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# Complete Hand Example #11

Player A – K♣, Q♣, J♣, K♠, 10♥, 9♣, 6♦, 5♥, 3♥, 3♠, 2♦
Player B – 6♣, 5♣, 4♣, 10♠, 9♦, 8♦, 8♠, 7♠, 2♣, A♦

Conditions: Knock card is the A♣, which means it is a must-gin hand. Player A has no count in any game. Player B has no count in the first game, a safe count of eight in the second game, and a safe count of 14 in the third game.

Play of the Hand:

Player A – Discards the K♠ as his safest card.

Player B – going to the deck, he pulls the 8♥ for six melded cards and throws the 10♠, since his first consideration is to get under the two counts that are available to him.

Player A – Draws the Q♦ from the deck and discards it as his safest card.

Player B – Obtains from the deck the Q♥ and discards it.

Player A – Picks the 9♠ and discards the 10♥.

Player B – Going to the deck, he obtains the J♠ and has a choice of discarding either it, which can be only used for Jacks, or the 9♦ which can only be used for 9’s. Since he holds the 8, and the Q♦ was just discarded by his opponent, he throws the J♠ as the higher of the two cards.

Player A – Picks the J♥ from the deck and discards it.

Player B – Obtains the 10♣ from the deck and has the choice of discarding it or the 9♦. Although the 10♥ and the 10♠ have been played and no 9’s have been played, the 10♣ is at this point the more dangerous card since the K♠, Q♥, J♠, J♥, 10♥, and the 10♠ have been played but none of the club suit has appeared. In addition, the entire club suit is open below the 10 and could tie up either the needed 8♣ or the 7♣ as well. Since the 9♦ has no more value than the 10♣ did in getting under the count, the 9♦ is discarded.

Player A – He takes the discard, which establishes his hand as 9’s. In picking this card, he has committed himself to win rather than play to the wall even though he has no counts. His choice of discard at this point is limited to either the 2♦ or the 6♦. The 2♦ could tie up the needed 3♦ possibly, whereas the 6♦ could not tie up any needed card. Therefore, he throws the 6♦.

Player B – Draws the 7♥ from the deck and throws the 10♣.

Player A – Takes the discard and throws the 2♦. He has now a committed gin hand.

Player B – He must take the 2♦, not just for its offensive possibility but because it can put him under the count of the last game by breaking his 7’s. He discards the 7♠.

Player A – Picks and throws the 10♦.

Player B – Selects the A♥ and discards the 7♥. He is now under in two games. He will not in any circumstances go over the count of 8.

Player A – going to the deck, he pulls the A♣ and since none of the cards that he holds are actually any safer, he throws the A♣.

Player B – Takes the discard and throws back the 2♣, even though he has just picked the 2♦. Although this move indicates that he is holding Aces, he is also forcing his opponent to hold a 3♦, 4♦, 5♦ against his pick of the 2♦.

Player A – Picks and throws the J♦.

Player B – Pulls the 4♠ from the deck and now must throw the 2♦ as the safest card. The 4♠ still leaves him under in both games.

Player A – Draws the 8♣ from the deck. He is aware that no 8’s have shown, nor have any of the clubs under the 8. He is further aware of the fact that, from his opponent’s throws, he is obviously under the count of 8. This means that his opponent either has nine melded, or, at worse, seven melded. If he has nine melded, the 8♣ would most likely gin him. If he has seven melded the 8♣ will not do any harm since his three-card combination must be 8 points or less. If he has seven melded, what could his three-card combination be? In view of the cards that have been played, he had picked an Ace which could give him a pair of Aces. Certainly it would not be with another 2 because he has just thrown back two Deuces. It could not be with a 3 because the two Deuces that were thrown were both under the two Three’s that the player is missing. So, if his opponent had either of the missing 3’s, he would have had gin. The only other combination available would be the two missing Threes with a Deuce. However since both Deuces have been played from the two missing Threes, it is most unlikely that he would hold this combination. The percentages therefore favor the fact that he has nine melded. The 8♣ therefore cannot be thrown. What about the 5♥? This appears to be a safe card, since his opponent cannot have three 5’s if he gives him credit for the missing run. He cannot have the 6♥, 7♥, 8♥ since the 7♥ has been played and he would not be holding a combination that would keep him over the count. If he gives his opponent three 8’s and 3 Aces, his missing run would have to be clubs between the 3 and 7. His likelihood of taking the hand to the wall is very slight since he is missing an Ace and either side of the lower club run. His only opportunity to win therefore is to throw his 5♥ and hope to buy a three. It is more unlikely at this point that his opponent has both missing Threes. He therefore throws the 5♥.

Player B – Picks and throws the K♦.

Player A – Going to the deck, he buys the A♠, which he cannot throw for the reasons expressed on the last pick. He can still win his hand by breaking his 3’s, which to him are definitely safe, and play for the single 2♠. He discards the 3♥.

Player B – Picks the 9♥. Although he knows that his opponent is holding 9’s, he is forced to throw it because of the count.

Player A – Counts the remaining cards in the deck and finds that there are 12 left. He then knows that he will not be giving up an extra pick to his opponent by picking this discard. The 9♥ is a most important pick to his hand in view of the fact that he is now holding two of his opponent’s most needed cards. If he were to buy another one, either the 3♣, 4♣, or 7♣, he would then be unable to win his hand. By picking the 9♥ now and discarding the K♣, he is still in the same position to gin. However, he has the 8♣ tied up, so that if he were to pick the 3♣, 4♣, or 7♣, he could then throw the Q♣ and still retain a possibility for gin. He then discards the K♣.

Player B – Draws the 2♥ from the deck, which he throws as a dead card.

Player A – Pulls from the deck the 3♣ and discards the Q♣. Not only has he picked a possible gin card for his opponent, but he has also given himself one extra opportunity for gin with the established 3♦.

Player B – Picks the 2♠. He now has a choice of throwing either the 2 or the 4, neither of which is safe at this point or he also could attempt to take the hand to the wall. He knows from the plays just made that his opponent is sitting with seven melded. What are the missing three cards? They must obviously include one or two of his needed cards since he has seen his opponent break a possible gin combination. He knows that his opponent has the 8♣ tied up by his pick of the 9♣ and throw off from the top end of his club run. His opponent must be holding the A♠, since Aces are a definite known run. What could he be holding with it though is the question. Obviously he is holding the 3♠, so that the 2 cannot be thrown. He could be holding the A♠, 3♠, 5♠ so that the 4♠ is a gin possibility, or he could be holding one of the two missing 3’s. Most likely it would be the 3♣ since he is missing all of the higher clubs. So, what are the chances of ginning him or of going to the wall? Actually they are 50/50. If he is holding the A♠, 3♠, 5♠, his hand is dead. If the dealer breaks his three Aces, which he can do and remain under the count, he has no problem in subsequent throws because he knows that the only way his opponent can gin now is with the remaining three, if this was his holding. If the dealer picks the three he will have the opponent dead and can throw anything, regardless of count. He also knows that if he decides to play for gin, his opponent must use the 2♠, but does not necessarily have to use the 4♠. Since he has a 50/50 chance of getting by with the 4♠, the determining factor of whether to throw this card or play to the wall are the odds in his favor. If he gets away with the 4♠ and gins his hand, he wins all three games, whereas if the 4♠ turns out to gin his opponent, he only loses one game. He therefore takes advantage of the odds in his favor and throws the 4♠.

Player A – Draws from the deck and gets a 5♦. He discards it.

Player B – Picks the 5♠ and discards it. He is now reasonably sure of his opponent’s holdings.

Player A – Draws the 7♣ from the deck. He now is in a perfect position. He knows his opponent’s holding and knows that he has him completely dead, whereas he has two ways to gin going for him. Since his opponent has nine melded he cannot be holding more than one of these two cards. He discards the J♣.

Player B – Going to the deck, he buys the Q♠ and discards it.

Player A – Picks and discards the 7♦.

Player B – On his last pick he obtains the 3♦. Now knowing that he has his opponent completely dead, he discards the A♦. The hand is then over.