting.
