Probability of X<Y for two independent geometric random variables, intuitively

Given two independent, geometric random variables $X$ and $Y$ with the same success probability $p\in(0,1)$, what is $\mathbf P(X<Y)$?

The answer can be computed formally using total probability and one gets $$\mathbf P(X<Y)=\frac{1-p}{2-p}$$

In the concrete example, where $X$ and $Y$ model six sided fair dice and $p$ is the probability to throw a six with one dice, the problem above is the probability that the first dice shows a six for the first time before a second dice does when both are thrown simulatanously in rounds.

Curiously, the probability $\mathbf P(X<Y)=5/11$, which is exactly the number of events in $\Omega=\{1,2,...,6\}^2$ where the first entry is six, divided by the number of results that contain at least one six.

This observation opens the question whether there is a heuristic proof of the above general formula?

EDIT: Let's reformulate this as a game between player A and player B. They both bet that some event occurs as a result of independent experiments for them before it occurs in the same type of experiments for the other player independently. There is a remis, if the event happens for both at the same time.

Examples could be that both players throw a dice each and bet on who gets 6 first. Or they throw 2 coins each and bet on who gets two heads first.

The probability that the event happens is $p\in(0,1)$. Then, we are in the case above, where the number of trials until first success for each is geometric with success parameter $p$; let's say the number of trials for a is $X\sim Geo(p)$ and for player B $Y\sim Geo(p)$. Thus, the probability that player A wins is $\mathbf P(X<Y)$.

Now, in the two examples that i gave, the probability to win is just the number of events in which A wins divided by the number of events that either one wins or there is a remis. Is that coincidence? It should not be considering the case that the $p$ is a probability in a Laplace experiment.

EDIT 2: Thanks to an answer below, we can make this observation precise. Let $A$ be the event that player A wins and $D_i$ be the event that the game ends in round i. Then, $$D_i=\{\min(X,Y)=i\}$$ Furthermore, $$\mathbf P(X<Y)=\mathbf P(A) = \sum_{i=1}^\infty \mathbf P(A|D_i)\mathbf P(D_i)$$

Since the geometric distribution is made up of independent Bernoulli trials, we have $\mathbf P(A|D_i)=\mathbf P(A|D_1)$ and thus $$\sum_{i=1}^\infty \mathbf P(A|D_i)\mathbf P(D_i)=\mathbf P(A|D_1)\sum_{i=1}^\infty \mathbf P(\min(X,Y)=i)=\mathbf P(A|D_1)$$

We now remember that $X=\inf\{n\geq 1: X_n=1\}$ where $(X_i)_{i\geq 1}$ is a sequence of independent Bernoulli random variables with success probability $p$ and the same for $Y=\inf\{n\geq 1: Y_n=1\}$ where $(Y_i)_{i\geq 1}$, $Y_i\sim Ber(p)$. That means $$\mathbf P(A|D_1)=\frac{\mathbf P(X_1=1,Y_1=0)}{1-\mathbf P(X_1=0,Y_1=0)} =\frac{p(1-p)}{1-(1-p)^2}=\frac{1-p}{2-p}$$

Let $q=\mathbb P(X<Y)$. We condition on the outcome of the first trial for players $A$ and $B$.

If player $A$ gets a success, then $X<Y$ iff $B$ fails. This event happens with probability $p(1-p)$.

If $A$ gets a fail, then we need $B$ to fail too. After this we are in the same state as initially. So in this case $X<Y$ with probability $(1-p)^2q$.

So $$q=p(1-p)+(1-p)^2q\implies q=\frac{1-p}{2-p}.$$

This can be thought of in terms of conditional probability. We have $X < Y$ if and only if, in the first round where either $A$ or $B$ rolls a six, $A$ rolls the six and $B$ does not. Focusing on that particular round $i$, we know that at least one six has been rolled, but we don't know by whom, or if both rolled a six. We can therefore evaluate:

$P$[$A$ wins] = $\mathbb{E}_i$ $P$[$A$ rolls six in round $i$ and $B$ does not | At least one six is rolled in round $i$]

This expected value is easy to compute: The distribution over which $i$ is the final round doesn't matter, since this conditional probability is 5/11 for all $i$.