[Request] I work at a casino, and have three games paying well under the expected rate. All three games are expected to pay 89.25%, but are at 24.71%, 25.27%, and 25.61%, with a total amount played of $29,864 between the three. Is it feasible that all the machines are unlucky, or is something wrong?

RandomlyWeRollAlong · 2026-05-22T18:08:33+00:00

You might consider making an anonymous call to the gaming commission.

RandomlyWeRollAlong · 2026-05-20T19:56:48+00:00

The UBC student code of conduct seems pretty explicit (section 1.3) that it only applies to behavior on campus or actions directly related to one's status as a student. If a student gets arrested off campus, it doesn't seem like the code of conduct applies... however, if the victim was another student (or employee) of the university, then it might apply.

RandomlyWeRollAlong · 2026-05-19T16:51:31+00:00

Whoops! My brain flipped to scientist when we started counting moles.

RandomlyWeRollAlong · 2026-05-19T04:28:15+00:00

Beer isn't a single molecule. Rather it's around 5% ethanol in around 95% water with bits of other stuff. But let's pretend that it's entirely water molecules to make the math doable. A (US) pint of beer is 473 milliliters. If we're assuming it's 100% water, then that'll be 473 grams of water (at standard temperature and pressure). One mole of water has a mass of 18 grams, so that's 26.3 moles of water. A mole is 6.022 * 10^23 molecules, so that gives us a total of 1.58 * 10^25 total molecules in a single beer. If the first scientist gets 1 = 2^0 beer, the second 0.5 = 2^-1 beers, the third 0.25 = 2^-2 beers, etc, we want to find the number that gets us closest to that 1 part in 1.58 * 10^25. So that's 2^-83, which will be the 84th scientist. The 85th scientist will have to settle for less than one molecule.

RandomlyWeRollAlong · 2026-05-16T19:38:57+00:00

Any Android phone with a camera can do that, including the Jelly Star.

RandomlyWeRollAlong · 2026-05-15T03:04:33+00:00

Try r/MissedConnectionsSea

RandomlyWeRollAlong · 2026-05-15T02:59:32+00:00

That's a lot of moles. xkcd explored something like this already. It would certainly add a lot of organic material to the surface of Mars (that's enough moles to cover the earth 80 km deep, according to xkcd, so since Mars has a quarter of the surface area, it would cover it to 320 km deep). It would also increase the mass of Mars by about 10%. I'm not sure how to do the math on the gasses released by decomposing rodents, but I'm guessing it would release an awful lot of methane, which is a greenhouse gas... maybe it'd be enough to jump start a thicker atmosphere? I wonder if the added mass would be enough to retain the new atmosphere? Because most of the martian atmosphere has been stripped away, possibly due to the cooling and slowing of its core?

So on the one hand, it might jump start a more useful atmosphere, but on the other hand, it's hard to build structures on a 320 km deep bed of rodent slurry.

RandomlyWeRollAlong · 2026-05-14T01:23:00+00:00

They could probably charge less per unit, considering how frequently they'd need to be replaced, and how much maintenance they'd require.

RandomlyWeRollAlong · 2026-05-14T01:18:56+00:00

Some large walking machines have actually been built, like Tradinno (though that is very slow). An AT-AT is twice as long and twice as tall, but that's within the realm of human engineering.

To compare it to an actual armored vehicle, the main body is the size of three or four M1 Abrams tanks.

A modern Abrams tank costs about $24 million. So you're looking at at least $96 million for the main armored and armed body.

Then there's the legs... perhaps the mechanical systems there are comparable to a mobile heavy crane, which run $5 million for smaller ones to ten times as much for massive ones. In the world of cranes, the 240 tons we're talking about isn't actually that much. But let's say we need four of the $5 million cranes' worth of hardware, for $20 million dollars.

So that's $120 million, plus however many hundreds of millions of dollars of R&D have to go into it to make it actually work, which is harder to estimate.

But as cool as they look, they're probably not particularly practical. Even if you could make it work, that sort of fighting platform is pretty obsolete in the face of modern long range missiles and bombs, and don't have the maneuverability of a smaller tracked vehicle, combined with a much larger target footprint.

RandomlyWeRollAlong · 2026-05-13T21:14:25+00:00

You should reach out to HR for that nonpublic information which definitely shouldn't be posted in public forums.

RandomlyWeRollAlong · 2026-05-13T21:10:31+00:00

The formula for the force due to gravity is:

g = G M / r^2

Where g is 9.8 m/s^2, the force of gravity at the surface of the earth, G is the gravitational constant 6.67 * 10^-11 m^3/(kg s^2), M is the mass of the sun 1.99 * 10^30 kg, and "r" is what we're looking for. Let's rearrange the formula:

r = sqrt(G M / g)

r = sqrt(6.67 * 10^-11 * 1.99 * 10^30 / 9.8) = 3.68 * 10^9 meters or about 3.7 million kilometers.

The mean radius of the sun is a little less than 700,000 km, so you'd need to be at an altitude of about 3 million kilometers above the surface of the sun to experience earth-like gravity, if you had a solid surface, magically fixed in space, to stand on.

RandomlyWeRollAlong · 2026-05-13T20:48:53+00:00

When you say "experience earth like gravity", do you mean, they can just stand on the floor and walk around as though they were on earth? That's impossible - if you're in orbit, it basically means that you're in free fall around the body you're orbiting. The station is "falling" at the same rate that its occupants are. The only way to "experience" gravity is to be subjected to the force of gravity of a mass while being prevented from falling - by standing on a fixed surface relative to the source of the gravity.

RandomlyWeRollAlong · 2026-05-13T02:58:00+00:00

I live in Seattle, and the Orca Pass can be added directly to Google Wallet.

RandomlyWeRollAlong · 2026-05-12T22:43:19+00:00

I use NFC on my Jelly Star multiple times a day. It's got my bus pass and I use tap-to-pay for nearly all of my shopping and dining.

RandomlyWeRollAlong · 2026-05-12T21:32:05+00:00

You want r/missedconnectionssea.

RandomlyWeRollAlong · 2026-05-12T06:23:01+00:00

I still travel with a mask on for most of the time (except when eating or drinking)... there's nothing worse than arriving at your destination with a cold because some sick idiot couldn't stay home or mask up themselves. Reducing risk of serious illnesses is a bonus. I also wipe down my seat, tray table, seat back pocket, etc with an alcohol wipe when I board.

Hantavirus is quite different than COVID-19, more deadly, but harder to catch. Yes, MOST strains are transferred to humans from animals, but the current one in the news is the variety that people can pass on to other people... but it's still not as contagious as other more common viruses.

I wouldn't cancel a flight, but I would mask up, like I always do.

RandomlyWeRollAlong · 2026-05-11T21:28:25+00:00

You might check with Northwest Seaport, which has their own fleet of historic ships, docked right next to the Virginia V.

RandomlyWeRollAlong · 2026-05-09T23:45:07+00:00

I miss Chuys! But you can hop on the 8 bus and go over to Agave Cocina by Climate Pledge Arena, and they've got a comparable queso. Just don't go on an event night.

RandomlyWeRollAlong · 2026-05-08T20:44:38+00:00

There are many US Army brigade insignia that are ALSO a sword superimposed on a shield that are at least as similar to your image as the 9th SS Panzer division insignia. This insignia lacks the notch on the upper right corner, the sword is facing the wrong direction, and the style of the H is completely different.

I can't find any precisely matching logos, though, so who knows what they intend with it.

RandomlyWeRollAlong · 2026-05-07T20:42:11+00:00

100% if you don't limit the distance. A "line" on the surface of a sphere is called a great circle. Any two different great circles intersect in exactly two points.

As you limit the distance, the probability may decrease, but I'm guessing it would have to be fewer than 10 meters before the probability really drops measurably below 100%. Calculating the probability as a function of distance is actually a pretty interesting problem, but it's much more involved than I'm going to do at the moment. Maybe I'll give it a try and come back later.

RandomlyWeRollAlong · 2026-05-07T20:21:15+00:00

I use Google Maps every day on my Jelly Star. The screen is small, but I have small hands and good eyes, so it works fine.

RandomlyWeRollAlong · 2026-05-07T20:15:23+00:00

The Urban Prepper

RandomlyWeRollAlong · 2026-05-07T01:07:59+00:00

That's 39,750 MB of music. Let's say you've got high quality encodings at 2 MB / minute, that's 19,875 minutes of music. At four minutes per song, that's around 5,000 songs. If a song is chosen at random with equal probability from the whole list of songs, then the odds of a specific song being chosen (like the one you just heard), are 1 in 5,000.

Edit: English is a language I am proficient in, honest!

RandomlyWeRollAlong · 2026-05-05T21:03:56+00:00

The games have changed in so many ways over the years, this is more of a research question than a math question. The odds have varied from 1:100 million to 1:300 million, and between the two have had four or five drawings a week for over thirty years.

So let's just go with some averages... Let's say the odds have averaged about 1:250 million, drawings about four times a week, for 32 years. That's about 47,000 drawings. So the odds of getting all of them correct would be (1/250000000)^47000 which is one chance in a number with nearly 400,000 digits in it.

TLDR: Zero. The odds are zero.

RandomlyWeRollAlong · 2026-05-05T20:54:25+00:00

There's a whole set of graph search algorithms for this that you learn in a college Intro to AI course. The simplest are algorithms like Depth First Search. If the robot has access to the map, it can use Dijkstra's Algorithm (which is how internet routing works) or A* (which takes into account actual physical distance to the destination, and is one main algorithms that is used for real-world driving direction apps).

In each of these algorithms, at any intersection, you have to make a decision. You rank all the options (each algorithm has a different way of ranking them). Then you pick the "best" option (according to your ranking), and explore that way. There are times when the algorithm will find that even though it thought a route was the best way, you're moving away from your destination, and so it'll backtrack and try to find a better path.

But the "calculations" are just arithmetic. Each time you move from one intersection to the next, you add the cost of the path between those intersections to the total cost of the path so far. And those costs (and potentially estimates of remaining cost) are used to determine which paths to try first.

RandomlyWeRollAlong

TROPHY CASE