[Request] Ok no I didn’t, I’m 30, but I had a math thought. Let’s say I broke into my parents vodka, I think “I can drink 10% of the vodka and then fill the rest with water, mix it all up, and they’ll never know.” I do this, but the next time I think the same thing, “I’ll drink 10%, fill it up with water, and they’ll never know.” Of course, I’m a big dumb dumb and it will taste less and less of vodka. Assuming the water and vodka blended perfectly (aren’t immiscible) at water point does it all become water? Or, does it? Would this be logarithmic?

Moreover (although I guess this is sort of a chemistry question), let’s say they keep their 80 proof (40% alcohol) in the freezer, at what point in my taking of their vodka, does that liquid freeze?

Ye I know it's a movie but what will actually happen(hope you cna use links here):
https://www.youtube.com/shorts/EtS7FQZ4Lcg

way smaller bullet hit centrally bigger bullet (I assume they have same speed)

Edit: 2 duodecillion in rubles. And also, it's debt only to TV channels, which I'm interested in.

Selling Price = (CoGS * (1 + Tariff%)) / (1 - Margin)

I was reading a fiction book (One Piece), and a certain situation that happens WAY LATE into the story intrigued me, and thinking about what could have happened led me to make this weird question.

If there was a hole big enough for the mass and area of an entire island (average) to fall through to the center of the Earth, considering drag in either water or air or magma, how long would it take? Because whenever I search it up, the quickest answer always discounts any resistance from any fluid at all, and example results that do take it into account wouldn't translate well into this particular question.

The Pokémon TCG Pocket game has a mechanic where you are shown 5 cards face down, in which there is usually only one desirable card, and let you pick one at random to keep

That's all well-and-good, but recently they added an event where you get to "peak" at one of the cards before you make your choice.

Now - if the one you peak is correct, you can just pick it, so that math ends there. However, if the card you pick is incorrect, does the logic of the Monty Hall problem apply?

I, and I think most others, mentally decide a card to pick before it's even time (I always pick the bottom right). If I reveal a different card and it's incorrect, is it statistically probable for me to forsake my mental guess and pick one of the other 3 cards? This feels wrong, the game didn't know that was my choice, it feels like it should now be no less likely than the other 3 cards.

However, wouldn't the logic of the Monty Hall problem apply to this, and say it is incorrect? That, logically speaking, my initial probability doesn't change from 1/5 despite the fact another one was eliminate (this is, to be clear, under the theoretical that the 'peaked' card is wrong, as the peaked card being correct ends the scenario). If there was a 1/5 chance my initial guess was correct, there is a 4/5 chance it was wrong. If a card is revealed, there is still a 4/5 chance I was initially wrong, but if there are only 3 possible cards to switch to, they split that 4/5 3-ways, making them each 26.66% likely to be correct (as opposed to my 20%), no?

This is my way of understanding the Monty Hall problem but practically speaking I don't feel like that can be incorrect? The game doesn't even know my initial 'mental' pick, so how could there be a statistical difference if I choose it or swap it.

And if any of the 3 swaps are really 26.66% likely in that scenario, wouldn't that mean mentally envisioning one and then swapping it is actually (very slightly) more likely than peaking at one at random and then picking one of the remaining 4 (25%)? Again, in both these scenarios just not factoring in the possibility that the peak is correct, which should apply to both scenarios equally anyway and not change the end result.

I don't know what the flaw in my logic is but I can't imagine that 26.66% correct. If the 1/5 chance the "peaked" card was incorrect is distributed to all 4 cards then it's 25%, but wouldn't that mean the Monty Hall problem results in a 50% chance instead of a 66% chance?

Can anyone help me break down the probability in this scenario? Is there a flaw in my understanding of the probabiltiy or do you really increase your chances by mentally choosing one and then refusing to take it?

Firstly, please excuse the odd numbering system - it matches up to a spreadsheet that I am using!

I am going away on a golf trip in a week or so's time. There are ten golfers and we are playing three rounds. For each round, we need two groups of three, and one group of four.

I would like to organise the groups in such a way as to ensure that we can all play with as many different golfers as possible. My best current attempt is in the image. I was hoping that somebody might be able to tell me a) whether there is a better solution and b) if there is, either what the solution is, or how best to go about calculating it.

I have attempted to use websites that separate into groups using simpler parameters, but they don't give me anything better than what I have got thus far.

Thanks in advance!

As the title says.

Imagine you had some structure, like a tower or space elevator that extended into space, well beyond the atmosphere (ignoring the difficulties of creating such as structure).

If you were standing on the Earth's surface, and assuming relatively level terrain (we'll say approximately "at sea level"), how far away would you be able to be and still see at least a glimpse of this structure?

Bonus question (not really math, more optics/physics): What would the "top" of the tower look like at the point where you can no longer see it going up? Would it just kind of fade to a "vanishing point", have some clear cut-off, or basically fade as if someone put a transparent gradient at the top?

Thanks.

I don’t mean like flapping their wings but - theoretically - could a wing suit skydiver find some perfect slope that works where they can gain altitude to a point of stall and land safely without a parachute? Alternatively could a glider do this?

Daredevil can hear a gun cock from blocks away.

Is he hearing hundreds of farts per minute all day long?

People are speculating on the expected payments for a huge anti-trust suit against many schools, who colluded to cook the books on financial aid calculations.

So the question is. If there is 200M dollars available, and 200k POSSIBLE claimants, and the claim amount varies by the number of years attended for undergrad... what is the expected claim amount if every possible person submits a claim... for the classes of people who attended for 1, 2, 3, and 4 years of the relevant window.

I think it is a pretty interesting question.

i'm assuming that they meant the flight time between the cities but can't do thing myself cuz i'm buzy laughing my butt off

The weight of each bit is just one electron and one electron weight around 9.1*10^-31.

(69kg)/((9.110^{-31)kg/electron)=7.58}10³¹ electrons

This is also the number of bits, to convert to bytes. We just divide by 8 and get

9.4810³⁰ bytes or 7.8 million yottabytes(8.610¹⁸ terabytes).

This is just an approximation and I was being generous in the calculations, please correct me if i got something wrong, it is very possible since I am just a child and English is not my first language.