In this game, we use a standard shuffled deck of $52$ cards. The suits of each card do not play any part.
Take a card from the deck one by one. If the card's rank is equal to the number of cards you have taken you win. You may take up to $10$ cards in a single game.
To clarify the rules here are some winning game examples;
What are the odds of winning a single game?
To make things easier, always draw 10 cards regardless if we win or not. We use inclusion-exclusion principle over the events $X_1,X_2,\dots X_{10}$ where $X_i$ is the event that position $i$ has a card of rank $i$.
We are interested in $\Pr(X_1\cup X_2\cup \dots X_{10})$ which expands as:
$$\Pr(X_1)+\Pr(X_2)+\dots+\Pr(X_{10})-\Pr(X_1\cap X_2)-\dots -\Pr(X_9\cap X_{10})+\Pr(X_1\cap X_2\cap X_3)+\dots \pm \Pr(X_1\cap X_2\cap \dots \cap X_{10})$$
Assuming $10$ is the only rank of card said to have value of $10$ and Jacks, Queens, and Kings do not count as working... massive simplifications can be made in the above thanks to symmetry.
$$10\Pr(X_1)-\binom{10}{2}\Pr(X_1\cap X_2) +\binom{10}{3}\Pr(X_1\cap X_2\cap X_3)\pm\dots\pm \binom{10}{k}\Pr(X_1\cap \dots \cap X_k)\pm\dots$$
The probability of $\Pr(X_1\cap X_2\cap \dots\cap X_k)$ will be that an ace is in the first spot, a 2 is in the second spot, on up until a $k$ is in the $k$'th spot. This expands by multiplication principle as $\frac{4}{52}\times \frac{4}{51}\times \cdots \times \frac{4}{52-k+1}$, noting that in the above we only care about hitting or not hitting the desired cards in the specified slots and any additional slots are ignored and can have any result including additional matches, making calculations easy.
We get then a probability of winning as:
$$\sum\limits_{k=1}^{10}(-1)^{k+1}\binom{10}{k}\frac{4^k}{52\frac{k}{~}}\approx 0.5481536$$ wolfram calculations
