Fiendish 52-card combinatorial query

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.

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

If you don’t mind I’ll leave it to you to to do the actual computation, but I’ll set them up for you. Assume we have q quads (four of a kind), t triplets p pairs and s singles. Notationally I’ll use <N;k> to be N choose k = N!/((N-k)!*k!).

Step 1: Determine which ranks fall into each category. First you to choose the ranks of the q quads. There are **<13,q> **ways of doing this. Then you need to choose the t triplets. You have already taken q of the ranks out of contention, so there are 13-q ranks to choose from. So the number of ways to do this is <13-q,t>. Similarly after taking out the fours and triplets you have 13-q-t ranks to choose from for the pairs so <13-q-t,p> ways. Finally you have <13-q-t-p,s> ways of choosing the singles. So as a final total you have <13,q><13-q,t><13-q-t,p>*<13-q-t-p,s> ways of choosing the ranks.

Step 2: Choosing the suits. There is only one way to choose the suits of the quads, but there are <4,3>=4 ways of choosing suits for each of the triplets,** <4,2>=6** ways of choosing suits for each of the pairs, and <4,1>=4 ways of choosing suits each of the singles. Thus there are *4^(t+s)6^p ways of choosing suits.

In all cases the denominator is <52,13> so for each case the final probability is
**<13,q><13-q,t><13-q-t,p>*<13-q-t-p,s>4^(t+s)6^p/<52,13>

Incidentally while you are correct that the least likely hand is the straight flush (only 4 ways), the full straight doesn’t come in second. There are 4^13=67,108,864 ways of getting a straight, but only <13,4>94=25,740 ways of getting a 3-4…0-3…0-2…1-X hand.

Other interesting things may also come out such as two pair being more likely than one pair.

If I were doing this, I would just pick the best five card hand out of each 13 cards and use the usual rankings. I’m not certain this is correct, but you keep stats to see how it works. I once worked out rankings for 7 card hands, but never used them. For one thing, a full house had to be four of one and three of the other. No 3-2-1-1, although 3-2-2 was a ranked hand. So the first thing you have to work out is what you consider a ranked hand. Then it can be calculated.

Incidentally, in three card poker, straights outrank flushes. IIRC, sets outrank straight flushes. I don’t know what happens in four card poker (not that I’ve ever heard of it as a game, but I once played 3-stud—a down card, an up card, a round of betting, then a last down card and another round of betting—quite a bit).

OT but my grandmother taught me a solitaire she called “Idiots Delight”. You shuffle the deck and turn over the first card saying “Ace”, the second card saying “two” and so on until the card you are turning and what you are saying are the same thing. You lose. I don’t know if I’ve ever beaten the game. How would you figure the odds on that?

I don’t know the precise answer to that question, but I do know that if you shuffle an arbitrary-sized deck of cards, the expectation value of the number of cards in the “right” place is always exactly 1.

Assuming all turns were independent, you would expect the probability to be (12/13)^52=0.0155 or about 1 in 64.

However since the cards are drawn from a finite set, there is probably some correlation or anti-correlation between the success of each draw. It probably very minor though, so I bet 1/64 is pretty accurate.

They’re not independent, but they are exchangeable, so that should make the computation a lot easier.

So do the letters q, t, and p stand for the number of cards in the rank (as in a quad is simply 4), the total number of cards for all ranks (thus, two quads would equal 8)? I can handle the math, but I’m not clear on the substitutions. But thanks for the formula.

What do you mean by “a ranked hand”? If n-card poker is going to be an analog to 5-card poker, then all hands have to be ranked. There are no sets of 5 cards that aren’t a poker hand of some sort, even if it just comes down to who has the highest extra cards.

Are you suggesting that 3-2-2 gets a name (uh, full house and granny unit), while 3-2-1-1 is just as worthless as any set of cards, and is ranked only by the high card? If so, that doesn’t fit with poker, either. Any pair or triple of cards actually matters, and puts the hand above any hand without one of those. So 3-2-1-1 should get credit for the three of a kind and the pair, even if it scores worse than three of a kind and two pair.

OK, despite the handicap of having had a public school education in Possum Hollow, SC, I think I have the solution. What was interesting is that two different sets of hands actually tied for probability. (Actually, this shows number of combinations, but the rankings would be the same.) Again, thanks for the assist. I can rest easier.
Straight Flush…4
3…0…0…1…11,440
1…3…0…0…183,040
2…1…1…0…205,920
2…1…0…2…2,471,040
2…0…2…1…5,559,840
0…4…0…1…6,589,440
1…1…3…22,239,360
1…2…1…1…29,652,480
0…3…2…0…29,652,480
2…0…0…5…36,900,864
2…0…1…3…39,536,640
Straight…67,108,860
1…2…0…3…105,431,040
1…0…4…1…266,872,320
0…1…5…0…320,246,784
0…3…1…2…632,586,240
1…1…2…2…711,659,520
1…0…0…9…749,731,840
0…3…0…4…984,023,040
1…1…0…6…1,180,827,648
0…2…3…1…1,423,319,040
1…1…1…4…2,214,051,840
0…0…6…1…2,241,727,488
1…0…3…3…3,321,077,760
0…1…0…10…3,598,712,832
0…0…1…11…3,925,868,544
1…0…1…7…5,060,689,920
0…2…0…7…6,747,586,560
1…0…2…5…7,970,586,624
…2…2…3…13,284,311,040
…1…4…2…14,944,849,920
…2…1…5…21,254,897,664
…0…5…3…35,867,639,808
…1…1…8…40,485,519,360
…0…2…9…40,485,519,360
…1…3…4…79,705,866,240
…1…2…6…106,274,488,320
…4…5…119,558,799,360
…3…7…121,456,558,080
Total…635,013,559,600

Could you explain your notation, there, Earl? Particularly the twinned hands: What’s a 1…2…1…1 , and what’s a 0…3…2…0 ?

Number of: Quads, trips, pairs, singles, and the total combinations. I plugged the formula above into Excel (except for the straight and straight flush), and coincidentally, the total of those hands match the total for 52 choose 13 combinations.

OK then, so your results say that a quad, two trips, and a pair is just as likely as three trips and two pairs. The way I like to compare hands is to start from the intersection of them: That is, if you look at the cards one at a time, what’s the most you could get where you could still get either of those hands. In this case, that’d be three trips and a pair, plus one other card, or in your notation, 0…3…1…1 . Now, I look at how many cards can complete that to either of the two hands in question. To convert that to 1…2…1…1 , I need to match one of the trips. Each trip has only one card remaining that would match it, and there are three trips, so I have three outs to make that hand. On the other hand, if I wanted to convert it to 0…3…1…1 , I’d need to find a match for that lone card. There are three matches for that card in the deck, so again, I have three outs. Since I have the same number of outs for each hand, the odds of the two hands are the same, just as you found.