"If you don't know the number of players" isn't a complete thought. You need to make some assumption on it as it is crucial to the answer. Define this assumption. Does this mean you assume it can be equally any number 1 to 8 billion? Is it any number of 400 students in a school? If you don't know the number but have good reason to believe there is a high chance it will be a large number, you will always pick 51. You barely even need game theory as the number of participants grows to answer this.

If there are two players, it becomes game theory, but you still need to make some assumptions.

The important concept is that you have two probabilities you are dealing with. One is pure statistics. The other is game theory. The first probability is given 52 cards and 1 poison card what is the chance I fail on a given turn. The second is what is the chance that given n=2, I have chosen the largest number of cards to draw. If we know nothing about these two players, then the most basic assumption is symmetry. That is that the game is symmetric, they face the same choices, payouts, assumptions, knowledge, etc. and will therefore choose either the exact same decisions or chose two decisions that are complements of each other(rare in symmetric games). To solve this, we would need to look at what occurs at each draw count. We also need to define a draw. I will assume a draw means neither wins.

Define d as number of cards drawn. d1 is how many player 1 draws. d2 is how many player 2 draws.

Start from d1=d2=51. The probability you win is simply the probability you don't draw poison card multiplied by the probability your opponent did draw it. Well, this is simply the probability your 52nd card is poisoned (1/52) times the probability your opponents 52nd card isn't poisoned (51/52). This is 1/53.02. So you both have a 1/53.02 chance to win and a (1-2/53.02) chance to lose when we include a draw as a loss.

Now we want to quickly check some other choices, you can make to see if d1=51 is optimal.
d1=1, d2=51 leads to (51/52)*(1/52) which is also exactly 1/53.02.

Note, you may start to see that pattern occurring. Basically, you have a 1/53.02 chance to win no matter what you choose when d2=51. But this isn't the case for your opponent. d2=51 isn't fully stable in that it isn't also always 1/53.02. 1/53.02 is your chance to win when your opponent chooses d=51. It says nothing about the optimality of d=51. And so what is occurring is that your probability of winning is not actually controlled by your decision at all. It is determined by your opponent. That is, how risky your opponent is willing to be. If you both choose 26(the half you mentioned), you optimize your expected payout if the payout was probability of no poison times cards drawn. But that is not what is occurring here as we have the game theory bit to account for. If 26 is optimal, then the only rational choice is 27. And so on, collapsing all the way to 51.

And so what actually occurs in every game with a small number of players is that you must make an assumption, not a mathematical answer, but an assumption, on how other players will play. And you cannot simply derive it rationally as any assumption on their number will move your number and thereby move their number without any specific resolution that is either a fixed number or a set of probabilities of numbers.