cards | RobertLovesPi.net

There are 52 rotating polyhedra below. Starting right now, 52 of anything can be called a “deck” of that thing. The derivation of this term is the the number of cards in a standard playing-card deck. The deck is the smaller “cousin” of the mole, or 6.02 x 10²³ of anything, from chemistry, as well as the “big brother” of the dozen, or twelve of anything. All three units are measures of specific quantities, and can be applied to any objects, at least in principle. One never encounters a mole of people because there aren’t that many of us, but a mole of people can still be imagined. The same thing applies to a deck of Earths.

This deck of virtual, rotating polyhedra was created using Stella 4d, a program you may purchase here. If you would like to see a larger image of any one of these polyhedra, simply click on it.

6 hexagons 12 octagons 8 hexagons 24 each of two rectangle types, and four dozen each of two triangle types 170 faces in all

non-convex polyhedron using reg triangles and pentagons 2

polyhedron featuring 12 regular octagons

non-convex polyhedron using reg triangles and pentagons 3

non-convex polyhedron using reg triangles and pentagons

Zonohedrified Icosidodeca stellated a lot

v f based zonish icosahedron stellated many times

pyritohedral version of Kepler's stellated dodecahedra

392 faces including pentadecagons and heptagons

Convex hull of trunc icosa and dual compound dual a twice-truncated tetrahedron

a particular twice-truncated dodecahedron

dodecagons and heptagons and other facesl

1412 and dual has 780 and both have 2190 edges

20 enneagons and 60 each of two types of pentagons

60 heptagons 20 enneagons and 60 triangles

bigslick

For one card, this is easy: the odds are one in thirteen, for there are four aces in 52 cards, and 4/52 = 1/13.

With a second card drawn at the same time, we must consider the 12/13ths of the time that the first card drawn is not an ace. When this happens, 51 cards remain, with four of them aces, so there is an additional 4/51sts of this 12/13ths that must be added to the 1/13th for the first card drawn.

Therefore, the odds of drawing at least one ace, in two cards drawn from a standard deck, are 1/13 + (4/51)(12/13) = (1/13)(51/51) + (4/51)(12/13) = (51 + 48)/[(51)(13)] = 99/663 = 33/221, or 33 out of 221 attempts, which is as far as the fraction will reduce. In decimal form, as a percentage, this happens ~14.93% of the time.

If I made an error above, please let me know in a comment. I do not claim to be infallible.

[Image credit: I found the image above here.]