Asimov card trivia

[Guest]

Just discovered this in Isaac Asimov's Book of Facts:

'There are fifty-two cards in an ordinary deck. The number of ways in which they can be arranged is just about 80,660,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000. If you had that many decks of cards, each arranged a different way, and if each deck weighed only as much as a single hydrogen atom (the lightest atom), all the decks together would weigh a billion times as much as the sun.'

Btw, I counted the zeros. It equals 80,660 x 10 to the power of 62. As it's an American book, I assume he means American billion rather than UK billion.

[Bill Mullins]

I think that Isaac assumed that every card was face up (or face down). If you allow the cards to be randomly oriented, ala a "triumph" shuffle, then multiply the above number by 4.50 x 10^15.

Big numbers here.

[Jonathan Townsend]

Originally posted by Bill Mullins:
... Big numbers here.

Pinky are you thinking what I'm thinking?

I don't know Brain, but where are we going to get a card index for that many decks?

{ie every possible desired state represented in a deck ready to swithch... great use of classic pass or deck shell ?? }

[Guest]

I make it 8.0658175170943878571660636856404 X 10 to the 67th, but while it's true tht the faceup face down possibilities are 450,359,962,737,0496, we should explain, for anyone who was absent that day in school, that this merely represents 2 to the 52nd power, and that that big number is called 52 factorial, which math guys write like this: 52!

You can do this with a pencil and paper, if you have time. It's 52 X 51 X 50 X 49 .... (et cetera)X 3 X 2 X 1.

[Michael Kamen]

Ahem (for anyone interested). . . If I'm not mistaken we have here 2^(8.07 x 10^67), which according to my prognostications is way larger than (8.07 x 10^67) x (2^52), too large in fact for my TI-85 calculator to evaluate. Maybe that is why John Keats said, "when old age shall our generation waste, only card tricks shall remain, a friend to man to whom thou sayest. . ."

[Mike Powers]

Lets get this right. As stated above, there are 52 factorial (52!) possible ways to arrange a deck. This is because there are 52 possible cards for the 1st position; 51 possible for the 2nd position etc. This comes out to 52x51x50x49x.x3x2x1 possible arrangements. That comes out to be 8.0658 x 10^67 (10^67 is 10 to the 67th power, or a 1 followed by 67 zeros). This is one whopping number.

If you want to get one deck in each possible arrangement and then measure the mass (weight) heres the deal: A deck weighs in at very close to 3 oz. so the weight of all the decks is 3 times 8.0658 x 10^67 = 2.4197 x 10^68. There are 35.274 oz in a kilogram (a kg is about 2.2 pounds). This leads to a mass of 6.86 x 10^66 kg. Our sun has a mass of about 2 x 10^30 kg. Division now tells us that the mass of the decks is equal to about 3.43 x 10^36 suns! To get a handle on this remember that there are about 100 billion stars in our Milky Way galaxy. This is 10^11 stars. Divide 10^11 into 3.42 x 10^36 and find that the mass of the decks equals the mass of about 3.43 x 10^25 galaxies!

This is about 34 million billion billion galaxies worth of mass. Since the mass of our entire universe is estimated to only be around 25 billion times the mass of our galaxy, tt would take the mass of a million billion entire UNIVERSES like ours (all stars everywhere) to make the needed decks.

Check the numbers for yourself. You can use the scientific calculator at: http://www.creativearts.com/scientificcalculator/

Next we can calculate the value of one dollar invested 2000 years ago at an average of, say, 5% interest compounded quarterly. What would it be worth today?

Mike

(BTW - in this analysis all the cards are facing one way. This is how cards are normally used. If you allow them to cover all possible combinations of face up and face down too.... yeow.)

[Guest]

2,391,102,204,613,620,000,000,000,000,000,000,000,000,000

:cool:

[Jonathan Townsend]

Yes Pinkie, this means we're going to vegas wearing cargo pants.

[Randy DiMarco]

Mike
It is good to see that you have found a way to pass the time now that you are retired.

[Bill Mullins]

If it was a one way deck, then each card has any of four different orientations . . .

[Mike Powers]

Once it turned out that more than the entire mass of our universe is required to construct the decks, I gave up on trying to increase the number of decks required! Remember, it takes a million billion entire universes worth of atoms to make these decks. That's more than enough decks for me.

Mike

Hey, if you had enough pockets to house all the decks, you could do "any arrangement of the deck" called for instead of any card called for. Maybe Kenton K can narrow down the number of decks needed...

[Mike Powers]

Hi WarlockDrummer,

I got 1.446 x 10^43 dollars for $1 at 5% compounded quarterly for 2000 years. The formula is FV = P(1+r)^n. Here P is 1 (dollar), r = .05/4 = .0125 (the quarterly rate) and n is 8000 since there would be 8000 quarters in 2000 years. My calculator gives me the number above from 1.0125^8000. Thats a load of money! Lets make it all one dollar bills and stack them up. How high would the pile be? The bureau of engraving and printing says: "Our present sized currency measures 2.61 inches wide by 6.14 inches long, and the thickness is.0043 inches."

Lets get the answer in miles. BTW our galaxy is about 100,000 light years across. (http://www.star.le.ac.uk/edu/mway/) Light travels at 186,000 miles per second so this distance is about: 365.25x24x60x60 sec in a year x 186,000 = 5.8697 x 10^12 or about 6 trillion miles. Thats one light year. Our galaxy is 100,000 light years across which makes it 5.8697 x 10^17 miles across. (It takes 100,000 years to cross our galaxy if you can travel at the speed of light 186,000 miles/second. Forger relativity for the moment.)

O.K. here we go weve collected 1.446 x 10^43 dollars from our $1 investment 2000 years ago. Lets stack the one dollar bills. One dollar is .0043 inches thick so our stack will be 6.2178 x 10^40 inches tall. Divide by 12 = 5.1815 x 10^39 feet tall. Now divide by 5280 ft/mile and get: 9.813 x 10^35 miles tall. But remember our galaxy is only 100,000 light years across. Thats 5.8697 x 10^17 miles across.

Thus the stack of ones would extend back and forth across the galaxy (not solar system the entire galaxy) 1.67 x 10^18 times. Thats across and back and across again etc over a billion billion times. Thats a lot of money.

Aint it fun to be retired!

Mike

[Guest]

Sorry--I used a 5% CAGR...and I believe that you meant a 5% ANNUAL rate (hence the 1.25% quarterly rate).

:cool:

[Randy DiMarco]

Mike
How many Flicker lighters would all that money buy? ;)

[Pete Biro]

:sleep: :sleep: :sleep: :sleep: :sleep:

[Luke]

:help: :help: :help: :help: :help: :help:

[Mike Powers]

Hi Randy,

Flicker lighters are a rare commodity these days. However I think you could buy all that I have in stock (which is pretty much all that were made) for a 6 pack of good beer.

P.S. somebody wake Biro, I think he's sleeping again..

Mike

[Pete McCabe]

Wait! Wait! Sorry!

I forgot to take out the Jokers. We'll have to run all the numbers again.


Pete

[Mike Powers]

Damn, now we're going to need 35 million billion billion galaxies worth of mass.

Mike

[Guest]

How many monkeys would it take overhand shuffling mixed decks and for how long to achieve new deck order?

[Jonathan Townsend]

Originally posted by Mike Powers:
we're going to need 35 million billion billion galaxies worth..

That's the layman mind working...

Magicians may wish to use the quantum mechanical approach and use a QDeck. Which has all cards in uncertain state. The cards each collapse only when turned face up. Then go back to uncertainty when face down again.

Or, could cross connect same deck as exists in MANY universes.

[Guest]

Now I'm thinking that there could be a great
"Schroedinger's Cat" card trick.

:cool:

[Jim Maloney_dup1]

Originally posted by WarlockDrummer:
Now I'm thinking that there could be a great
"Schroedinger's Cat" card trick.

:cool:

Honestly, that's something I've thought about before. I haven't been able to figure out exactly what "Schroedinger's Card" would be like. I'll get it one day.

-Jim

[Jeff Hinchliffe]

Hmm... here's an idea...

Have a blank faced card gaffed by using a red backed card, and use Fako sheets to put the face of the card over the back of the card. You now have a blank-faced card with a red back and the face on the same side. The presentation could be that you have a card selected, and placed into the box, and then the spectator takes the box, throws it up into the air, and lets it hit the ground. At this point, the card could be face-up, or it could be face-down. In essence, it is both face-up and face-down at the same time! Have the spectator remove the card to find the double-faced(??) card.

Just a thought...

Jeff

[Jonathan Townsend]

Originally posted by Jim Maloney:
... "Schroedinger's Card" would be like.

Used to use Mylar backed cards for some such tricks. Wesley may have some good ideas for you. My thoughts go towards Derek's Chameleon card and Hofzinser's routine.

Looking for help using photographs to produce some special cards. Know anyone?

[Mike Powers]

I've had an item called "Schrdinger's Dream" in Top Secret Stuff. It has been updated in my new lecture notes. It's a card through table routine. The connection with Schrdinger is that quantum mechanics allows for barrier penetration...

Mike

[Guest]

Originally posted by Jonathan Townsend:
Pinky are you thinking what I'm thinking?

I don't know Brain, but where are we going to get a card index for that many decks?
}

I believe the technique used in this case is called the Palm of the Almighty.

[Guest]

The technique might be the palm of the Almighty but the Effect would have to be "Out of this Universe"!

[David Neighbors]

Hi Gang,
I see Lots Of People Have Time On There Hands!!! Hay You could All Ways Lurn A Few Coin Tricks! :) !

Best David Neighbors
The Coinjurer