Let's look at how Super Mario Bros. 3 uses random numbers to influence how the roulette bonus game works, and how poorly it shuffles cards in the card matching bonus game.
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On the front cover of The Lara-Su Chronicles Beginnings by Ken Penders (top-right)
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in reply to @rgmechex's post:
My favorite part was showing how the eight card patterns that actually exist come to be, it's very funny how ubiquitous the assumption (including me before this video) that those eight patterns are hardcoded is
When I was younger, we had the Nintendo Power guide that just laid out all the possible grids, and then the trick was just to take the top corner, and the second from the left on the bottom row, and compare to the grid. I still do this by instinct whenever I play even though I've long forgotten all the layouts.
The roulette is insidious. Even though there's no real reason to believe it's coded "fairly", there's something about almost always getting the first two-thirds of the star if your timing is good that tells your brain "just do that again and 5-Up is yours". And then it usually fails, and you wonder if you missed the window, or you're supposed to be hitting the button at a different moment.
Yeah, so the card-matching game, assuming the shuffles are actually random, has only 1215 possible positions, all of which are accessible by random rotation + random triple swaps once. The issue is that the group generated by rt₀ = t₁r, rt₁ = t₂r, rt₂ = t₃r, rt₃ = t₄r, rt₄ = t₀r, t₀t₁t₂t₃t₄ = r⁵, t₀³ = ε, and the commutators of the ts is just the same group as <r, t | [t, t^r], [t, t^r²], t³, r⁵t>.
In other words, the problem is that the permutations (123456789ABCDEF), (16B), (27C), (38D), (49E), (5AF) generate a much smaller group than you thought, because any product of them can be written with 0-4 copies of the first followed by 0-2 copies of each of the remainder, in that order.
In other other words, the idea is that triple swap followed by rotate is the same as rotate followed by different triple swap.
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