More generally, suppose that there are $n$ cards total in the deck, where that $k$ of the cards are considered to be good, while the remaining $n-k$ cards are bad. The goal is the same; the deck is shuffled, and as the cards are flipped from the deck, you have the option to call "stop" at any time. You win as long as the next card dealt is good. What is the optimal strategy?
Furthermore, let us modify the rules to say that, when there is only one card left, you must call stop. A player who does not call stop before the last card automatically loses, so this modification makes no difference to an optimal player.
It turns out that every strategy is the optimal strategy.
Theorem: Every strategy for this game has a probability of winning of $k/n$.
Therefore, to answer your question, the probability of winning is the number of primes in $\{1,\dots,N\}$, divided by $N$.
Proof 1: Clever insight
Consider this modification of the game. Suppose that, when you call stop, instead of winning when the next card dealt is good, you win if the bottom card of the deck is good. This is a silly game, because it does not matter when you call "stop." You will always win if and only if the bottom card is good. For the modified game, the probability of winning is $k/n$.
However, this game has the exact same probability for winning as the original. The point is that, after calling "stop", both the next card and the last cards have the same same random chance of being good, because of the symmetry of the shuffled deck. Therefore, it follows the original game has the same probability of winning.
Proof 2: Induction
We prove, by induction on $n$, that for all $k\in \{0,1,\dots,n\}$, the probability of winning is $k/n$ for all strategies in an $n$-card deck with $k$ good cards.
The base case of $n=1$ is obvious. Assuming that the statement is true for $n$, where $n\ge 1$, we prove it for $n+1$ as follows:
$$
\begin{align}
P(\text{win})
&=P(\text{win}\mid \text{first card is good})P(\text{first card good})
\\&\quad +P(\text{win}\mid \text{first card is bad})P(\text{first card bad})
\\&=\frac{k-1}{n-1}\cdot \frac{k}{n}+\frac{k}{n-1}\cdot \frac{n-k}{n}=\frac{k}n.
\end{align}$$
This completes the proof by induction.
Proof 3: Martingale
For each $k\in \{0,1,\dots,n-1\}$, let $X_k$ be the proportion of good cards among the unseen cards after $k$ cards have been seen. So, $X_0=k/n$, and $X_1$ is either equal to $k/(n-1)$ or $(k-1)/(n-1)$.
Imagine the player is following a particular strategy. Let $T$ be the turn number that the player calls stop, under that strategy. The proportion of good cards at the time of the stop call is $X_T$, so $X_T$ is the probability the player wins (conditional on $T$). We just need to evaluate $\mathbb E[X_T]$, which is the unconditional probability of the strategy winning.
The sequence $X_0,X_1,\dots,X_{n-1}$ is a martingale, with respect to the natural filtration. To prove this, we must show that $\mathbb E[X_{k+1}\mid X_k]=X_k$ for all $k$. Indeed, given $X_k$, there are two possibilities; either you next draw a good card with probability $X_k$, in which case $X_{k+1}$ decreases to $\frac{(n-k)X_k-1}{n-k-1}$, or you draw a bad card with probability $(1-X_k)$, in which case $X_{k+1}$ increase to $\frac{(n-k)X_k}{n-k-1}$. Therefore,
$$
\mathbb E[X_{k+1}\mid X_k]= X_k\cdot \frac{(n-k)X_k-1}{n-k-1}+\left(1-{X_k}\right)\frac{(n-k)X_k}{n-k-1}={X_k}.
$$
Since the sequence is a martingale, and since $T$ is bounded, we can use the optional stopping theorem to conclude that $\mathbb E[X_T]=\mathbb E[X_0]=k/n$. Therefore, every strategy has a probability of success of $k/n$.
