tag:blogger.com,1999:blog-5730391.post1669609258730884010..comments2018-05-24T20:15:06.804-07:00Comments on Blaine's Puzzle Blog: GeekDad Puzzle of the Week: When Are the Odds Even?Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-5730391.post-3486470019133817082012-06-20T07:32:53.679-07:002012-06-20T07:32:53.679-07:00I was gonna post a comment showing how the Pell nu...I was gonna post a comment showing how the Pell numbers and half-companion Pell numbers are involved in the answer, but since I can neither employ <br><br /><FONT face="Courier">, nor use <TABLE> tags,<br />I therefore just post the following URL:<br /><a href="http://users.az.com/~jwaters/Puzzles/GeekDad_2012-06-17.htm" rel="nofollow">http://users.az.com/~jwaters/Puzzles/GeekDad_2012-06-17.htm</a>, which will direct anyone interested to an HTML document which <b>DOES</b> use those tools to show the correlation. Enjoy.Enya_and_Weird_Al_fanhttp://profile.typepad.com/6p014e89252582970dnoreply@blogger.comtag:blogger.com,1999:blog-5730391.post-7915849684405459792012-06-17T09:31:02.932-07:002012-06-17T09:31:02.932-07:00I sensed right away that this puzzle involves one ...I sensed right away that this puzzle involves one of my favorite series, the half-companion Pell numbers over the Pell numbers, i.e. the rational approximations to the square root of 2:<br /><br />(*SIGH* How I wish I could use <font face="Courier">)<br /><br />1 1 3 7 17 41 99 239 577 1393 3363<br />- - - - -- -- -- --- --- ---- ----<br />0 1 2 5 12 29 70 169 408 985 2378<br /><br />(I started with the illegal fraction 1/0 only because it gives 1st case of<br>numerator^2 = 2*denominator^2 +/- 1)<br /><br />Of course I was right. This series DOES INDEED play into the answer to this puzzle.Enya_and_Weird_Al_fanhttp://profile.typepad.com/6p014e89252582970dnoreply@blogger.comtag:blogger.com,1999:blog-5730391.post-85307210439600640892012-06-16T19:16:36.091-07:002012-06-16T19:16:36.091-07:00The puzzle is asking you to think of a set of marb...The puzzle is asking you to think of a set of marbles in a bag... say 3 marbles. Can you make the proportion of black and white marbles such that the chance of drawing 2 black marbles is exactly 50%? If you had 2 black marbles and 1 white marble, the probability of drawing 2 black marbles would be 2/3 x 1/2 = 1/3. So that's not one of the possible sets. So basically it is asking for all the possible amounts of marbles (from 3 marbles up to 1 million marbles) is it possible to proportion the marbles so that the chance of drawing two black marbles is exactly 50%? Is there only one set (85 black marbles out of a total of 120) or are there more? And if so, what are those sets? Finally, if you figure that out, can you answer the puzzle for drawing *3* black marbles from the bags? The answer will be revealed soon, and I'll add my own thoughts.Blainehttps://www.blogger.com/profile/06379274325110866036noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-65181667810763248292012-06-16T18:34:57.674-07:002012-06-16T18:34:57.674-07:00Answer? LOL! I can handle math, but have no idea...Answer? LOL! I can handle math, but have no idea what the conditions of this puzzle really are. Is the 1,000,000 total to be made up of many sets of 120 marbles each? Are we free to put any number of marbles in a "set"? How many colors are to be used? I think I'll just wait until the answer is posted to find out what the question really is.dumpsterdiveladhttps://www.blogger.com/profile/05296636733423295606noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-30321262350379752422012-06-16T17:39:29.839-07:002012-06-16T17:39:29.839-07:00Blaine!
Have you lost your marbles?Blaine! <br />Have you lost your marbles?skydiveboyhttps://www.blogger.com/profile/17174073226290431753noreply@blogger.com