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 Poker Precept (Posted on 2013-04-19)
In a game of poker, one of the hands of five cards had the following features:
1. There was no card above a 10 (an ace is above a 10 in poker).
2. No two cards were of the same value.
3. All four suits were represented.
4. The total values of the odd and even cards were equal.
5. No three cards were in sequence.
6. The black cards totaled 10 in value.
7. The hearts totaled 14.
8. The lowest card was a spade.
What was the hand?

 No Solution Yet Submitted by K Sengupta Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 solution Comment 4 of 4 |
By clues 1, 7 there are at least 2 hearts.

By 3 there are at most 2 hearts.  Thus there are 2 hearts which by 7 can only be 4T, 59, or 68.

By 6 and 8, SC are 28, 37, or 46.

SC37 and H59 requires the impossible card D24 by clue 4.

SC37 and H4T forces D4 by 4, impossible by 2.

SC37 and H68 forces either an odd diamond = 4 or an even diamond = -4.

Matching the CS combos with even cards against the H combos with even cards requires D=24.

SC46 and H59 forces D4 by 4, impossible by 2.

The remaining option satisfies the problem:  S2, H5, H9, D4, C8.

 Posted by xdog on 2013-04-19 15:45:27

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