This game is a two-player symmetric (almost) zero-sum finite-horizon simultaneous game with perfect-information. The set of subgame-perfect equilibria (SGE) can be derived via backward induction. Solving this particular game will take quite a bit of computation.
Setting up the game: It is useful to describe your game in slightly greater generality, to see how it can be solved in general, from an arbitrary point. Since the game is symmetric we can consider it from the perspective of a single “acting” player in an arbitrary game state, which can be described by the following variables:
$$\begin{matrix}{}
\text{Remaining rounds} \quad \quad & & & k \in \mathbb{N}, \\
\text{Wealth of acting player} & & & a \in \mathbb{N}, \\
\text{Wealth of opponent} \quad \text{ } \text{ } & & & b \in \mathbb{N}. \end{matrix}$$
In each round, each player simultaneously chooses an action $\mathbf{p}_{k,a,b}$, which is a probability-vector over the allowable bets $\{ 1, 2, ..., w \}$. The chosen action will depend on the number of remaining rounds and the present wealth for the acting player and his opponent.$^\dagger$
A strategy for the game is a set of probability vectors for all possible values of these arguments:
$$\mathscr{S} = \{ \mathbf{p}_{k,a,b} \in \Delta_{a,w} | k \in \mathbb{N}, a \in \mathbb{N}, b \in \mathbb{N} \},$$
where $\Delta_{a,w} \equiv \{ (p_1, \cdots, p_w) | \sum p_i = 1, (\forall i): p_i \geqslant 0, (\forall i > a): p_i = 0 \}$ is the simplex
of all probability-mass functions on the allowable bets. (This is always taken to be a vector with length $w$, but the player is not able to bet more than his current wealth, so any values over this amount must have zero probability.)
You have not specified what happens in the event of a draw (i.e., both players have the same wealth once $k=0$), so I will assume that this constitutes a loss for both players (i.e., they both get nothing). This means that each player will choose a strategy to maximise the probability of winning the game. (It is not quite a zero-sum game because of the way I am treating a draw.) Since the game is symmetric with respect to the two-players, there will be a set of optimal strategies (subgame-perfect equilibria), which are optimal reactions to all other optimal strategies. These will be optimal in the sense that they maximise the probability of winning the game, under the assumption that the other player also plays optimally.
Solving the game: Solving the game requires us to calculate the “value function” for any game-state $(k,a,b)$ (up to some specified maximum values for a particular implementation of the game). The value is for the acting player, so it is the probability that the acting player wins, under optimal play (by both players) from a particular game-state:
$$\upsilon (k,a,b) \equiv \mathbb{P}( \text{Acting player wins} | k,a,b).$$
To solve the game using backward induction we start at the termination points of the game, which occur when $k=0$ or $a=0$ or $b=0$. At any of these points the value for the game is:
$$\begin{array}
\upsilon \upsilon (0,a,b) = \mathbb{I}(a>b), \\[6pt]
\upsilon (k,0,b) = 1, \\[6pt]
\upsilon (k,a,0) = 0, \\[6pt]
\end{array}$$
To find the remaining outcomes of the value function we proceed recursively, by moving away from these “edges”. To do this, let $\mathbf{p} \in \Delta_{a,w}$ be the action chosen by the acting player, and assume that the opponent takes the optimal action from his perspective, which is $\hat{\mathbf{p}}(k,b,a)$. This gives us recursive equations for the optimal action:
$$\upsilon (k,a,b) = \max_{\mathbf{p} \in \Delta_{a,w}} \upsilon_\mathbf{p} (k,a,b),$$
where:
$$\begin{equation} \begin{aligned}
\upsilon_\mathbf{p} (k,a,b)
&\equiv \mathbb{P}(\text{Acting player wins} | \mathbf{p}, k,a,b) \\[6pt]
&= \sum_{x = 1}^{\min(a,w)} \sum_{y=1}^{\min(b,w)} p_{k,a,b}(x) \hat{p}_{k,b,a}(y) [ \tfrac{1}{6} \upsilon (k,a+5x,b+5y) + \tfrac{5}{6} \upsilon (k,a-x,b-y) ].
\end{aligned} \end{equation}$$
These recursive equations can be combined with the base cases to obtain the value function for each game-state $(k,a,b)$. This is a complicated non-linear programming problem, so it will require some finesse. I will leave it to you to look at how to program this optimisation problem, but in any case, once you have the value function at each game-state of interest, you will easily be able to obtain the optimal strategy, and the probability that each player wins (or the probability of a draw) when both players play optimally.
$^\dagger$ Technically the game-state and actions may also depend on the actual round number (not just the number of remaining rounds) and the wealth levels of the players in past rounds, but there is no reason to take this into account, since the optimal strategy will be "memory-less" in this game. Hence, we restrict attention to the class of actions that ignore this irrelevant information.
