Hold'em Posers Revisited

A few issues back, I posed four situational questions about hold'em hands. I supplied answers for three of them and left the fourth as an exercise for the reader.



You can find that column in Vol. 19/No. 13, and also online.



The third question was:



3. The following question was posed and discussed on the BARGE mailing list. (BARGE is the Big August Rec.Gambling Excursion, an annual convention held in Las Vegas by members of rec.gambling.poker – that is, online poker players – featuring a no-limit tournament, other less-organized events, and much hilarity, including "must-toke" games such as Chowaha.) What is the minimum number of hands such that at least one of them has no pot equity? That is, such that at least one cannot win or tie (including board ties). Specify the hands. Related to that question, what is the maximum number among that group that can have no pot equity?



I provided this answer:



3. The minimum number of hands such that at least one of them has no pot equity is eight. One set of possibilities uses all of the aces, tens, sixes, and fives, as, for example, A A, A A, 10 10, 10 10, 6 6, 6 6, 5 5, and 5 5. Here, all of the hands are pairs, and they preclude the possibility of any tie involving five boardcards that constitute a straight (because such cannot be made). The only possible straights, those with four cards on board, give winners to either the aces or tens. Any flush or four-flush on board gives a winner to one of the ace hands. Any four of a kind or full house does the same, as do three of a kind, two pair, and one pair. This gives all four lowest pocket-pair hands zero equity. So, the answer to what is the maximum number of hands among that group that can have no pot equity is four.



You can change that to just two hands having zero equity by exchanging cards such that there are two 6-5 hands, as in A A, A A, 10 10, 10 10, 6 6, 6 5, 6 5 , and 5 5. Or, you can guarantee that only one hand has zero equity by also changing two hands to A-10, as in A A, A A, 10, 10, 10 10, 6 6, 6 5, 6 5, and 5 5.



More possibilities exist. The eight hands must consist of all the aces and tens and either the sixes or fives, plus four more cards of the same rank. Thus, A A, A A, 10 10, 10 10, 5 5, 5 5, 2 2, and 2 2 is one of another set of possibilities.



As it turns out, more possibilities than these exist, as reader Jack M. informed me:



The following set of cards involves no tens: A A, A A, J J, J J, 7 7, 7 7, 6 6, and 6 6. In this matchup, both pocket-sevens hands have zero equity.



Now, here's that fourth question:



4. And now let me leave you with a question that also was posed on the BARGE mailing list. You hold pocket aces. What is the minimum number of opponents needed to give you zero equity on the pot, and what would be their hands?



You may consider the answer to be a bit of cheating, because the solution is not possible in the normal ninehanded or 10-handed game. In fact, no solution exists even among 11 hands. A number of BARGE members provided solutions involving 13 hands, and concluded therefrom that the minimum number of opponents needed to give the holder of pocket aces zero equity on the pot is 12. Poker player and theorist extraordinaire (he has won two bracelets in the 2006 World Series of Poker as I write this) and math Ph.D. Bill Chen (co-author of the forthcoming The Mathematics of Poker) provided a solution involving just 12 hands (that is, 11 opponents). One hand must be the A A, and the other hands must be A-A, K-10, Q-10, J-10, 10-9, 8-5, 7-5, 6-5, 5-4, 3-3, and 2-2. Bill explained: "You need at least eight other spades and hearts, the two other aces, a Broadway diamond and club, and a baby diamond and club to block the straight flushes; hence, at least 11 other hands." The following set of hands meets the criteria: A A, A A, K 10, Q 10, J 10, 10 9, 8 5, 7 5, 6 5, 5 4, 3 3, and 2 2.



With these cards, all winning hands for the A A hand are blocked, and no set of five cards can be flopped that would cause any play-the-board 12-way ties. The A A wins a few pots with flushes. The hand with the highest equity is the pair of threes, followed by the deuces.



Twelve hands is not really cheating, either. Hold'em can theoretically be played by as many as 23 players. Even considering the burn cards and that the last card is customarily not dealt, 21 hands are possible.



You can test these answers with the poker odds calculator at www.Twodimes.net. For most hand comparisons, it generates all possible boards. It does not work on a Monte Carlo simulation – that is, a dealing out of millions of randomly generated hands – as do many poker odds calculators, such as the one at www.CardPlayer.com



Michael Wiesenberg's The Ultimate Casino Guide, published by Sourcebooks, is available at fine bookstores and at Amazon.com and other online book purveyors. Send suggestions, sticklers, and suppositions to [email protected].