As Ross rightly points out, the cuts are just to seemingly make it random though it doesn't affect anything. Irrespective of the cuts, note that there are $15$ cards between the first card the contestant places and the second card the contestant places. Similarly, irrespective of the cuts, note that there are $15$ cards between the second card the contestant places and the third card the contestant places.
Let us label the cards as follows. Let $a_k^{j}$ be the $k^{th}$ card from top in the hand of the performer after the $j^{th}$ up-down phase. Initially, i.e. after the contestant places the cards and before the first up-down starts, $j=0$.
Now the cards in the last pile i.e. the pile containing $9$ cards (the only pile on which the contestant doesn't place any card) be $a_1^{0}, a_2^{0}, \ldots, a_9^{0}$ starting from the top most card. Let the third card the contestant places on the third pile be $a_{10}^{0}$. Then there are $15$ cards followed by the second card the contestant places on the second pile. Accounting for the $15$ cards in between, the second card is $a_{26}^{0}$. Now there are $15$ cards followed by the first card the contestant places on the first pile. Accounting for the $15$ cards in between, the first card is $a_{42}^{0}$. Hence, now the contestant cards are $\color{red}{a_{10}^{0}, a_{26}^{0}, a_{42}^{0}}$.
Now the performer moves $4$ cards to the rear. Hence, now the contestant cards are $a_6^{0}, a_{22}^{0}$ and $a_{38}^{0}$.
$$\color{red}{\{a_{10}^{0}, a_{26}^{0}, a_{42}^{0}\} \to \{a_{6}^{0}, a_{22}^{0}, a_{38}^{0}\}}$$
Now in the first up-down phase all the odd number cards are eliminated i.e. $a_{2k-1}^{0}$ gets eliminated. However, on the pile with the cards closed, the order has reversed i.e. $a_2^{0}$ is the bottom most card, followed by $a_4^{0}$ and so on and the top-most card is $a_{52}^{0}$. Now reordering the card so that the topmost card is now $a_1^{1}$, we find that the card $a_{2k}^{0}$ gets mapped to $a_{27-k}^{1}$. Hence, the contestant cards are now at $a_{24}^{1}, a_{16}^{1}$ and $a_8^{1}$. $$\color{red}{\{a_{6}^{0}, a_{22}^{0}, a_{38}^{0}\} \to \{a_{24}^{1}, a_{16}^{1}, a_8^{1}\}}$$
There are now $26$ cards left.
Now in the second up-down phase all the odd number cards are eliminated i.e. $a_{2k-1}^{1}$ gets eliminated. As before, on the pile with the cards closed, the order has reversed i.e. $a_2^{1}$ is the bottom most card, followed by $a_4^{1}$ and so on and the top-most card is $a_{26}^{1}$. Now reordering the card so that the topmost card is now $a_1^{2}$, we find that the card $a_{2k}^{1}$ gets mapped to $a_{14-k}^{2}$. Hence, the contestant cards are now at $a_{2}^{2}, a_{6}^{2}$ and $a_{10}^{2}$.
$$\color{red}{\{a_{24}^{1}, a_{16}^{1}, a_8^{1}\} \to \{a_{2}^{2}, a_{6}^{2}, a_{10}^{2}\}}$$
There are now $13$ cards left.
Now in the third up-down phase all the odd number cards are eliminated i.e. $a_{2k-1}^{2}$ gets eliminated. As before, on the pile with the cards closed, the order has reversed i.e. $a_2^{2}$ is the bottom most card, followed by $a_4^{2}$ and so on and the top-most card is $a_{6}^{2}$. Now reordering the card so that the topmost card is now $a_1^{3}$, we find that the card $a_{2k}^{2}$ gets mapped to $a_{7-k}^{3}$. Hence, the contestant cards are now at $a_{6}^{3}, a_{4}^{3}$ and $a_{2}^{3}$.
$$\color{red}{\{a_{2}^{2}, a_{6}^{2}, a_{10}^{2}\} \to \{a_6^3,a_4^3, a_2^3\}}$$
There are now $6$ cards left.
Hence, the last up-down has the open cards as $a_1^{3}$, $a_3^{3}$ and $a_5^{3}$; the closed cards being $a_2^{3}$, $a_4^{3}$ and $a_6^{3}$, which are precisely the contestant cards.
EDIT
Below is an attempt to explain this pictorially. The document was created using $\LaTeX$ and below is a screenshot.
Let's skip to where all of your cards are on the table. You pick up all but three decks, and there's decks $S$, $T$, and $U$ on the table. If the top cards are $s,t,u$, then you have $(14*3 - s - t - u)$ cards on the table and $52 - (42 - s - t - u) = 10 + s + t + u$ cards in your hand. You have to put $(10 + s + t)$ cards on the table, leaving you with $(10 + s + t + u) - (10 + s + t) = u$ cards in your hand. So the number of cards you have will match the top of the turned-over deck.
Given this, here's a fun extension of the problem: why do you have to shuffle kings back in?
Best Answer
Suppose that, when you first lay the cards on the table, the card I choose is at position $x$ in its column. You don't know $x$, but you know that $1\leq x\leq 7$.
Now, when you pick up the cards, my card will be at position $7+x$ in the full stack. The second time you lay the cards on the table, my card will appear at position $p_1=\lceil\frac{7+x}{3}\rceil$ in its column.
The second time you pick up the cards, my card will be at position $7+p_1$ in the full stack. The third time you lay the cards on the table, my card will appear at position $p_2=\lceil\frac{7+p_1}{3}\rceil$ in its column.
Finally, when you pick up the cards for the third time, my card will be at position $7+p_2$ in the full stack. Putting this all together, my card will be at position $$7+p_2=7+\left\lceil\frac{7+\lceil\frac{7+x}{3}\rceil}{3}\right\rceil$$ in the full stack. The trick is that this is equal to $11$ for all $x$ in the range $1\leq x\leq 7$.
For a proof of this last statement, as jpmc26 mentions, one can apply the identities $\lceil\frac{m+\lceil x\rceil}{n}\rceil=\lceil\frac{m+x}{n}\rceil$ and $\lceil n+x\rceil = n+\lceil x\rceil$ (for real $x$, integer $m$, and positive integer $n$) to show that $$7+\left\lceil\frac{7+\lceil\frac{7+x}{3}\rceil}{3}\right\rceil = 7+\left\lceil\frac{7+\frac{7+x}{3}}{3}\right\rceil = 7 + \left\lceil 3 + \frac{x+1}{9}\right\rceil = 10 + \left\lceil\frac{x+1}{9}\right\rceil \enspace,$$ which is clearly equal to $11$ for $1\leq x\leq 7$.