What poker hand of 5 cards can meet these criteria:
There are no picture cards, and an ace counts as 1.
No two cards had the same denomination.
All four suits were present.
The total of the odd cards equalled the total of the even cards.
No three cards were in sequence in the sense of a straight.
The black cards' values totaled 10.
The values of the hearts totaled 14.
The lowest card was a spade.
1) since there are 5 cards and each suit is present then that means one of suits is represented twice and the others only once.
2) since the hearts total 14 and no single non face card can equal 14 then hearts must be suit with 2 cards. Thus we now know the hand to be HHDSC with H hearts, D diamonds, S spades, and C clubs.
3) there are only 2 possible value combinations for the hearts and that is either 5,9 or 6,8 Now in both cases they are either both even or both odd thus the total for even/odds is at least 14. Same can be argued for the total of the black cards, and since that only leaves the diamond then the hearts and the black cards can not both be even or both be odd. Thus the total of the odds/evens is 14 and that makes the diamond a 4. Which then says that the black cards must both be even to contribute to the even sum and thus the hearts must be odd and thus the hearts are 5 and 9.
4) Now there are 3 possible pairs for the blacks 1,9 2,8 or 3,7 but only one is even and that is 2,8. Thus the blacks must be 2,8 and since the spade is the smallest value it must be the 2.
5) that gives our hand as 5H 9H 4D 2S 8C
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Posted by Daniel
on 2009-03-08 21:55:30 |
solution explanation
