Probability of getting the same 10 cards after a 104 card shuffle Consider this situation:
You have two standard decks merged together (104 cards). You give 10 cards to 4 players. What is the probability that the next time you deal (after the deck is randomly shuffled), everyone gets the same hand they got last turn? Just one person gets the same hand they got last turn?
My approach was something along 104 choose 40 (for everyone getting the same card) or 104 choose 10 (for one person getting the same card), but in my approach I'm not considering the duplicate cards in my deck. I don't think it's 52 choose 40 or 52 choose 10 because then I wouldn't be considering that there is more than one card.
I am very curious about this question and I would love to get an answer. I know its a combinatorics question, but I am unsure of the details.
P.S. If the question is confusing or non-descriptive, please ask for clarification instead of down voting.
This is not a complete answer, but provides some suggestions on how to proceed.

Merge two $n$ card decks, and deal one hand of $k$ cards.  The merged deck has $2n$ cards, but only $n$ kinds of card.  A hand can contain the same kind of card twice.  When a card appears in a hand along with the other card of the same kind, we call it a double.  Otherwise, we call it a single.

There are $\binom{2n}{k}$ ways of dealing a $k$ card hand.  There are $\binom{n}{d}\binom{n-d}{k-2d}$ kinds of hand with $d$ doubles and $k-2d$ singles.  There are $2^{k-2d}$ different ways of dealing a given $k$ card hand with exactly $d$ doubles, giving a probability of $2^{k-2d}\big/\binom{2n}{k}$ of getting such a hand.  There are $2^{k-2d}\binom{n}{d}\binom{n-d}{k-2d}$ ways of dealing some $k$ card hand with exactly $d$ doubles.  If we sum over all possible values of $d,$ we should get $\binom{2n}{k}.$  It can be confirmed that, indeed,
$$
\sum_{d=0}^{\lfloor k/2\rfloor}2^{k-2d}\binom{n}{d}\binom{n-d}{k-2d}=\binom{2n}{k}
$$
by evaluating the $x^k$ term in $(1+x)^{2n}=(1+2x+x^2)^n$ in two different ways.

Define $g(n,k,d)$ to be the probability that a $k$ card hand contains exactly $d$ doubles:
$$
g(n,k,d):=\frac{2^{k-2d}\binom{n}{d}\binom{n-d}{k-2d}}{\binom{2n}{k}}.
$$
Define $h(n,k,d)$ to be the probability of getting a particular $k$ card hand with exactly $d$ doubles:
$$
h(n,k,d):=\frac{2^{k-2d}}{\binom{2n}{k}}.
$$

Then
$$
\begin{aligned}
&amp;\Pr[\text{I get the hand I was dealt last time}]\\
&amp;\quad=\sum_{d=0}^{\lfloor k/2\rfloor}\Pr[\text{hand last time had exactly $d$ doubles}]\\
&amp;\quad\quad\cdot\Pr[\text{I get a particular hand with exactly $d$ doubles}]\\
&amp;\quad=\sum_{d=0}^{\lfloor k/2\rfloor}g(n,k,d)\,h(n,k,d)\\
&amp;\quad=\sum_{d=0}^{\lfloor k/2\rfloor}\frac{2^{k-2d}\binom{n}{d}\binom{n-d}{k-2d}}{\binom{2n}{k}}\frac{2^{k-2d}}{\binom{2n}{k}}.
\end{aligned}
$$
By means of further manipulations, this expression can be written in terms of Gauss's ${}_2F_1$ hypergeometric function:
$$
\begin{aligned}
&amp;\Pr[\text{I get the hand I was dealt last time}]\\
&amp;\ =\frac{1}{\left[\binom{2n}{k}\right]^2}\sum_{d=0}^{\lfloor k/2\rfloor}\frac{4^{k-2d}n!}{d!(k-2d)!(n-k+d)!}\\
&amp;\ =\frac{\binom{n}{k}}{\left[\binom{2n}{k}\right]^2}\sum_{d=0}^{\lfloor k/2\rfloor}\frac{4^{k-2d}k!(n-k)!}{d!(k-2d)!(n-k+d)!}\\
&amp;\ =\frac{4^k\binom{n}{k}}{\left[\binom{2n}{k}\right]^2}\sum_{d=0}^\infty\frac{4^{-2d}k(k-1)(k-2)\ldots(k-2d+1)}{d!(n-k+1)(n-k+2)\ldots(n-k+d)}\\
&amp;\ =\frac{4^k\binom{n}{k}}{\left[\binom{2n}{k}\right]^2}\\
&amp;\ \ \cdot\sum_{d=0}^\infty\frac{4^{-d}\left(-\frac{k}{2}\right)\left(-\frac{k}{2}+1\right)\ldots\left(-\frac{k}{2}+d-1\right)\cdot\left(-\frac{k}{2}+\frac{1}{2}\right)\left(-\frac{k}{2}+\frac{3}{2}\right)\ldots\left(-\frac{k}{2}+d-\frac{1}{2}\right)}{d!(n-k+1)(n-k+2)\ldots(n-k+d)}\\
&amp;\ =\frac{4^k\binom{n}{k}}{\left[\binom{2n}{k}\right]^2}{}_2F_1\left(-\frac{k}{2},-\frac{k}{2}+\frac{1}{2};n-k+1;\frac{1}{4}\right).
\end{aligned}
$$

Now let's suppose that $k$ card hands are dealt to two players.  What is the probability that if this is done twice, both players get the same hands on both deals?  We need to generalize the functions defined in the previous analysis.  Let $G(D,S,k,d,e,i)$ denote the probability that a $k$ card hand dealt from a deck with $D$ doubles and $S$ singles contains $d$ doubles, $e$ singles with no matching card left in the deck, and $k-2d-e$ singles with a matching card left in the deck.  Then
$$
G(D,S,k,d,e)=\frac{2^{k-2d-e}\binom{D}{d}\binom{S}{e}\binom{D-d}{k-2d-e}}{\binom{2D+S}{k}}.
$$
Let $H(D,S,k,d,e)$ be the probability of getting a particular $k$ card hand with exactly $d$ doubles and $e$ singles with no matching card left in the deck.  Then
$$
H(D,S,k,d,e)=\frac{2^{k-2d-e}}{\binom{2D+S}{k}}.
$$

Then the probability that both players get the same hand on the second deal as on the first is
$$
\begin{aligned}
&amp;\sum_{d_1=0}^{\lfloor k/2\rfloor}\sum_{d_2=0}^{\lfloor k/2\rfloor}\sum_{e_2=0}^{k-2d_2}G(n,0,k,d_1,0)G(n-k+d_1,k-2d_1,d_2,e_2)\\
&amp;\qquad\qquad\qquad\times H(n,0,k,d_1,0)H(n-k+d_1,k-2d_1,d_2,e_2).
\end{aligned}
$$
This doesn't appear to simplify.  The upper limits on the sums can be replaced by $\infty$ since the summand will be zero beyond those limits.

To find the probability that exactly one of the two players gets the same hand on the second deal as on the first, subtract the probability that both players do from the probability one player does.  Then double the result.  To find the probability that at least one of the two players gets the same hand, add the probability that both do to the probability that exacty one does.

This doesn't answer your questions, but the method can be extended to handle the four-player situation.
1st Question: Of the $52$ distinct cards in a deck, choose $40$. Given $2$ decks, this is the only way we can duplicate the hands of all players in the second deal. In the second deal, we need the same $40$ cards chosen in the first deal. However, each card can come from either deck, giving us $2^{40}$ ways to choose them. Finally, in dealing groups of $10$ out to our players, we choose $10$ from $40, 30, 20$ in succession, giving the remainder to the final player. In doing so, there is only one way to designate the correct groups of $10$ to each player according to our previous deal. So the probability that all get their original hands would be:$$\frac{\binom{52}{40}}{\binom{104}{40}}\cdot\frac{2^{40}}{\binom{104}{40}}\cdot\frac{1}{\binom{40}{10}\binom{30}{10}\binom{20}{10}}$$