a coupling probability problem and random walk game

Let $\ R_{ni}\ $ be the number of times player $\ i\ $ gets chosen and tosses tails, and $\ R_{n\,i+3}\ $ the number of times he or she gets chosen and tosses heads. The random variables $\ R_{nj}\ $ are multinomially distributed with parameters $\ n\ $ and $\ \frac{1}{6}, \frac{1}{6},\dots, \frac{1}{6}\ $: \begin{align} P\big(R_{n1}=n_1,R_{n2}=n_2,\dots,&R_{n6}=n_6\big)\\ =&\cases{\displaystyle \frac{n!}{\big(n_1!n_2!\dots n_6!\big)6^n}&if $\ \displaystyle n=\sum_{i=1}^6n_i$\\ \hspace{3em}0&otherwise, } \end{align} and \begin{align} y_n&=\sum_{j=1}^3\big(R_{nj}-R_{n\,j+3}\big)\\ x_{in}&= 4\big(R_{n\,i+3}-R_{ni}\big)+ \sum_{j=1}^3\big(R_{nj}-R_{n\,j+3}\big)\ . \end{align} Therefore, \begin{align} P\big(y_n>\max_{i\in\{1,2,3\}}x_{in}\big)&=P\big(R_{n1}>R_{n4}, R_{n2}>R_ {n5},\ R_{n3}>R_{n6}\big)\\ &=\frac{1}{8}P \big(R_{n1}\ne R_{n4}, R_{n2} \ne R_ {n5},\ R_{n3}\ne R_{n6}\big) \end{align} by symmetry. Now \begin{align} P\big(R_{n1}\ne R_{n4},\ R_{n2} &\ne R_ {n5},\ R_{n3}\ne R_{n6}\big)\\ =1-&3P\big(R_{n1}=R_{n4}, R_{n2}\ne R_ {n5},\ R_{n3}\ne R_{n6}\big)\\ -&3 P\big(R_{n1}=R_{n4}, R_{n2}= R_ {n5},\ R_{n3}\ne R_{n6}\big)\\ -& P\big(R_{n1}=R_{n4}, R_{n2}= R_ {n5},\ R_{n3}= R_{n6}\big)\\ =1-&3\big(P\big(R_{n1}=R_{n4}, R_{n3}\ne R_{n6}\big)\\ &\hspace{2em}-P\big(R_{n1}=R_{n4}, R_{n2}= R_ {n5},\ R_{n3}\ne R_{n6}\big)\big)\\ -& 3 P\big(R_{n1}=R_{n4}, R_{n2}= R_ {n5},\ R_{n3}\ne R_{n6}\big)\\ -& P\big(R_{n1}=R_{n4}, R_{n2}= R_ {n5},\ R_{n3}= R_{n6}\big)\\ =1-&3P\big(R_{n1}=R_{n4}\big)+3P \big(R_{n1}=R_{n4}, R_{n3}= R_{n6}\big)\\ -& P\big(R_{n1}=R_{n4}, R_{n2}= R_ {n5},\ R_{n3}= R_{n6}\big)\ ,\\ P\big(R_{n1}=R_{n4}\big)=&\sum_{j=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{n!}{(j!)^2(n-2j)!}\left(\frac{1}{6}\right)^{2j}\left(\frac{2}{3}\right)^{n-2j}\ ,\\ P \big(R_{n1}=R_{n4},\ R_{n3}&= R_{n6}\big)\\ =\sum_{j=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor-j} &\frac{n!}{(j!)^2(k!)^2(n-2(j+k))!} \left(\frac{1}{6}\right)^{2(j+k)} \left(\frac{1}{3}\right)^{n-2(j+k)}\ ,\\ P\big(R_{n1}=R_{n4}, R_{n2}&= R_ {n5},\ R_{n3}= R_{n6}\big)\\ =&\cases{0&if $\ n\ $ is odd\\ \displaystyle\frac{1}{6^{2r}}\sum_{j=0}^r\sum_{k=0}^{r-j}\frac{(2r)!}{(j!)^2(k!)^2((r-j-k)!)^2}&if $ n=2r$}\ . \end{align} Putting all this together, we have \begin{align} g(n)= \frac{1}{8} &\Bigg(1-3 \sum_{j=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{n!}{(j!)^2(n-2j)!}\left(\frac{1}{6}\right)^{2j}\left(\frac{2}{3}\right)^{n-2j}\\ +&3 \sum_{j=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor-j} \frac{n!}{(j!)^2(k!)^2(n-2(j+k))!} \left(\frac{1}{6}\right)^{2(j+k)} \left(\frac{1}{3}\right)^{n-2(j+k)}\Bigg) \end{align} if $\ n\ $ is odd, or \begin{align} g(2r)= \frac{1}{8} &\Bigg(1-3 \sum_{j=0}^r\frac{(2r)!}{(j!)^2(2(r-j))!}\left(\frac{1}{6}\right)^{2j}\left(\frac{2}{3}\right)^{2(r-j)}\\ +&3\sum_{j=0}^r\sum_{k=0}^{r-j} \frac{(2r)!}{(j!)^2(k!)^2(2(r-j-k))!} \left(\frac{1}{6}\right)^{2(j+k)} \left(\frac{1}{3}\right)^{2(r-j-k)}\\ -&\frac{1}{6^{2r}}\sum_{j=0}^r\sum_{k=0}^{r-j}\frac{(2r)!}{(j!)^2(k!)^2((r-j-k)!)^2}\Bigg)\ , \end{align} if $\ n=2r\ $ is even.

The values $\ g(n)\ $ for $\ n=1,2,\dots,10\ $ are given in the following table $$ \begin{array}{c|c|c|c|c|c|c|c|} &1,2&3,4&5,6&7&8&9&10\\ \hline \text{exact}&0&\frac{1}{36}&\frac{55}{1\,296}&\frac{2\,401}{46\,656}&\frac{7\,217}{139\,968}&\frac{97\,147}{1\,679\,616}&\frac{16\ 235}{279\,936}\\ \hline \text{approx.}&&0.028&0.042&0.05146&0.05156&0.0578&0.05800\\ \hline \end{array} $$