Solution to the ladybug clock puzzle

Last month, I posted a puzzle. A ladybug starts on the 12 of a clock. She takes a series of random steps, either clockwise or counterclockwise, each with random chance. Each number she touches gets colored red. And the puzzle is, what is the probability that the last number to get colored is a six. If I simulate it many times in a row, recording which number happened to be the last one colored, here's what it looks like. Running this many hundreds of times, a somewhat surprising fact starts to make itself clear. All the numbers between 1 and 11 seem to be about equally likely to be the last one touched. I say this is surprising because initially it feels like ending in a six should be less likely than ending on the numbers closer to the initial position like the 1 or the 11. And yet empirically that just doesn't seem true. And there is a key insight to understand why. Rather than calculating the probability of ending on a six from the initial position, start by letting the ladybug wander until the first moment that she hits one of the two neighbors of that six, the five or the seven. And then from there, ask the probability of ending in a six. That is a different question. But the reason it doesn't change the answer is because for any run of the simulation, eventually she's going to hit the five or the seven at some point. That's bound to happen with probability one. waiting for this middle condition does nothing to change the probability of that final outcome. So from this state where she's on say the seven, what's required to end on the six? Well, she has to reach that untouched five before ever touching the six. Meaning she has to eventually end up 10 steps away clockwise on net before ever ending up one step away counterclockwise. If that happens, if she ends up that far around coloring all the other numbers, it's a done deal that six is going to be the last. Phrased another way, it's like if we straighten things out and you're taking a random walk where each step is plus one or minus one with a 50/50 chance, and you want to know the probability of ending up at + 10 before ever touching the negative numbers. Again, this is a hard question to answer directly, but the point is that you don't actually have to answer it directly. There's nothing special about the number six. What's the probability of ending on the three? Well, let the simulation run until the first moment she hits a two or a four, which again is bound to happen eventually. And then from there, it's the probability of taking on net 10 steps around one way before ending one step around the other way. Because calculating this probability is identical for all the numbers between 1 and 11, the answer must be 1 and 11.