If you got people to listen to every one of those CDs and make a guess as to the musician, the result would be that over 99.99999999% of them would be labelled “Merzbow”.
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Wait! How many atoms are on Earth?!?
First it blows me away that ‘we’ know the number - but does ON Earth mean the Earth itself, everything on the surface, and things in the atmosphere?
Or is it just things on the surface?
Whatever the case; What’s the number?
Also, I apparently totally don’t comprehend the size/weight of an atom becaue Id’ve guessed there were 80,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 atoms in my couch.
I feel duty bound to point out that you can achieve this number of combinations by reordering any 52 objects. They need not be marked hearts, spades, diamonds or clubs.
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Be awful if it did; cut the deck once, and suddenly, it’s all your fault.
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According to astrophysicist Arthur Eddington (1882~1944) there are exactly 15 747 724 136 275 002 577 605 653 961 181 555 468 044 717 914 527 116 709 366 231 425 076 185 631 031 296 protons in the universe and the same number of electrons.
Modern estimates are closer to 10^80…
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Why, if I had a Bitcoin for every way you can shuffle a deck of cards…
Hey guys, keep the game on the right thread…
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Ah yes, I did misunderstand, apologies. Based on the approximation listed on wikipedia of 0.5sqrt(.25-252!*ln(.5)) it is about 10^34 to have a better than 50-50 chance of two shuffled decks landing on the same permutation. The original source of that approximation is at: A Generalized Birthday Problem on JSTOR
Though, it might be useful to note that if you have your own shuffled deck, the probability that any other shuffled deck so far matching your particular deck is actually lower than any two randomly chosen shufflings having already occurred. The approximation for that seems to break down for a very large sample space such as 52!. I went from p(d) = 1-(d-1/d)^n => -.5=-(d-1/d)^n => ln(.5)=nln(d-1/d) => n=ln(.5)/ln(d-1/d)
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Let’s pretend that your couch is made out of only carbon atoms.
Each carbon atom weighs 0.00000000000000000000001994 grams. Your couch probably weighs 500 Kg. So 500000g / (1.994 x 10^-23)g = 2.5 * 10^28 atoms.
So there are
25,000,000,000,000,000,000,000,000,000 atoms in your couch, compared to
80,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 atoms in the world.
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You think my couch weighs 1100 pounds? I’d guess 200 myself.
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If it’s made from solid carbon, it could easily weigh in excess of 500kg
1m^3 of carbon weighs ~3500kg, choosing the most appropriate form of carbon for interior decoration.
A 500kg (or, using the preferred measurement, 2,500 000 carats) sofa would cost appx. $12,500,000,000 assuming diamonds of a reasonable quality, and would be extremely uncomfortable, impractical and fragile, despite looking amazing.
If you bought this a few years back, you could have swapped it for Motorola. 
I don’t know how many atoms are in my couch, but then again it has been stuck in the stairwell ever since I asked Dirk to help me move.
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Treasury Genies should be looking at their bugzilla by now. Seriously, holding back nickel folding for a major revision of nickels?
I thought I’d knock 10 digits off on the way to the end of reading the objects, too, thanks! ttttttttttllllllllllllll::ddddddddddrrrrrrrrrr
Slip out of the pack, Jack?
You don’t need to be coy, Roi?
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The odds that a hand would be duplicated aren’t what we would expect from that number. Think of the bit about how many people must be in a room to have a 50% chance that they share a birthday.
So how many hands need to be dealt to have a 50% chance that the same hand got dealt twice? And how many hands have been dealt?
On the strength of that line alone, I nominate you for winning the internet 
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I enjoyed that line also. It makes me think that black holes may be the result of over-ambitious hedge fund managers. Maybe dark matter is the universal equivalent of bank accounts in the Cayman Islands.
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See above re: birthday paradox. I overcame some precision issues when trying to calculate the numbers as here’s the final outcome:
If one had a pool of previously shuffled decks to choose from, the chances that any pair of those decks are the same rise above 50% would require a sample space size of approximately 10^34. This follows from the issues in the birthday problem.
However, in the context of shuffling your own deck and then checking if your deck matches a deck that was previously shuffled is a different probability and one that requires a much larger sample space to match against to have a higher than 50% probability of finding a match. Given the approximation n=ln(.5)/ln((52!-1)/52!), the required n is still on the order of 10^67 or more than half of the total available sample space.
52! => 80658175170943878571660636856403766975289505440883277824000000000000
n=ln(.5)/ln((52!-1)/52!) ~ 55907986708849934223332477802005437614613669480434146509789214142941
So the nuanced answer is that, if you wanted a 50-50 chance to find any duplicate shuffles, you’d need to search through 10^34 shuffles
If you wanted a 50-50 chance to figure out if a particular shuffle matched another shuffle, it’s still a 10^67 search space.
This makes sense, as it is related to the birthday problem when you want to find out how many people have to enter a room before there is a 50% chance that someone in that group has the same birthday as you do: ln(.5)/ln((365-1)/365) ~ 253
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This all assumes God doesn’t know basic sleight of hand techniques, of course. I don’t know if the deity has a hobby so I can’t say.
