@raganwald The same holds true, of course, for Obergefell. What the Supreme Court gives with one hand, it can take away with the other, as people *should* understand by now after Dobbs v. Jackson if they didn't before.
mjd@mathstodon.xyz
Posts
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Like Trevor Noah, I was born a crime. -
Like Trevor Noah, I was born a crime.@raganwald I think people don't understand what shaky ground this is on, legally. The Constitution doesn't give Congress the power to regulate marriage laws, and marriage laws were traditionally understood as a central example of exactly the sort of thing left up to individual state regulation. The court in _Loving_ had to argue from the Equal Protection clause of the 14th amendment, which is broad but not obviously applicable here. The racist states argued that there was no violation of equal protection because blacks and whites alike were equally prohibited from interracial marriages.
My own marriage would have been illegal in many states before 1967, and I made sure our kids are aware of this.
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Today I needed to write a function that would select a random set of N=5 items, containing at most one special item, from a set of R=24 regular and S=4 special items.This reminds me of the time that Reddit's `r/programmingCircleJerk` described one of my blog posts as “Haskaller too smart to get anything done”.
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And that's how the bunnies invented Bunny-Karate, which is a lot like regular Karate, except fluffier and cuter 😜😂@NickEast_IndieWriter @amenonsen Paw-to-paw combat, surely.
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Today I needed to write a function that would select a random set of N=5 items, containing at most one special item, from a set of R=24 regular and S=4 special items.Today I needed to write a function that would select a random set of N=5 items, containing at most one special item, from a set of R=24 regular and S=4 special items.
I calculated that the number of sets with no special items is \( \binom{R}{N} =\binom{24}{5} = 42504 \).
And I calculated that the number of sets with one special item is \( \binom{R}{N-1} \cdot S = \binom{24}{4} · 4 = 42504 \) — the same, so the probability of getting a set with one special item is exactly ½.
Then I spent a lot of time trying to figure out where my mistake was. Surely not a bug in my venerable binomial-coefficient-computing program?
No, they are equal, and it is a pure coincidence, an effect of the fact that \( R-N+1 = N\cdot S\).
If any of the three numbers had been different, the two groups would have been different sizes.