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Subject:
Card Playing Probability
Category: Miscellaneous Asked by: darren35-ga List Price: $10.00 |
Posted:
09 Oct 2006 03:50 PDT
Expires: 08 Nov 2006 02:50 PST Question ID: 771926 |
We play a card game once a week which is a variation on whist/trumps, nomination whist using a standard deck of cards without jokers. The game has a total of 20 rounds. During one round the object is not to win any of the seven available tricks. The penalty for gaining a trick is 5 points deduction per trick taken. However, if a person chooses to call three or more ticks and gets the correct number of tricks called, points (10 points) are gained instead. At the start of each round the top card of the pack is turned which determines the trump suit for the round. The person with the lead, on this occasion, Andy, called three tricks. Trumps were declared as hearts with the 7 of hearts turned. His hand was: 10 diamonds 2, 3, King, Queen of clubs Ace, King 3 hearts (trumps) There were six players in total, each having seven cards. (Each group of six cards is one trick ie. one card played by each player.) The person on the left of the lead player has to play their card next and so on until everyone has played one card each. The winner of that trick takes the played cards and claims it as a trick. He then has to play the next lead card and play continues as described above. Andy was the only person to call three or more tricks. It should be noted that the penalty for not getting exactly three tricks on this hand is a loss of many more points -approximately 10 points per trick taken. It has therefore been known for other players to deliberately prevent the calling player to get their call correct by either forcing them to win one more or lose one more trick than they have called. Ie. over-trumping to take a trick away. It should also be noted that if you have a card in the suit that is laid by the lead player you have to play it. If you do not have the suit you can put any other suit on the trick. Eg. Diamonds are trumps; If a club is played and you do not have a club you can either play a card of another suit to reduce your chances of winning with that card on a future occasion if you do not wish to win any tricks. Similarly you can ?trump in? with a diamond and take the trick if you do need the trick. Having the lead meant that Andy could lead any card he wished to. Winning a trick means that you have to lead the next card. I need to know in which order Andy should have played the cards in order for him to win exactly three tricks. I am sure it is possible to work out statistically which is the safest order to play them in but I need a GAR to work it out for me! Please let me know if you need anything clarified as having read this through it does appear to be a little confusing! Playing the game is easier than writing down how it is played! |
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There is no answer at this time. |
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Subject:
Re: Card Playing Probability
From: stanmartin1952-ga on 09 Oct 2006 16:26 PDT |
You should probably bump it up to $100. |
Subject:
Re: Card Playing Probability
From: jack_of_few_trades-ga on 10 Oct 2006 05:41 PDT |
As an avid card player (yet never having played this particular game), I have some input. Andy has 2 sure winners (A,K hearts) and some very low cards (2,3 clubs & 3 hearts) that his task is not a hard one. With 2 sure winners, he only needs to win 1 more trick and avoid taking extras by tossing his low cards before other people are out of those suits. If I were leading, I would start with the 10 diamonds. That is the most iffy card he has (could win, but will probably lose). If the 10 wins, then leading the 2 clubs is an obvious move. From there on, it is simply a matter of tossing off the highest card possible that doesn't win a trick... and taking the last 2 tricks with the A,K hearts. If the 10 loses (the more likely outcome since there are 4 diamonds that beat it and the possibility that 1 of the 6 players would trump it), then Andy should try desperately to take a trick by trumping with the 3 of hearts or winning with a high club. Then after successfully taking a trick with 1 of those 3 cards, play to lose all the tricks as described above. Statistically, I'm not sure any conclusion can be made because it is unknown how other players would play even if we know what cards they have. Some players would probably be more aggressive in trying to make Andy take fewer than 3 tricks while others would give up hope of that very soon and try to bag him with extra tricks (even given the same cards)... people are very hard to predict in cards unless you already know their playing style. |
Subject:
Re: Card Playing Probability
From: statsman-ga on 25 Oct 2006 00:57 PDT |
Stastically this cannot be done with certainty as you cannot gaurantee all forced moves. The way the second player plays the card would affect the way the third plays etc. If the second player, plays to win the first trick that does not stop the third player if they had both higher and lower in the same suit, from letting the second player take the trick or winning it themselves. Unfortunately this one has the 'human element' on it, so I do not think it can be done. |
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