The Dirty Dozen - SOLD
 

Not that dirty but predominantly mid grade

Attachment 339482

Mueller, Bolling & Schmitz - All EX-MT 6

Attachment 339483

Kuenn, Keely & Trice - All EX 5

Attachment 339484

Jacobs, Kinder & Galan - 5.5, 3.5 & 3.5

Attachment 339485

Smith, Antonelli & Fain - 2, 1.5 & 60 (5)

Would prefer to sell as a lot and will include a raw Boyd to make it a Baker's dozen.

$65 plus actual shipping for the Baker's Dozen.

Will entertain offers for individual groups of cards.

I believe 10 (or eleven with Boyd) will fit in a small flat rate box.

Smith and Antonelli could be cracked to ship in the small flat rate box, which would result in lower shipping cost for the whole lot.

Will wait 48 hours for a complete lot sale, before entertaining smaller lot offers.

Will respond to smaller lot offers in the order received.

Will crack slabs to reduce cost and shipping.

Have at it.

How many combinations of 5 cards could be made from this group of 12?

I believe it's 792 combos. Statistically, 12 choose 5, which would be 12!/[(5!)(7!)].

Indeed. Dividing by 7! accounts for only choosing 5 items of the 12, and dividing by 5! accounts for the fact that within the group of 5 sequence doesn't matter.

I appreciate the responses so much, having majored in Mathemagic, that I will pose another question.

How many combinations of the 12 cards shown can be made to form a Baker's Dozen?

Answer correctly and you can purchase all 12, or is it 13, for $40 shipped.

Please answer only if you intend to purchase.

I’m gonna go with this answer. It depends on whether you are seeking combinations or permutations.

Permutations = 13! (13 - 12)! = 6,227,020,800
Combinations = 13! 12! × (13 - 12)! = 13

Best
SGL

Not quite right.;)

If I were to answer I would answer 0.

None, since a bakers dozen is 13 and there are only 12 cards available...

Well done gents. I reversed the numbers. I concur

I'm 0 for 1 picking the home team to win in the Wild card games. So are you, both in the wild card games and in answering my question.

Read the first post again and then the question again. Second chance answers are allowed.

I stand by my answer. No combinations of 12 cards will yield a group of 13 cards. I reject in advance whatever weirdness you're going to come back with. :eek:

You're free to stand where you want.

There may be some word play here but if so I do not see it. Like Peter, I stand by my answer, though it is not the one you are seeking

Still a fun thread!!!

My next guesses
1
3.14159
12
13
52 (the number of cards in a deck don't you know)
2,673,186 (with Frank anything is possible)