Markov chain magic trick (Kruskal Count) 
I am trying to understand why this works.
We have a Markov chain here with 10 states which are the numbers 1,..,10.
There is a none zero probability to pass between any two states, so we can show that the Markov chain is irreducible and aperiodic.
This means that with any starting distribution, the distribution after n steps will converge to the stationary distribution.
This doesn't mean that all runs will converge to the same state. For example, if the stationary distribution is Uniform, we will get that when simulating the run as described, you can get to any ending card with equal probability (at least that is what you are converging to).
So why are you likely to end up on the same card?
 A: As already mentioned, once you have shuffled the deck it is a deterministic graph. When two trajectories meet they will continue along the same path.
Now, having simulated the game on a computer I am not so sure that I would recommend it to show off among friends.
Given 52 cards taking values from 2 to 10 (with each face card  having the value 10), and if two players start independently at random among the first 10 cards the success rate for ending at the same card is about 62 %
If you eliminate the three royal cards and take ace to have the value 1 (so you have 40 cards taking values from 1 to 10) and if you pick as your initial choice the 1st card you increase the success rate to about 76 %
If you want better odds take two decks of cards from the start. Success rate becomes about 85 % with two decks of 52 cards, and 95 % in the case of two decks with 40 cards (with values 1 to 10).
An explanation of how to provide a rough estimate, using a kind of Markov approximation, for the convergence of two trajectories is given in the very nice article (link provided by @awkward, thanks!): The Kruskal Count. In that article they let the royal cards count as five and ace as 1, which yields a success rate of about 85 %.
