Safety rules are written in blood. When you spend at least 8hours/day doing the same thing, even if that thing has a very small chance of generating an accident, that’s a lot of time spent doing something risky. Everyone has bad days, any one of those could kill you/severely injure you if you don’t take safety protocols seriously.

You don’t need to have wired across the room. You can put them through the wall like every other cable. If the wire tubes are not full, it isn’t very complicated. I put my Ethernet wires in the wall.

PowerShell, because of autocomplete and shift+arrows select.

Fellow pythonistas, how can I make this code more pythonic?

I’m not sure if it fixes your problem, but it fixed mine.

Pin the tab.

In my case, Whatsapp web didn’t get loaded when I opened firefox, so notifications didn’t reach me unless I opened the tab at least once.

If the tab is pinned, however, it will load when you open firefox, I’m unsure if the tab stays loaded.

Do you need those features? If not, go LTS. LTS means you’ll have to update the distro less frequently than latest.

If you want those features, go non-LTS, there’s no other choice. If you don’t want them, go LTS, it’s less of a hassle.

IPv6 addresses are practically endless, therefore their value is practically 0. ISPs justify charging extra for static IPv4 because IPv4 addresses do have a value.

If ISPs charge for static IPv6, then one of them could just give that service for free (while keeping the rest of the prices the same as their competitors). That would get them more customers while costing them nothing.

EDIT: I can’t give you an example of an ISP that offers free static IPv6 because there are no ISPs in my country that offer IPv6.

50%?? I fucking wish. In Spain we are at 5%. I finally got IPv6 in my phone this year, but I want it in my home, which is still only available as IPv4 even if they’re the same ISP.

In the world of computers, why would remembering numbers be the stop for new technologies?

Do you remember anyone’s public key? Certificate?

I don’t even remember domain (most) names, just Google them or save them as bookmarks or something.

The reason IPv4 still exists is because ISPs benefit from its scarcity. Big ISPs already paid a lot of money to own IPv4 addresses, if they switched to IPv6 that investnywould be worthless.

Try selling static IPv6 addresses as they do now with IPv4. People would laugh at them and just get a free IPv6 address from an ISP that wants to get new users and doesn’t charge for it.

The longer ISPs delay the adoption of IPv6, the longer they can milk IPv4 scarcity.

Favourite? Windows terminal.

Favourite Linux compatible? Vs code integrated terminal.

The one I actually use on Linux because viscose integrated terminal is not standalone? Konsole, just because it’s the default In kde.

EDIT: may I know why I got downvoted? Just because I have an unpopular preference? Is this reddit?

I don’t think I’ve ever noticed Firefox updating. The only sign I get that it updates is that when it does a special tab opens telling me about the new features.

Well, that implies that it is made with that community in mind, not that everyone in that community needs it.

I tried to answer but idk why Lemmy failed to post it, so I’ll make a tldr instead.

TLDR:

Instead of reasoning I used actual statistics equations and you are correct: the chance in the coins case is 1/3.

However, I was misguided assuming that both the “girl and boy” problem and “coins” problem are the same, when in fact they are not.

In the “coins” case, the statement “at least one of them is heads” has a probability of 3/4. In the “girl and boy” case, the statement “the child that opened the door was a boy” has a probability of 1/2.

You didn’t eliminate BG and GB where a girl opens the door though. If you do that, then the answer is 50%. Because you remove half the probability from BG and GB and you remove none from BB.

I know you didn’t eliminate those cases because you said “That leaves us with 3 possibilities with equal probabilities”. That would be false, BB is twice as likely.

You can’t just end the experiment if the randomly chosen child doesn’t “fit the parameters”, by doing that you aren’t accounting for half the girls in the whole event pool. Half of the girls have siblings that are girls.

Being 2 girls was a possible event at the start, you can’t just remove it. This time it happened to be a boy who opened the door, but it could’ve been as likely for a girl to open it.

If it was phrased like “there are 2 siblings, only boys can open doors. Of all the houses that opened their doors, how many have a girl in them?”, then it will be 2/3. In this example, there is an initial pool of events, then I narrowed down to a smaller one (with less probability). If you “just” eliminate the GG scenario, then the set of events got smaller without reducing the set’s probability.

If you simulate it like that, it leads to a contradiction.

According to the problem, 2 coins are flipped, and we all agree that it leads to an event pool of {TT,TH,HT,HH}, where all 4 events are equally as likely.

In the simulation, however, you just ignore the “TT” situation, which leads to a total event pool of {HT,TH,HH}. Where all events are equally likely.

The way the problem was phrased was “2 coin flips happen, and I have a machine that tells me either if both are Tails or not, this time it turns out that there is at least 1 heads”. But the way you simulated it is “I will make coinflips until I have at least 1 heads”.

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Again, you are assuming that every occurrence has the same chance. When in fact, they have not. There are 3 random events happening here:

1. Flip of one coin (50% chance each)
2. Flip of the other coin (50% chance each)
3. The coin that you told me (let’s say it’s 50% nickel 50% dime for simplicity’s sake)

Also, I am assuming that these 3 events are completely unrelated. That is, the result of a coin flip won’t determine whether you tell me the nickel or dime. A complete list of events is as follows:

T T N

T T D

H T N

H T D

T H N

T H D

H H N

H H D

After telling me that one of them is heads, the list is as follows:

H T D

T H N

H H D

H H N

H H is 50% chance, and the sum of HT + TH is the other 50%

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You assume that the probability of TH = HT = HH

When In fact, the probabilities are as follows:

P(HT)+P(TH) = 50% P(HH) = 50%

For all the probabilities being equal, you’d have to consider 4 cases:

HT, TH, HH (1) and HH (2).

The difference between HH (1) and HH (2) is which one you were told that was heads.

Then P(HH) = P(HH (1)) + P(HH (2)) = 2/4 = 50%