9 results for “mapologies”
Apologies to anybody running a Fediverse instance in China, but this morning I've added ASN blocks for
- CHINA UNICOM (57gb)
- China Mobile (30gb)
- Chinanet (17gb)
- Tencent (135gb)
Traffic volume is the last 14 days and accounting for more than 50% of total
As huge numbers of IP addresses from these ranges were just constantly hammering my instance and has burned through my normal monthly usage in 10 days
I'll blog about how I did this a little later
Letra da música “All Apologies” de Nirvana
#Nirvana #AllApologies
https://daletra.com.br/nirvana/letra/all-apologies.html
2/2 My apologies! I forgot to update that the #puppies of #PuppyLive were born early hours on Saturday, and you can watch them here - where ever you are!
https://areena.yle.fi/1-78104441
In #Finland we live stream #dogs getting puppies, in #USA they live stream people getting executed.
EDIT: Article in English: https://yle.fi/a/74-20251119
#koira #koirat #dog #hund #hundar #chien #chiens #dogstodon #mastodogs #dogsoffedi #dogsofmastodon #pentu #pennut #valp #valpar #puppy #chiot #chiots #PentuLive
Today’s comic. With apologies to Carole King and the Chiffons.
#inktober #inktober2026
Which one are you? 1, 2, or 3? 😅
Three dad jokes, zero apologies.
Don't Panic. You're just Dadding.
Send this to a dad who tells all three 👀
#dadjokes #dadjokesdaily #dadlife #parentinghumor
Analyst Connor Orr apologizes for his comments on Steelers Offense: “I watched this Steelers offense, and I have nothing but apologies for the fans. It was a blast. It was fun to watch. This team had no reason to be good. Ben Skowronek went down. Michael Pittman Jr.’s down. They’re down to four healthy wide receivers, and that was the most fun-looking the offense has been all year. The defense cut Jonathan Taylor off at the knees. If you give me this team every Sunday, an absolute blast. So, I’m sorry. I’ll say it again. I’m very, very sorry.” #Steelers #NFL
Sorry we made you grind LeetCode
https://www.hackerrank.com/writing/why-were-sorry
Comments: https://news.ycombinator.com/item?id=50050762
#HackerNews #LeetCode #HackerNews #TechIndustry #Apologies #CodingCommunity #DeveloperExperience
Where to find #FrightClub tonight!
#SnakeOuttaCompton (10ET/9C/7P)
Tubi: tubitv.com/movies/48642...
ty @[email protected] for the pick!
(Our apologies. Only on Prime w/third-party service.)
#Cooties (11:30ET/10:30C/8:30P)
Tubi: tubitv.com/movies/34809...
YouTube Free: youtu.be/W7eWOFGIT8s
Snake Outta Compton (2018)
@skewray (2/2)
An example more relevant to your proposed extra rule is the following. Take a 3D spherical octant, with radius r, latitude φ, longitude λ as coordinates:
r ∈ [0, 1], φ ∈ [0, π/2], λ ∈ [0, π/2]
and the probability density
p(r, φ, λ) dr dφ dλ = 2 cos(φ)/π dr dφ dλ .
Let's take the 1D set B defined by {φ = π/2} (line from centre to north pole), and the 2D set A defined by {λ = 0}. These sets satisfy your requirement B ⊆ A. It is easy to verify that both have zero measure.
What's the value of p(A|B)? A priori undetermined. Let's calculate it through some limit. For instance, we could define B as the limit of the region π/2 - ε < φ ≤ π/2, and A as the limit of the region 0 ≤ λ < δ. We find
p[A(δ) | B(ε)] = p[A(δ) ∩ B(ε)] / p[B(ε)] =
p(0 ≤ r ≤ 1, π/2 - ε < φ ≤ π/2, 0 ≤ λ < δ) / p(0 ≤ r ≤ 1, π/2 - ε < φ ≤ π/2, 0 ≤ λ < π/2) =
2 δ/π
and in the limit δ→0 we get p(A|B) = 0.
You may object that this particular choice of limit way of defining B and A is arbitrary; and you're completely right. We could use other limiting ways, and you can check that you could get other limit values for p(A|B), also different from 0 or 1.
This is exactly the point: what's "arbitrary" or not depends on the specific context. In some applications the limits above can make sense, and therefore p(A|B) = 0 makes sense there; in other applications, they may be unnatural. This is why I like that p(A|B) (with B ⊆ A, both zero measure) is a priori undefined; it means "please provide more information or a context, to get a definite answer".
Apologies if my discussion sounds patronizing, that's really not my purpose! The reason why I got engaged with your initial post is that I was also, for quite some time, completely unsatisfied with the indeterminacy of conditional probability in situations like these. I didn't get why it should be so, since in some cases some definitions seemed natural to me. So I get where you're coming from! :) One funny thing is that in the case you're considering, my intuition told me "p(A|B) = 0" was "right" instead. It's when I started to do concrete applications that I realized things can go in very different directions, for very natural reasons in every specific application (shock waves, shells in continua, condensation points, sets in manifolds with generic Riemannian metrics...), that's when I learned to appreciate the indeterminacy, as it was warning me "get more info first".