2005-03-30, 20:22 | #1 |
Apr 2004
Copenhagen, Denmark
2^{2}×29 Posts |
Poker puzzle
It's by birthday today, and one of my gifts was a poker set. Sweeeet
Of course we couldn't resist trying to see if we could handle the chips like the pros on TV. More specific we tried to take two stacks of chips of equal height, and then "fold" them together into one stack using only one hand. We were'nt that good... But, we found an interesting problem in the process: Find (is possible) a closed form for the Poker Number P(n). The Poker Number is defined as follows: Take two differently coloured stacks (here: Red(R) and Blue(B)) of chips of equal height, n, and apply the following shuffle procedure: 1) Place the R stack in your left hand and the B in your right hand. 2) Take one chip from the left stack and place it on the table in fromt of you. 3) Take one chip from the right stack and place it on top of the stack in front of you. 4) repeat 2) and 3) until all chips are in the stack in front of you. 5) Take the top n chips and make that your new right stack and let the bottom n chips be your new left stack. Steps 2)-5) is defined to be one operation. The Poker Number P(n) > 0 is defined as the minimum number of operations needed for all the chips in the left stack to be red again and all the chips in the right stack to be blue. Here are the first few values: P(n)= 1,2,4,3,6,10,12,4,8,.... Is there a simple formula for the Poker Number? Is seems that P(2^n)=n+1. -- Cheers, Jes |
2005-03-31, 09:57 | #2 |
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}·3·641 Posts |
See the similar sequence A002326, "Number of riffle shuffles of 2n+2 cards required to return a deck to initial state", in the OEIS at http://www.research.att.com/cgi-bin/...i?Anum=A002326
A002326 starts with a first term of 0 instead of 1 as your sequence begins. Last fiddled with by cheesehead on 2005-03-31 at 10:01 |
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