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Universe multiplication, revisited.
Okay, at the beginning of the series, Quinn says that no one knows how many universes there are, and tosses out the number 6. So we'll do these calculations with 6 as the base point even though it becomes obvious that there are far more than 6, because it's easier to see the orders of magnitude if you start with a small number.
Before anyone ever slid, even alternate Quinns, there were 6 universes.
When you slide, you have the possibility of hitting any of the universes in existence. But you can only hit one of them.
However, the reason there are more than one universe in the first place is that the universe divides over possibilities such as this.
So the multiverse divides into 6 bundles, each containing identical copies of the 6 original universes up until the point of slide. And the sliders themselves are divided into 6 copies of themselves, sliding to a different universe in each bundle. In Bundle 1, the sliders reach Universe A. In Bundle 2, the sliders reach Universe B. In Bundle 3, the sliders reach Universe C. In Bundle 4, the sliders reach Universe D. In Bundle 5, the sliders reach Universe E. In Bundle 6, the sliders reach universe F. Which means one (un?)lucky set of sliders will land in their home universe.
Now there are 36 (6^2) universes in the multiverse. 1/6 of the have been visited by sliders, which means that the next time a slider slides, he has a one in 6 chance of hitting a universe that a copy of his sliding self is in or has just been in. Our slider also has a 1 in 6 chance of hitting any of the 6 copies of his home universe--although one will be occupied by the version of him that didn't go anywhere, so make that 5 (6-1) in 36.
Here's the catch: The next time anyone in any of the universes slides, 36 bundles of 36 universes each will be created to handle all the possiblities, resulting in 1, 296 (6^2^2 or 6^4) universes.
But it won't just be one set of sliders sliding. All 6 sets of sliders in the multiverse of 36 universes will slide (except possibly the crazy kids who landed in their home universe on the first slide and figured it was a machine that did nothing). If all 6 sets of sliders slide simultaneously, the resultant number of universes will be 2, 176, 782, 336 (6^12). If they do it sequentially, it will be 4, 011, 991, 914, 547, 600, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000 approximately (6^128). (If the one set who didn't go anywhere doesn't try to slide again, and only 5 (6-1) sets of sliders slide, it's only 60, 466, 176 (6^[(6-1)*2]) and 63, 340, 286, 662, 973, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000 approximately (6^[2^(6-1)*2]), respectively).
Fortunately, 1/6 of them will still be copies of the home universe, so while the multiverse has been growing dramatically, the chances of hitting home remain slightly less than 1 in 6.
This does not account for InterQuinnocity, ie, the effect of a slider teaching non-sliding selves to slide independently. There is no easy way to calculate how often sliders will meet versions of themselves who have the necessary background to understand sliding, or how often sliders will willingly pass on the knowledge, given what they should know about the drastic increase in the number of universes when any slider or set of slider slides.
Now, go back to the beginning and replace all bolded 6s with infinity, which is a much more likely starting number of universes.
My favorite resulting sentence is: "Fortunately, 1/∞ of them will still be copies of the home universe, so while the multiverse has been growing dramatically, the chances of hitting home remain slightly less than 1 in ∞." Considering that 1 in ∞ is, for practical purposes, zero, the sliders are, as I've said, fucked up the ass.
Before anyone ever slid, even alternate Quinns, there were 6 universes.
When you slide, you have the possibility of hitting any of the universes in existence. But you can only hit one of them.
However, the reason there are more than one universe in the first place is that the universe divides over possibilities such as this.
So the multiverse divides into 6 bundles, each containing identical copies of the 6 original universes up until the point of slide. And the sliders themselves are divided into 6 copies of themselves, sliding to a different universe in each bundle. In Bundle 1, the sliders reach Universe A. In Bundle 2, the sliders reach Universe B. In Bundle 3, the sliders reach Universe C. In Bundle 4, the sliders reach Universe D. In Bundle 5, the sliders reach Universe E. In Bundle 6, the sliders reach universe F. Which means one (un?)lucky set of sliders will land in their home universe.
Now there are 36 (6^2) universes in the multiverse. 1/6 of the have been visited by sliders, which means that the next time a slider slides, he has a one in 6 chance of hitting a universe that a copy of his sliding self is in or has just been in. Our slider also has a 1 in 6 chance of hitting any of the 6 copies of his home universe--although one will be occupied by the version of him that didn't go anywhere, so make that 5 (6-1) in 36.
Here's the catch: The next time anyone in any of the universes slides, 36 bundles of 36 universes each will be created to handle all the possiblities, resulting in 1, 296 (6^2^2 or 6^4) universes.
But it won't just be one set of sliders sliding. All 6 sets of sliders in the multiverse of 36 universes will slide (except possibly the crazy kids who landed in their home universe on the first slide and figured it was a machine that did nothing). If all 6 sets of sliders slide simultaneously, the resultant number of universes will be 2, 176, 782, 336 (6^12). If they do it sequentially, it will be 4, 011, 991, 914, 547, 600, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000 approximately (6^128). (If the one set who didn't go anywhere doesn't try to slide again, and only 5 (6-1) sets of sliders slide, it's only 60, 466, 176 (6^[(6-1)*2]) and 63, 340, 286, 662, 973, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000 approximately (6^[2^(6-1)*2]), respectively).
Fortunately, 1/6 of them will still be copies of the home universe, so while the multiverse has been growing dramatically, the chances of hitting home remain slightly less than 1 in 6.
This does not account for InterQuinnocity, ie, the effect of a slider teaching non-sliding selves to slide independently. There is no easy way to calculate how often sliders will meet versions of themselves who have the necessary background to understand sliding, or how often sliders will willingly pass on the knowledge, given what they should know about the drastic increase in the number of universes when any slider or set of slider slides.
Now, go back to the beginning and replace all bolded 6s with infinity, which is a much more likely starting number of universes.
My favorite resulting sentence is: "Fortunately, 1/∞ of them will still be copies of the home universe, so while the multiverse has been growing dramatically, the chances of hitting home remain slightly less than 1 in ∞." Considering that 1 in ∞ is, for practical purposes, zero, the sliders are, as I've said, fucked up the ass.