The problem can be mapped on a one-dimensional discrete random walk. Imagine that you start at 0 and you make a step +1 (for the outcome HT) with probability $1/4$, a step -1 (for the outcome TH) with probability $1/4$, and a step 0 (for the outcomes TT and HH) with probability $1/2$. The question about the number of coin tosses after which both have accumulate the same number of heads is then equivalent to asking after how many steps the random walk returns to the origin.
Let us introduce the random variable $Z_i$ at step $i$ with value +1 (for HT), -1 (for TH), and 0 in the other cases. As described above, it holds that $P(Z_i = 1) =1/4$, $P(Z_i = -1) =1/4$, and $P(Z_i = 0) =1/2$. Furthermore, introduce $S_n = \sum_{i=1}^n Z_i$. $S_n$ denotes the number of heads which player 1 has obtained minus the number of heads which player 2 has obtained after $n$ steps. The sequence $\{S_n\}$ is called random walk. The question about both players having the same number of heads corresponds to the problem of the first return to the origin.
Let us first solve a simpler problem: what is the probability $P_n$ to return after $n$ steps? We should have the same number of $Z$'s having +1 and -1. Each step we have 3 outcomes $Z= \pm 1,0$. In order to return to the origin, we need to have $m$ times $+1$, $m$ times $-1$, and $n-2m$ times $0$ ($0 \leq m \leq n/2$); this event has probability $$ \frac{n!}{m! m! (n-2m)!} P(Z_i = 1)^m P(Z_i = 1)^m P(Z_i = 0)^{n-2m} = \frac{n!}{(m!)^2 (n-2m)! 2^{n+2m}} .$$ The probability to return after $n$ steps is thereby given by
$$ P_n = \sum_{m=0}^{\lfloor n/2 \rfloor} \frac{n!}{(m!)^2 (n-2m)! 2^{n+2m} }= {2n \choose n}2^{-2n}.$$ What remains to be done is to obtain the probability for the first return.
There is a quite general relation between the probability $P_n$ for a return after $n$ steps and the probability $\pi_n$ for the first return,
$$ P_n= \pi_1 P_{n-1} + \cdots + \pi_n P_0.$$ The interpretation is simple: to return at step $n$ either you have the first return after one step and then return again after $n-1$ steps, or you return the first time after the second step and then again after $n-2$ steps, ..., or you return the first time after $n$ steps ($P_0$ = 1).
The right hand side is a convolution of two probability distributions. Therefore, it is convenient to introduce the generating functions
$P(x) = \sum_{n=0}^\infty P_n x^n$ and $\Pi (x) = \sum_{n=0}^\infty \pi_n x^n$ (we set $\pi_0=0$). Multiplying the aforementioned equation by $x^n$ and summing over $n$, we obtain the relation $P(x) = 1 + P(x) \Pi(x)$ (remember $\pi_0 =0$ and $P_0 =1$). We can solve this equation for $$\Pi(x) = \frac{ P(x) -1 }{P(x)}.$$ All what remains to be done is to evaluate $P(x)$. This can be done (binomial theorem) with the result $$ P(x) = \frac{1}{\sqrt{1-x}}$$ and therefore we obtain $$\Pi(x) = 1- \sqrt{1-x}.$$ Expanding this result in a Taylor series, we obtain the probability for the first return (the probability that the game stops after $n$ rounds)
$$ \pi_n = \frac{{2n \choose n} }{(2n -1) 2^{2n}} \quad n\geq1.$$
The expected number of rounds can be also calculated $$
\bar n = \sum_{n=1}^\infty \pi_n n \to \infty,$$ i.e., it diverges. The reason is that the asymptotic behavior of $\pi_n$ is given by
$$\pi_n \sim \frac{1}{2 \sqrt{\pi} n^{3/2}}.$$
Edit:
The question has arisen about the evaluate the sum leading to $P_n$. Borrowing from the solution of Mike there is an alternative way to get $P_n$ (and thereby proving that the sum really is what I wrote). Instead of looking at the game as having $n$ rounds, we can think that the game has $2n$ rounds where we split the tosses of player 1 and player 2. If player 1 has H then we move $+1/2$ for tail $-1/2$. If player 2 has T then we move $+1/2$ for tail $-1/2$. After two rounds this leads to the same random walk as I described above (HH and TT no move, HT move $+1$, TH move $-1$). The probability $P_n$ to return after $n$ rounds (of the original game) is therefore equivalent to return after $2n$ rounds in the new game. There are $2n \choose n$ possible paths which return after $2n$ rounds because we can choose which $n$ steps which move up (H for player 1 and T for player 2) and in the rest of the steps we have to move down. In total there are $2^{2n}$ paths of length $2n$. The probability to return at the origin after $2n$ steps ($n$ steps in the original game) is given by
$$ P_n = {2n \choose n} 2^{-2n}.$$
You set up your calculation well, but it is a little lengthy. There is a quicker way.
The coins that did not land heads in the first round will feel bad about being left out of the second round. So let's change the game a little, and toss the first coin twice, and the second, and the third, and so on. The probability a coin lands Head then Head is $(0.4)^2$, so the probability this happens $5$ times is $\binom{15}{5}((0.4)^2)^5(1-(0.4)^2)^{10}$.
Let random variable $X$ be the number of heads in the second round. You can find the distribution of $X$, that is, $\Pr(X=k)$, in the same way.
Best Answer
The probability till Mike and Dean get the same results at most 5 rounds:
The final results can be... $${HH}\ or\ {TT}$$ The probability that Mike and Dean get the same results in a round $$={0.6}\times{0.1}+{(1-0.6)}\times{(1-0.1)}=0.42 $$ The probability that Mike and Dean do not get the same results in a round $$={1-0.42}=0.58$$
Let X be the number of rounds until Mike and Dean get the same results. $X\sim Geo(0.42)$
$$P(X\leq5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)$$ $$P(X\leq5)=0.42+0.58\times0.42+(0.58)^2\times0.42+(0.58)^3\times0.42+(0.58)^4\times0.42\approx0.9344\ (corr.to\ 4\ d.p.)$$