Oh Hell! a Gamesman Puzzle

Oh Hell! A Gamesman Puzzle Vaughan Clarkson 5th January 2018 Problem Oh Hell! is a popular card game in my family, maybe in yours too. It’s a trick-taking game with trumps and contracts, played using a standard 52- card deck. Any number can play. An interesting feature of Oh Hell! is that the number of cards dealt into each hand varies from one round to the next. This problem is about what happens in the first round of an Oh Hell! game. In the first round, each player is dealt just one card, face down. The dealer then turns face-up the next card in the deck. That card determines the trump suit. In a clockwise direction around the table, starting from the left of the dealer, each player states how many tricks he or she will contract to win. In the first round, this must be either zero or one. Once each player has made a bid, the player to the left of the dealer reveals his card, i.e., he leads. Again in a clockwise direction, each other player plays his card too. Being a single-trick round, players have no choice about which card to play. They must reveal the card they have been dealt. The highest card in the led suit wins the trick, unless a trump is played, in which case the highest trump wins. The highest ranking card in any suit is the ace, followed by the king, queen, etc. The scoring system is such that players are rewarded if they win exactly the number of tricks they contract to take and, conversely, they are penal- ised if they win either too few tricks or too many. (a) For the player to the left of the dealer, what is the strategy for making a bid in the first round that will give the highest probability of fulfilling his contract (the “optimal strategy”)? (b) For a three-player game, what are the optimal strategies for the re- maining players in the first round? Assume that: • all other players are likewise using their optimal strategies and • the screw the dealer rule is in play, in which the total number of tricks bid must not equal the number of cards dealt. Solution (a) The player to the left of the dealer—let’s call him Larry—will lead. Two cards are known to Larry when he makes his bid: the card in his hand 1 and the trump-suit-setting card—the “face-up card”. Of the 50 other cards, at least 26 are inferior to the card in Larry’s hand, being those cards belonging to the other two suits. Therefore, in a two-player game, the probability that Larry will win the trick is at least 26=50, which is greater than 50%. That is, Larry should bid to win one trick, regardless of the card he is dealt. To continue, let’s assume without loss of generality that the trump suit is hearts. If Larry was dealt a card that isn’t hearts, let’s assume, again without loss of generality, that it is clubs. If he is dealt 2|, only 26 of the 50 unseen cards will lose to it. If he is dealt A~, all 50 cards will lose to it. Therefore, it’s possible to assign numbers from 26 to 50 to each of the cards 2|;:::;A|; 2~;:::;A~, from which sequence the face-up card has been removed. This number, k, is the number of unseen cards which will lose to Larry’s card. It is the card’s rank. The probability that the player to the left of Larry has a losing card is k=50. The probability that the player to the left of her also has a losing card is (k − 1)/49. For an n-player game, then, the probability, p, that a card with rank k will win the trick for Larry is k k − 1 k − n + 2 k! (50 − n + 1)! p = × × · · · × = : 50 49 52 − n 50! (k − n + 1)! When n = 2, we already know that k = 26 is sufficient to make p > 0:5. By evaluating this function for different values of n and k, we discover that the optimal strategy for Larry is to bid to take one trick only when the card in his hand is at least as good as the card in the following table. Number of players, n Minimum card rank, k Name of card 2 26 2| 3 36 Q| 4 40 3~ 5 43 6~ 6 44 7~ 7 45 8~ 8 46 9~ 9–11 47 10~ 12–15 48 J~ 16–26 49 Q~ > 26 50 K~ Note that the name of the card in the table above must be increased by one if it would rank equal to or higher than the face-up card. (b) For a 3-player game, the other players can infer that Larry, if he bids to win the trick, has either a trump or he has a non-trump not lower than a queen. Conversely, if Larry does not bid to win the trick, he must have a non-trump lower than a queen. Suppose Larry bids to win the trick. The next player to bid, the player to the right of the dealer—let’s call her Rachel—can infer that the led 2 card will be one of the 12 unseen trump cards or one of 9 unseen non-trump cards of rank queen, king or ace. If Rachel does not have a trump, the led card will beat it with probability at least 12=21 > 0:5. When Rachel has a trump card of rank k > 38, it reduces the number of unseen cards from which Larry’s card is drawn to 20. The probability, q, that she will win is then k − 30 k − 1 q = × : 20 49 For q > 0:5, we therefore need k2 − 31k + 30 > 10 × 49, or k2 − 31k − 460 > 0, from which we conclude that k ≥ 42. That is, if Larry bids to win the trick, Rachel should only bid to win the trick too if she holds a card at least as good as 5~ (or 6~ if the face-up card is a 2, 3, 4 or 5 of hearts). Suppose Larry does not bid to win the trick. If Rachel has a trump, she will certainly beat the led card and has a better than even chance of beating the dealer’s card. If Rachel does not have a trump then, of the 30 unseen non-trump cards that might be led, at least twenty of them will beat Rachel’s card, being of a different suit. So, given that Larry does not bid to win the trick, Rachel should bid to win it if and only if she has a trump card. Finally, what should the dealer do? Let’s call him Dan. If the total tricks bid is equal to zero, Dan must bid one, under the screw the dealer rule. If the total tricks bid is equal to one, Dan must bid zero. So Dan only has a choice when both Larry and Rachel have bid to win the trick. Dan can infer that Rachel must be holding one of nine unseen cards, at least 5~. If Dan’s card has rank k ≤ 42, he will surely lose to Rachel. If Dan’s card has rank k > 42, this reduces the number of unseen cards from which Rachel’s card is drawn to 8. Likewise, the number of unseen cards from which Larry’s card is drawn is reduced to 19. The probability, r , that Dan will win is then k − 42 k − 31 r = × : 8 19 For r > 0:5, we need k2 − 73k + 1302 > 76, or k2 − 73k + 1226 > 0, from which we conclude that k ≥ 47. Therefore, when the round is already “overbid”, Dan should only bid to win the trick if he has 10~ or higher (or at least J~ if 10~ or lower is the face-up card). 3.

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