Let $S$ be the set of all $100$-digit numbers whose digits are from $\{ 1,2,3,4,5,6\}$. For each $x\in S$, let $x_{k}$ denote the $k$th digit of $x$, counting from left to right. Let the two players shout the number $f(t)$ where $t$ is the $100$-digit number each of them get from their dice-rollings and $f$ is a function from $S$ to $\{ 1,2,\dotsc ,100\}$ defined by $$f(x)=\begin{cases} \min \{ k\in \{ 1,2,\dotsc ,100\}\mid x_k=6\}, & \text{ if the set is not empty} \\ 100, & \text{ otherwise} \end{cases}$$for all $x\in S$. To show that this strategy works, we need to show that $$\sum_{a\in S}\sum_{b\in S} 1_{a_{f(b)}=6,b_{f(a)}=6} >\frac{6^{200}}{36}.$$This is true because \begin{align*} \sum_{a\in S}\sum_{b\in S} 1_{a_{f(b)}=6,b_{f(a)}=6} & \geqslant \sum_{a\in S}\sum_{b\in S}\sum_{k=1}^{99}{1_{f(a)=k,f(b)=k}}\\ & = \sum_{k=1}^{99}\left( 5^{k-1}\cdot 1 \cdot 6^{100-k}\right)^2 \\ & = \frac{6^{200}}{25} \sum_{k=1}^{99} \left( \frac{25}{36}\right)^k > \frac{6^{200}}{36}. \quad \blacksquare \end{align*}

Let $S$ be the set of all $100$-digit numbers whose digits are from $\{ 1,2,3,4,5,6\}$. For each $x\in S$, let $x_{k}$ denote the $k$th digit of $x$, counting from left to right. Let the two players shout the number $f(t)$ where $t$ is the $100$-digit number each of them get from their dice-rollings and $f$ is a function from $S$ to $\{ 1,2,\dotsc ,100\}$ defined by $$f(x)=\begin{cases} \min \{ k\in \{ 1,2,\dotsc ,100\}\mid x_k=6\}, & \text{ if the set is not empty} \\ 100, & \text{ otherwise} \end{cases}$$for all $x\in S$. To show that this strategy works, we need to show that $$\sum_{a\in S}\sum_{b\in S} 1_{a_{f(b)}=6,b_{f(a)}=6} >\frac{6^{200}}{36}.$$This is true because \begin{align*} \sum_{a\in S}\sum_{b\in S} 1_{a_{f(b)}=6,b_{f(a)}=6} & \geqslant \sum_{a\in S}\sum_{b\in S}\sum_{k=1}^{99}{1_{f(a)=k,f(b)=k}}\\ & = \sum_{k=1}^{99}\left( 5^{k-1}\cdot 1 \cdot 6^{100-k}\right)^2 \\ & = \frac{6^{200}}{25} \sum_{k=1}^{99} \left( \frac{25}{36}\right)^k > \frac{6^{200}}{36}. \quad \blacksquare \end{align*}

Each player's strategy is as follows. If their first number is a 6, they shout 1, and if their first number isn't a 6, they shout 2. This wins if either both start with 6, or with both start with $x6$ with $x\neq 6$, so we can guarantee a winning probability of at least \[\left(\frac16\right)^2 + \left(\frac56 \cdot \frac16 \right)^2 > \frac{1}{36} \]and so we are done. $\blacksquare$ Remark: I have no idea what the upper bound is, I feel like you extend this quite a bit.