Competition between two geometric distributions

The general problem I'm trying to solve is obtaining a closed-form solution for

$$\mathrm{P}(aX > bY)$$

where $a$ and $b$ are constants; $X$ and $Y$ are geometrically distributed random variables with dissimilar p.

Here is a motivating example that I've come up with:

Alice and Bob intend to play a game of chance. The proposed rules are that Alice flips a coin until she gets tails. Her score is the number of heads she flipped. Bob in turn rolls a die. He only gets to re-roll if he rolls a 1. His score is the number of 1's he rolled.

Bob realizes that this game is clearly in Alice's favour and proposes to multiply the scores by an appropriate factor. They do some research and find that their scores are modeled by the geometric distribution $$\mathrm{P}(X = k) = (1 - p)^kp$$

where Alice's $p = 1/2$ and Bob's $p = 5/6$. In this form of the geometric distribution, the expected value is $\mathrm{E}(X) = \frac{1 - p}{p}$. Alice's expected value is $1$ while Bob's expected value is $1/5$. Bob thus suggests that his score be multiplied by $5$. Alice accepts. Who is more likely to win?

Naively, you might assume that they are equally likely to win. However, my simulations have shown that Alice wins with about 41.9% probability, Bob wins with about 16.2% probability, and a draw occurs with 41.9% probability.

In my research, I've not been able to find mention of comparing geometric distributions with dissimilar $p$, only identical $p$.

Does a closed-form solution exist for the general problem of $\mathrm{P}(aX > bY)$? If not, does one exist for simpler cases like my example, where $p$ is rational?

For $i = 1, 2$ let $$X_i \sim \operatorname{Geometric}(p_i), \quad \Pr[X_i = x] = (1-p_i)^x p_i, \quad x = 0, 1, 2, \ldots.$$ Then \begin{align*}\Pr[aX_1 > bX_2] &= \sum_{x=0}^\infty \Pr\left[X_1 > \tfrac{b}{a}x \mid X_2 = x\right]\Pr[X_2 = x] \\ &= \sum_{x=0}^\infty (1-p_1)^{1+\lfloor bx/a \rfloor} (1-p_2)^x p_2. \end{align*} If $a$ divides $b$, i.e. $b = ka$ for some positive integer $k$, then this sum has a closed form: $$\Pr[X_1 > kX_2] = \frac{(1-p_1) p_2}{1-(1-p_1)^k (1-p_2)}.$$ But if not, then this sum is not likely to have a closed form. In your case, your choice of $a$ and $b$, or more importantly, the fraction $b/a = k$, is based on the expectations of $X_1$ and $X_2$, which are in turn based on $p_1$ and $p_2$.