tag:blogger.com,1999:blog-1124742229498315339.comments2017-12-26T10:04:28.211-05:00Card ColmMathematical Association of Americahttp://www.blogger.com/profile/10559021045290192742noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-1124742229498315339.post-75132973014755933922014-01-17T01:59:45.586-05:002014-01-17T01:59:45.586-05:00Forgive me if I am mistaken, but is it possible fo...Forgive me if I am mistaken, but is it possible for a poker player to get two cards from the D list and for those two cards to be 7D 8D? Then a straight flush of 45678D is possible.Grant Fikeshttps://www.blogger.com/profile/09060596353844196026noreply@blogger.comtag:blogger.com,1999:blog-1124742229498315339.post-70844257190256069872013-12-31T16:53:42.247-05:002013-12-31T16:53:42.247-05:00For the ABC version, how hard is it to mentally ch...For the ABC version, how hard is it to mentally check for a too long cycle? Here, a cycle of length 7 or more is too long, and one of length 6 or less is ok. If we can verify that 6 or more elements are in ok cycles, that tells us there can be no too long cycle. So, in the method described above, say you first find a 4 cycle, and then a 3 cycle. Then there are only 5 elements remaining and there cannot be a too long cycle. Or, if you find a 3 cycle and a 2 cycle, the next cycle must either be too long (and can be broken as explained) or not, in which case no cycle is too long. The worst case scenario is finding cycles covering just 5 elements, and then hitting a 7 cycle. That would require you to follow the complete cycle structure of the permutation. But most of the time you will either have a few short cycles (2,3, or 4) that are quick to follow, or hit a long cycle on the first or second go. I expect that with a little practice, one could survey the array of cards for a too long cycle and correct it with a transposition quite quickly.dan kalmanhttp://dankalman.netnoreply@blogger.comtag:blogger.com,1999:blog-1124742229498315339.post-49630245086603222182013-08-28T09:19:44.382-04:002013-08-28T09:19:44.382-04:00The fact that a sum of 16 arises in two possible w...The fact that a sum of 16 arises in two possible ways when using three value chosen from 1, 2, 3, 5, 8, 13 is addressed in <br>Additional Certainties http://www.maa.org/community/maa-columns/past-columns-card-colm/additional-certainties<br />See http://wordplay.blogs.nytimes.com/2013/07/29/sum-2/?_r=0 for other examples. Card Colmhttp://www.cardcolm.orgnoreply@blogger.comtag:blogger.com,1999:blog-1124742229498315339.post-28908825958439312272013-08-26T21:38:33.643-04:002013-08-26T21:38:33.643-04:00Not in the first set for three numbers: 13+1+2 = 3...Not in the first set for three numbers: 13+1+2 = 3+5+8 = 16Riebelhttps://www.blogger.com/profile/11707932932368521889noreply@blogger.comtag:blogger.com,1999:blog-1124742229498315339.post-27460933651428335352012-12-26T19:39:42.378-05:002012-12-26T19:39:42.378-05:00I was enchanted with Kruskal's Principle when ...I was enchanted with Kruskal's Principle when it appeared in MG. I ran some computer simulations and shared the trick with my students. For me, the operative clause in the MG article was:<br /><br />"Once this happens, the chains will be identical from that point on."<br /><br />I visualize the deck state as a graph, where the terminus is an attractor. The majority of the chains lead to it as the root of a subtree of the graph.John Millerhttps://www.blogger.com/profile/04475504380293438058noreply@blogger.comtag:blogger.com,1999:blog-1124742229498315339.post-58526457320718818542011-11-01T16:25:48.421-04:002011-11-01T16:25:48.421-04:00@JohnAllenPaulos ( https://twitter.com/#!/JohnAlle...@JohnAllenPaulos ( https://twitter.com/#!/JohnAllenPaulos ) tweeted a very nice way of expressing the basic binary Gilbreath Principle from 1958:<br /><br />"2 car lanes, 1 Black/White, B/W, etc, 1 W/B, W/B, etc. However they merge into 1-lane tunnel, 1st, 2nd, nth pair emerging are opp. colors."<br /><br />Of course this generalizes (a la Gilbreath 1966) as follows: 2 lanes of cars, one repeating Red, White, Blue, Green, White, Orange, the other repeating Orange, White, Green, Blue, White, Red, however they merge, will yield a single stream of cars in blocks of six, each block consisting of one each Red, Blue, Green and Orange, and two White (in some order within each block).<br /><br />My "Big Question" from August above then becomes, what kind of regularity can we impose of 3 or more merging lanes of cars, so that the single merged stream will have some predictability, regardless of how the merging is done?<br /><br />I only know the answer in the case of Faro or perfect merging.<br /><br />Does anyone have any ideas here?<br /><br />Maybe it's not a coincidence that Gilbreath 1966 sounds like a model of car: the word "auto" did turn up, automagically as it were, in the August Car Colm:-)Colm Mulcahyhttp://www.spelman.edu/~colm/noreply@blogger.com