Symmetry in Probability (AMC 12A 2023)

Question

Flora the frog starts at $0$ on the number line and makes a sequence of jumps to the right. In any one jump, independent of previous jumps, Flora leaps a positive integer distance $m$ with probability $\frac{1}{2^m}$. What is the probability that Flora will eventually land at $10$? (AMC 12A 2023/17)

Solution 1 says:

At any point, the probabilities of landing at $10$ and landing past $10$ are exactly the same. Therefore, the probability must be $\frac{1}{2}$.

If you apply any of solutions 2(recursion), 3 (combinations), or 7(induction), then solution 1 follows. But is there some elaboration of solution 1 that does not include 2, 3, or 7. Or some really elegant way to solve the question?

Answer 1

The idea in Solution 1, although not explicitly stated, is that if the frog has not reached or passed $10$, irrespective of how much further Flora needs to go, the probability that the next jump causes Flora to land exactly on $10$, is the same as the probability that Flora lands on a number strictly greater than $10$, because $$\frac{1}{2^k} = \sum_{m=k+1}^\infty \frac{1}{2^m}$$ for any nonnegative integer $k$. As a result, no matter the current position, the state of the frog is such that there is always an equal probability of stopping on $10$ or stopping beyond $10$. So for the initial position, before any jump has been made, the same must be true, so it must be $1/2$.

Answer 2

Based on comments:

There is nothing special about $10$ or even $\frac12$.

In the process which is Flora's progress rightwards, each position has a probability $p$ of being landed on and a probability $1-p$ of not being landed on, independently of which previous or later positions are landed on. The number of hits in an interval has a binomial distribution, and has a Bernoulli distribution for a single position. (This is the discrete equivalent of a Poisson process, where the number of hits in an interval has a Poisson distribution.)

This is equivalent to the geometric distribution for the lengths of jumps with probability $p(1−p)^{m−1}$ for a jump of length $m$, independent of other jumps. (In a continuous Poisson process you would get an exponential distribution for the length of jumps.)

In this question $p=1-p=\frac12$, so this is the probability of landing on position $10$ and also the probability of landing on any other particular position.