Conference rusure::math

    Re .0:
    
    Calculating the odds at that point is not a simple task.  First, if I
    understand the rules (from Hoyle), the player to the left of the dealer
    had the option to make the suit of the card exposed on the bone pile
    the trump suit, but they declined.  That gives us information about
    their hand -- if they had had lots of clubs, they would have accepted. 
    Similarly, the fact that the next player, the one opposite the dealer,
    accepted the suit gives us information about their hand.  Figuring out
    how this information affects the probabilities would requiring figuring
    out each player's best play -- i.e., solving the game for the best
    strategy.  And again, when the player led the ace of hearts, that is
    more information, and that the dealer discarded a heart is information
    -- not just about where the heart is, but about what is in their hand
    that would lead them to discard a heart under the circumstances.
    
    I doubt that computing the correct probability that the dealer still
    has a heart given all this information is feasible.  We could ignore
    all that and consider the probability that the dealer still has a heart
    given:
    
    	The top card of the bone pile is a club.
    	The player opposite the dealer has no hearts.
    	That player has at least two clubs.
    	The player to the left of the dealer had the ace of hearts.
    	The dealer had at least one heart.
    
    This would still be complicated, but could probably be done if somebody
    wanted to spend the time on it.  I'm not likely to do that, but you can
    get an even rougher approximation by considering only how many unknown
    cards are left.  We start with 5 spades, 6 hearts, 7 clubs, and 6
    diamonds.  (The 5 and 7 are because the jack of spades is counted in
    the clubs.)  We learn the locations of three clubs (the top of the
    discard pile, which moves into the dealer's hand, and two cards in the
    dealer's partner) and two hearts (the first player's lead and the
    dealer's discard).  That leaves 5 spades, 4 hearts, 4 clubs, and 6
    diamonds -- 19 cards.
    
    The dealer has 4 unknown cards in their hand.  There are 3876 ways to
    choose 4 cards from 19 (that's 19!/(4!*15!)).  There are 15 cards that
    are not hearts, and there are 1365 hands of 4 cards that can be made
    from 15 cards, so 1365 of the dealer's possible 3876 hands have no
    hearts.
    
    Those hands are not all equally probable, because of the other
    information available we have not used, but considering this as an
    approximation, the probability the dealer has no hearts remaining is
    1365/3876, or a bit more than 1/3.
    
    
    				-- edp

        Now how does that quote go...?
        
        	"I have not calculated the odds, Captain, but
        	surely they must be astronomical."
        
        Dan

    How many Vulcans does it take to change a lightbulb?
    
    Approximately 1.000000, Captain
    
    Andrew

    RE: .1
    
    Thanks for the try. Not supprising that I got lost I guess.
    
    I think I try simplifying the problem as much as possible.  No
    stratigy, odds of 1 hand having 2 hearts vs. 1 heart. 
    
    Scott