So the only consequence of not doing a 5 is that you can’t do a 1 after it? It’s the same with 31. If you don’t do a 3, you can’t do a 1 after it. I don’t see any consequences of skipping the 5 that skipping the 3 doesn’t also have.
Here’s something that hasn’t been mentioned yet, that will be relevant to thinking about how 51 works:
In 55500, there is a time when there is no incoming object towards a hand that is already holding an object, because both of your hands are empty. But does that mean the juggling phenomenon is not present at that time, and that pattern is not part of the specific activity of juggling? There is still something about the pattern at that time that means you are still juggling: There is a ball following behind another ball, that forces it to be thrown within a limited time, though it’s not possible to throw it yet. If you wait long enough without throwing, you will have to catch two balls in one hand or drop. So the situation of an incoming object forcing another object to be thrown doesn’t have to consist of a ball in a hand and a ball in the air; it can exist while they are both in the air.
In 51, when you’ve just passed the red ball across from left to right as a 1, and you haven’t caught the (green) incoming 5 yet, the only ball that will definitely have to be thrown within a limited time, based on the current configuration of objects, is the green one. This is a situation like in 55500: The green ball is still in the air, but it will be forced to be thrown soon because of the blue ball following behind it.
After you catch the green ball, the situation changes to the usual “incoming object towards a hand that is already holding an object”. The green ball will have to be thrown soon, after you do something with the red ball. There is nothing forcing you to throw the red ball right now, but you choose to throw it.
After you throw the red ball as a 5, there are now two balls that will be forced to be thrown within a limited time. The green ball is going to need to get out of the way of the blue ball, and the blue ball is going to need to get out of the way of the red ball. You’ve just created another all-in-the-air problem like the one in the first picture.
When you pass the green ball across as a 1, you will be eliminating one of those two “problems” without simultaneously creating a new one, as you do in the cascade. But since you created a new problem before eliminating one, you are continuously maintaining the existence of at least one problem. And like in the cascade, the new problem involves one of the same balls that was part of the old problem (the blue one).