Re: Honeymoon Whist
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My first impression is that one ought to take any of the 26 best cards in the deck
[/ QUOTE ]It seems unlikely to me that any optimal strategy will ignore the information you've seen already. For instance, if we continue under the assumption that the cards can be ordered, I think a better strategy than "Accept the top 26" would be "Accept any card above the current middle (median) card left in the deck."
Here's a scenario I want to consider, though. Let's say you know the deck contains {2[img]/images/graemlins/diamond.gif[/img], 3[img]/images/graemlins/diamond.gif[/img], 4[img]/images/graemlins/diamond.gif[/img], 5[img]/images/graemlins/diamond.gif[/img], 6[img]/images/graemlins/diamond.gif[/img], J[img]/images/graemlins/spade.gif[/img], Q[img]/images/graemlins/spade.gif[/img], K[img]/images/graemlins/spade.gif[/img], A[img]/images/graemlins/spade.gif[/img]}. You draw 6[img]/images/graemlins/diamond.gif[/img]. Although this card has an above average rank, I think you should toss it because it's so far below the expected value.
The problem with writing this into a strategy is that you need to know the value of each card. I don't think this is easy, and I don't think this is independant of you other cards. For instance, the 9[img]/images/graemlins/spade.gif[/img] is worth a whole trick if you have all the higher spades, but less if you don't.
-Sam
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