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.
Your approach won't work because:
Let's suppose that the descks have $13-a$, $13-b$, $13-c$, $13-d$ cards respectively. You can assign the rest of the cards in $\frac{(a+b+c+d)!}{a!b!c!d!}$ ways. If you assign the card to the first deck, you can assign the rest in $\frac{(a+b+c+d-1)!}{(a-1)!b!c!d!}$ ways. For the second deck, it is $\frac{(a+b+c+d-1)!}{a!(b-1)!c!d!}$. $\frac{(a+b+c+d-1)!}{a!b!(c-1)!d!}$ for third deck and $\frac{(a+b+c+d-1)!}{a!b!c!(d-1)!}$ for fourth deck. Therefore, you can't assign the next card with equal probabilities. You can easily figure out that the ratio of the probabilities has to be $a:b:c:d$. The easiest way to choose one deck with these probabilities is choosing a random integer from $1$ to $a+b+c+d$ and then look in which deck it falls. The upper bound for your random generator starts at 52 and drops by 1 each time you assign a card to a deck. Now the bits of entropy is: $\sum_{i=1}^{52}log_2(i) = log_2(52!)$. So if you fix your algorithm this way, you are back at the naive generator.
You can do it this way:
- generate a random number $k$ between $1$ and $\frac{52!}{13!13!13!13!}$.
- set $a,b,c,d:=13$, $u:=\frac{52!}{13!13!13!13!}$
- if all cards are assigned to a deck, end.
- set $g:=\frac{u}{a+b+c+d}$
- if $k \leq ga$ put card to first deck, set $a:=a-1$, $u:=ga$, return to step 3
- else if $k \leq g(a+b)$ put card to second deck, set $b:=b-1$, $u:=gb$, $k:=k-ga$, return to step 3
- else if $k \leq g(a+b+c)$ put card to third deck, set $c:=c-1$, $u:=gc$, $k:=k-g(a+b)$, return to step 3
- else put card to fourth deck, set $d:=d-1$, $u:=gd$, $k:=k-g(a+b+c)$, return to step 3
It is basically the same with the exception that you don't generate a new number in each step, but you decide according to a number which you generate at the beginning. It is basically an unrank function for permutations with repetitions.
Best Answer
Importantly, note that a card cannot decrease in height. If $\tau_i$ is the hitting time for the card $i^{\text{th}}$ from the top to reach the top, then you want $\mathbb E(\tau_{51})$. Use $\tau_0\equiv 0$ and the recurrence $$\mathbb E(\tau_i)=1+\frac{i}{52}\mathbb E(\tau_{i-1})+\frac{52-i}{52}\mathbb E(\tau_i)$$ to represent the two possibilites on each round. This simplifies to $\mathbb E(\tau_i)=\frac{52}{i}+\mathbb E(\tau_{i-1})$, which can be solved with a summation.
A nice side effect of formalizing this approach is that the follow-up question is simply asking for $\mathbb E(\tau_I\mid I)$, where $I$ is discrete uniform over $\{0,\ldots, 51\}$. More simply, this is $\frac{1}{52}\sum_{i=0}^{51}\mathbb E(\tau_i)$.