Baseball just completed theirs — congrats to the Chicago Cubs, as I eat my previous words — and Poker seems to have one every time I turn on my TV. Now is the time for Bridge to enter the post-season, so grab yourself a cold beer and take a front-row seat.
Please direct your attention to the infield as I throw out the first pitch.
Bridge and poker are vastly different games but share some common terms: open, call, raise, pair, to name a few. Another connection is the enduring pastime of selecting the best 5-card poker hand out of a 13-card bridge hand, and that is the subject of this puzzle.
There are 7462 distinct poker hands, most of which can never be best in a bridge hand. For example, every “one pair” is impossible, because the remaining 11 cards must form another pair, three-of-a-kind or a straight.
Test your bridge-poker ingenuity on the following seven questions. Choose the more likely poker hand, or ‘Equal’ if chances are the same. Most answers can be determined logically, but hey! It’s multiple choice, so a good guess scores just as well.
1. Which straight flush is more likely to be best in a bridge hand? A. QJT98 B. T9876 C. Equal
2. Which four-of-a-kind is more likely to be best in a bridge hand? A. 6666K B. 5555K C. Equal
3. Which full house is more likely to be best in a bridge hand? A. AAA88 B. KKK88 C. Equal
4. Which flush is more likely to be best in a bridge hand? A. AQT98 B. KQJT8 C. Equal
5. Which straight is more likely to be best in a bridge hand? A. KQJT9 B. JT987 C. Equal
6. Which three-of-a-kind is more likely to be best in a bridge hand? A. KKKAQ B. 333AK C. Equal
7. Which two pair is more likely to be best in a bridge hand? A. KKQQ9 B. QQJJ9 C. Equal
Contest tiebreaker: You should guess wright on this…
8. How many of the 7462 poker hands can be best in a bridge hand? A. 1492 B. 1903 C. 2573 D. 2576 E. 2729 F. 3184
Quit
Congratulations to Charles Blair, Illinois, who besides tying for the winning score, was the only person to hit the tiebreaker on the dot. For the past 16 years, Charles has been my most frequent participant, also winning the 2010 puzzle Two-Way Finesses, and three play contests (2003 2004 2005) against huge fields. Probability theory is his forte, which I can attest first-hand from trying to solve some of his own tricky posers over the years.
Also submitting the winning score were Tim Broeken, Netherlands, who was just two points short in the tiebreaker from achieving his 11th puzzle-contest win; and Jean-Baptiste Courtois, France, our new kid on the block.
Ranking is by the number of correct answers (ABBBCCC perfect), closest to the correct tiebreaker, and date-time of entry, in that order of priority. In the original contest the tiebreaker was not multiple choice, so solvers had to enter that number.
Determining the number of bridge hands where each poker hand is best is a daunting task, as there are 635+ billion bridge hands, and each must be evaluated to determine its best poker hand.* I have done this for verification, but it isn’t really necessary for comparisons. Probability logic can deduce the more likely best hand with little calculation (except Question 4 required some math).
*Actually, only generic patterns need to be examined, since suit identity is irrelevant in poker. Each unique generic hand must occur 4, 12 or 24 times via suit permutations. This reduces the task to about 62 billion evaluations, but daunting nonetheless.
The easiest way to approach this is to consider each poker hand as the first five cards dealt. (Obviously this will rarely be the case, but it’s theoretically sound for probability analysis.) Next consider which cards can be dealt to spoil the original five cards from remaining the best poker hand. The hand less likely to be spoiled is more likely to be best.
Only one specific card in the same suit (king or jack, respectively) will spoil either, so in that regard they’re equal. The decider is the possibility of a straight flush in another suit, where T9876 will be spoiled by JT987 or QJT98, while QJT98 will be unaffected. Since QJT98 is less likely to be spoiled, it is more likely to be best. Actual bridge hands: QJT98 (1,043,425,500) T9876 (1,043,188,380) or a difference of 237,120.
Tim Broeken: There are cases with two different straight flushes, which gives QJT98 an edge.
Either hand is spoiled by an ace (kicker) so equal there. The difference seems to be that 5555K is spoiled by four sixes, while 6666K is not spoiled by four fives. True, but there is another subtle difference in straight flushes: 6666K is spoiled by a suited T987, 9875, 8754, 7543 or 5432 (5 cases); 5555K is spoiled by a suited 9876, 8764, 7643 or 6432 (4 cases). Note that 5555K is not equally spoiled by a suited 432A, because the ace alone already spoiled it. Since there are four ways to get four specific cards in the same suit, and only one way get four sixes, 5555K is less likely to be spoiled and more likely to be best. Actual bridge hands: 5555K (428,222,052) 6666K (427,608,560) or a difference of 613,492.
Tina Denlee: 5555K is hampered by four sixes (1 chance); 6666K is hampered by an extra straight flush (4 chances).
With regard to rank sets, there is no difference: KKK88 is spoiled by the case king or a pair of aces; AAA88 is spoiled by the case ace or a pair of kings; either is spoiled by a pair of nines thru queens, or any four-of-a-kind. The difference exists in the formation of a straight flush around the top card: KKK88 is spoiled by a suited AQJT, while AAA88 is spoiled by a suited KQJT or 5432; hence KKK88 is less likely to be spoiled and more likely to be best. Actual bridge hands: KKK88 (2,016,396,772) AAA88 (2,012,610,080) or a difference of 3,786,692.
This was the most difficult to decide. As far as rank sets, there is no difference; each consists of five separate ranks, so the chances to form a full house or four-of-a-kind must be identical (specific ranks don’t matter since any such hand spoils the flush). Within the same suit, AQT98 is spoiled by the king, jack or 5432; KQJT8 is spoiled by the ace, nine or 7654 (no difference yet); but only AQT98 is spoiled by 76. Alas, this is offset by the possibility that in another suit, any ace-high flush spoils KQJT8, but it takes a flush at least AKxxx or AQJxx to spoil AQT98.
So which is more likely: (1) 76 in the same suit that spoils AQT98, or (2) a second flush that spoils only KQJT8? Using ballpark combinatorics (ignoring overlaps) and excluding the one-card spoilers of each: Case 1 ~ 43c6 = 6,096,454. I counted 293 flushes that are ace-high but not AK-high or AQJ-high, and three suits to pick from, so Case 2 ~ 293 × 3 × 32c3 = 4,359,840. While a crude estimate (Case 2 would be higher since I ignored extra low cards in the second flush) it convinces that a same-suit 76 is more likely. Hence, AQT98 is more likely to be spoiled, so KQJT8 is more likely to be best. Actual bridge hands: KQJT8 (476,291,232) AQT98 (474,216,756) or a difference of 2,074,476.
Tim Broeken: AQT98 cannot have 76 of the same suit, which is more likely than a two-flush hand ruining only KQJT8.
As with flushes, there is no difference in rank sets; five separate ranks must have the same chances to form a full house or four-of-a-kind. Logically, the chances to form a flush or straight flush must also be identical. Otherwise only one card rank (ace or queen, respectively) will spoil either, so their chances of being best poker hands are identical. Actual bridge hands: KQJT9 or JT987 (6,351,778,020).
Per the same logic, every straight from king-high thru nine-high is equally likely. An eight-high straight, however, is less likely because it is spoiled not only by a nine but by any AKQJT; likewise for lower straights. An ace-high straight cannot be spoiled by a single card rank, so it’s more likely by far — in fact it’s the most likely best poker hand, with 17,259,852,120 bridge hands.
Trivia question: What is the least likely (but possible) best poker hand? (answer below)
Three-of-a-kind as the best bridge poker hand is rare. In fact every such hand must consist of 11 specific card ranks: AKQJ9876432, with one of them tripled. The missing 10 and five are mandatory, as any other two vacant ranks would leave a straight; and of course no pair can exist, or you’d have a full house. Either hand must receive eight specific ranks not to be spoiled. This also shows that the kickers must always be the two highest ranks (besides the triplet). Consequently, only 11 three-of-a-kinds are possible, and all are equally likely. Actual bridge hands: KKKAQ or 333AK (1,302,000).
Tina Denlee: After triplets are set, the rest is automatic. Either hand needs 11 ranks without a five or 10 to avoid a full house or straight.
As far as kickers, KKQQ9 is spoiled by a jack, QQJJ9 by a king, and either is spoiled by an ace or 10; so in that regard they’re equal. Chances to form a full house are also equal, as KKQQ9 is spoiled by a king, QQJJ9 by a jack, and either by a queen. Flush chances logically must also be the same. Straight chances are irrelevant, as any card above a nine already spoils either, and low cards have equal effect. No difference! Actual bridge hands: KKQQ9 or QQJJ9 (88,024,800).
Trivia answer: The least likely (but possible) best poker hand is 22225.The other eight cards must be two fives, three fours and three threes.
Were you savvy to my “You should guess wright on this” remark before the tiebreaker? The misspelling was intentional. What does the word ‘wright’ suggest to you? My inkblot reaction is the Wright brothers, who invented and flew the first airplane. Yes, I’m aware of the controversy and counter claims, but this is what I was taught in school. And in what year did it happen? I was taught that too: 1903, which is the answer to the tiebreaker.
Of the 7462 poker hands, only 1903 can be best in a bridge hand, as summarized below:
We are Wilbur and Orville, and we approved this message
Every straight flush, full house, flush or straight is possible; and every one pair or high card is impossible. All that remains is to determine which four-of-a-kinds, three-of-a-kinds and two pairs are valid. The main criterion is the rank of the kickers.
Valid four-of-a-kinds cannot have a two or three kicker (what are the other eight cards?) which excludes 24 hands; or a four kicker with quad twos or threes, which excludes two more. Note that four fives with a four kicker is possible, as the remaining cards can all be twos, threes and fours; likewise for higher quads.
Valid three-of-a-kinds were explained in Answer 6. Any triplet except fives or 10s is possible, but each is mandated to a single distinct hand, since the kickers must be A-K (or A-Q with three kings or K-Q with three aces).
Valid two pairs are harder to decide. The lowest possible kicker is a seven, but this varies inversely upward with the lower pair, and the incidental formation of straights causes subtle anomalies. Based on the lower pair: Twos demand an ace kicker or AA22K, but TT22A-5522A are curiously impossible (10 of 132); threes demand a king or ace kicker, or AA33Q-KK33Q (22 of 121); fours demand at least a queen kicker, or AA44J-KK44J-QQ44J-KK44T-QQ44T (32 of 110); fives demand at least a 10 kicker, or KK559-QQ559-JJ559-TT559, but 6655J-AA55T-6655T are impossible (41 of 99); sixes demand at least a nine kicker, but 9966T-8866T-7766T-TT669-88669-77669 are impossible (36 of 88); sevens demand at least an eight kicker (42 of 77); anything higher, at least a seven kicker (126 of 231).
Charles Blair: Which of these first five cards is less likely to improve in a game of Florida hold ‘em?
Grant Peacock: Anyone for a quick game of pot limit 13-card draw high-low?
The Donald: This stuff is old hat, as it was covered completely in Poker 401 at Trump University. Now, as President-erect, I can only emphasize that in any full boat, the bigger the pair, the more likely she’ll be best. Living proof is the hostess on my yacht.
My Donald: Okay, so it was a bad year for ducks.
© 2016 Richard Pavlicek
by Richard Pavlicek by Richard Pavlicek