Main Puzzle 8N13 by Richard Pavlicek
Bridge and poker are vastly different games, but they share some common terms: open, call, raise, 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 (click here for a list in ranking order) 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. Most answers can be determined logically without calculation and hey, its multiple choice, so a good guess can score just as well.
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.
|4||Dan Gheorghiu||British Columbia||CABBCCC||5||1929|
*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 its 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.
Tim Broeken: There are cases with two different straight flushes, which gives QJT98 an edge.
Tina Denlee: 5555K is hampered by four sixes (1 chance); 6666K is hampered by an extra straight flush (4 chances).
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.
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 its more likely by far in fact its 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 at end)|
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.
Of the 7462 poker hands, only 1903 can be best in a bridge hand, as summarized below:
|Poker Hand Type||Distinct Hands||Valid in 13 Cards|
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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).
|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.
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 will just emphasize that in any full boat, the bigger the pair, the more likely to be best. Living proof is the hostess on my yacht.
My Donald: Okay, so it was a bad year for ducks.
© 2016 Richard Pavlicek