Assuming each player is given a random $26$ cards of $13$ ranks ( $4$ suits/possible cards for each rank ) out of a $52$-card deck, the basic probability of $n$ consecutive ties would be $\left(1/13\right)^{n}$:
I am not quite sure how to solve it past this point. Can anyone give me some $\mbox{guidance}\ ?$.
The first tie is $\frac 1{17}$ because the first player picks a card, leaving $51$ cards for the second player of which $3$ match the first card. You can ignore the face down cards because we don't look at them. The first player picks a card, having $\frac 2{50}$ chance to get the same rank as the first time and $\frac {24}{25}$ chance to get a different rank. If the first player picks the same rank, the second has $\frac 1{49}$ chance to match it, while f the first player picks a different rank the second has a $\frac 3{49}$ chance to match. Therefore, assuming you have a war the chance of a second is $$\frac 1{25} \cdot \frac 1{49}+\frac{24}{25}\cdot \frac 3{49}=\frac{73}{1225}$$ The next one will be more work but follows the same principle. You have to start with the chance the first two matches are the same or different ranks.
I will compute the probability that the very first play of the game results in $n$ consecutive ties, for each $0\le n\le 26$. This method will not apply to finding the probability of $n$ ties at a later point in the game, since each player will no longer have a uniformly random deck of $26$ cards.
Let $P_n$ be the probability of seeing $n$ consecutive pairs on the first play of the game. Then $$ \boxed{P_n=\frac{2^n\cdot n!\cdot (52-2n)!\cdot [x^n](1+6x+3x^2)^{13}}{52!}.} $$ The notation $[x^n]f(x)$ means the coefficient of $x^n$ in the polynomial $f(x)$. To apply this formula, you must first compute the polynomial $(1+6x+3x^2)^{13}$, which can be done using any computer algebra system. $$ (1+6x+3x^2)^{13}=1+\color{blue}{78}x+\color{green}{2847}x^2+\color{purple}{64584}x^3+\dots $$ Here are some examples for using this formula: $$ \begin{align} P_1&=\frac{2^1\cdot 1!\cdot 50!\cdot \color{blue}{78}}{52!}&&=\frac1{17},\\ P_2&=\frac{2^2\cdot 2!\cdot 48!\cdot \color{green}{2847}}{52!}&&=\frac{73}{20825},\\ P_3&=\frac{2^3\cdot 3!\cdot 46!\cdot \color{purple}{64584}}{52!}&&=\frac{207}{978775}. \end{align} $$
After the deck is shuffled, but before the deck is dealt out to the two players, consider the top $2n$ cards of the deck. The top two cards of the deck will be the first pair of cards pitted against each other, the third and fourth cards from the top will be the second pair, and so on. Therefore, we will see $n$ ties in a row if and only if the top $2n$ cards consist of $n$ consecutive pairs.
There are $52!$ equally likely ways to shuffle the deck, so this is our denominator. For the numerator, we count the number of shuffles for which the first $2n$ cards come in $n$ consecutive pairs. First, we choose the set of unordered pairs that we see. For example, if $n=3$, we could choose $\{\{\spadesuit A,\heartsuit A\},\{A\clubsuit,A\diamondsuit\},\{\heartsuit 7,\clubsuit 7\}\}$, where the order within each pair does not matter, and the order of entire set of pairs does not matter. The number of ways to do this is $[x^n](1+6x+3x^2)^{13}$ (see claim below). Next, we assign an ordering to each of these $n$ pairs in $2^n$ ways, and then we choose an ordering of all the pairs in $n!$ ways. This determines the ordering of the first $2n$ cards of the deck, and the last $52-2n$ cards can be ordered arbitrarily in $(52-2n)!$ ways. Putting this altogether, you get my claimed formula for $P_n$.
Claim: The number of ways to choose an (unordered) set of $n$ unordered pairs of cards, where each pair is the same rank, is $[x^n](1+6x+3x^2)^{13}$.
Proof sketch. We can divide the process of choosing a set of $n$ pairs into two steps. In the first step, we choose a list of $13$ nonnegative integers $a_1,a_2,\dots,a_{13}$, such that $a_1+\dots+a_{13}=n$. In the second step, we choose $a_1$ pairs of aces, $a_2$ pairs of twos, and so on, up until $a_{13}$ pairs of kings.
Let us think about the number of ways to complete the second step. First, look at the aces. If $a_1=0$, then there is nothing to choose. If $a_1=1$, we must choose a single pair of aces, in $\binom 42=6$ ways. If $a_2=2$, we must choose a set of two pairs of aces. There are $3$ ways to do this; there are three choices for the ace to pair with the ace of spades, and that choice determines the other pair. Now, given the whole list $a_1,\dots,a_{13}$, the number of ways to complete the second step is the product of $13$ numbers, where the $i^\text{th}$ number is $1$ if $a_i=0$, the $i^\text{th}$ number is $6$ if $a_i=1$, and the $i^\text{th}$ number is $3$ if $a_i=2$.
On the other hand, consider the following method for computing the coefficient of $x^n$ in $(1+6x+3x^2)^{13}$. There are $13$ factors being multiplied. When expanding the product of these factors using the distributive property, we choose one the summands from each factor, either $1,6x$, or $3x^2$, such that the sum of the chosen powers of $x$ is equal to $n$. Say that from the $i^\text{th}$ factor, we choose the power of $x^{a_i}$, for some list $(a_1,\dots,a_{13})$. We require $a_1+\dots+a_{13}=n$. The contribution for that factor is then the products of the coefficients for each $x^{a_i}$, which is $1$ if $a_i=0$, $6$ if $a_i=1$, and $3$ if $a_i=2$.
After squinting a bit, you can see that these two computational procedures are exactly the same! $\square$
