I sometimes play some random Solitaire type card games before I go to bed. Here’s one that I thought of some years ago (and there might already be some name for it). Basically, it’s a combination of bridge and poker; i.e., poker with four people playing13 cards each:

So. . . Deal all 52 cards to four people.

Person with the highest poker-type hand wins.

It would then appear that the ranking of hands would be:

Full suit (2-A, same suit)

Full straight (2-A, not all same suit).

Then it gets tricky. In poker, the hands are ranked by how mathematically hard they are to obtain. So, a pair is low since it’s common, and a straight flush is high because it’s not.

So these are the various combinations that are possible. In each pair of numbers, the first represents the quantity, and the second, the number of cards in that grouping. In the first case, the lucky person would have three four of a kind, no trips, no pairs, and one miscellaneous.

The GQ: **What I don’t know, though, is how these would actually rank by combinations.** I have no doubt that having 3 four of a kind would trump 6 pairs, but I don’t know that [1-4, 9-x] trumps [4-3, 1-x].

Anyone???

I think these are the possible hands, where “X” represents miscellaneous leftovers that don’t team up with other cards. Normally, I’d like to try to figure this out myself, but this is beyond my knowledge of combinatorials, and I would try to do it by brute force except: There are over 53 octillion (27 zeroes) combinations, and I’m leaving early today to spend the weekend in Charleston.

# 3-4…0-3…0-2…1-X

2-4…1-3…0-2…2-X

2-4…0-3…2-2…1-X

2-4…0-3…0-2…5-X

1-4…3-3

1-4…2-3…1-2…1-X

1-4…2-3…0-2…3-X

1-4…1-3…3-2

1-4…1-3…2-2…2-X

1-4…1-3…1-2…4-X

1-4…1-3…0-2…6-X

1-4…0-3…4-2…1-X

1-4…0-3…3-2…3-X

1-4…0-3…2-2…5-X

1-4…0-3…1-2…7-X

1-4…0-3…0-2…9-X

# 4-3…0-2…1-X

3-3…2-2

3-3…1-2…2-X

3-3…0-2…4-X

2-3…3-2…1-X

2-3…2-2…3-X

2-3…1-2…5-X

2-3…0-2…7-X

1-3…5-2

1-3…3-2…4-X

1-3…2-2…6-X

1-3…1-2…8-X

1-3…0-2…10-X

6-2…1-X

5-2…3-X

4-2…5-X

3-2…7-X

2-2…9-X

1-2…11-X