tag:blogger.com,1999:blog-8231784566931768362.post7173107661786771248..comments2021-06-16T05:56:38.112-04:00Comments on MathNotations: Dorothy Revisited -- Another View...Dave Marainhttp://www.blogger.com/profile/13321770881353644307noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-8231784566931768362.post-62043335246780571282009-07-13T21:56:46.365-04:002009-07-13T21:56:46.365-04:00The Tower of AION
...The Tower of AION<br /> <br /><br /> Aion is one of the most anticipated online games in 2009, http://www.rmt-wm.com/ NCsoft absorb of the "World of Warcraft" game and other forces in the camp set up on the successfulexperience,designedfor"Asmodians","Elyos"and"Balaur", http://www.rmt-wm.com/ The story centers around the three to start the war,Players can achieve large-scale "race vs race" . can become a Asmodians or Elyos, http://www.rmt-wm.com/ while Balaur is the monster and NPC.Aion schedule in Japan.http://www.rmt-wm.com/games/aion.html the official website of the public April 17, 2009internal testing operators June 5, 2009to start internal testing June 12, 2009limited public beta staff 6 At the end of 2009public beta June9 2009 .<br /> http://www.rmt-wm.com/games/aion.html<br /><br /> At that time，Rmtwm http://www.rmt-wm.com/ will provide to players aion Virtual game coin,this will reduce players hard time to fight coin.players through the reality of dollars to buy virtual coin to upgrade their level of equipment, Rmtwm http://www.rmt-wm.com/ as an operating a virtual coin company,In addition Aion rmt http://www.rmt-wm.com/games/aion.html，we are also operate on dozens of Frequently games.<br />http://www.rmt-wm.com/games/aion.html Low prices and convenient service，players can purchase peace of mind.<br /><br /> http://www.rmt-wm.com/games/aion.htmlThe Tower of AION The Tower of AION The Tower of AION The Tower of AION The Tower of AIONThe Tower of AIONhttp://www.rmt-wm.com/games/aion.htmlnoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-57222662013832833212009-07-02T12:00:08.880-04:002009-07-02T12:00:08.880-04:00I almost always have to do a permutations/combinat...I almost always have to do a permutations/combinations question more than one way in order to be confident of my answer. ;-) So it's good to know lots of ways to do it. I do still get temporarily stumped from time to time by a kid who will propose a reasonable-seeming method that does not come out with the correct answer -- figuring out what is wrong with an approach can be very tricky -- a possible challenge problem to pose to students sometimes.mathmomhttps://www.blogger.com/profile/05869925405540832241noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-56227173946421256682009-07-02T06:43:25.332-04:002009-07-02T06:43:25.332-04:00Nice!
I don't know if anyone else is compelled...Nice!<br />I don't know if anyone else is compelled by this discussion but this problem has opened up many important avenues of instruction for solving combinatorial problems.<br /><br />Your explanation applies equally well to demonstrating that the probability of drawing the red card on any other draw is also 1/n. Since there are n-1 ways to win, the winning probability is therefore (n-1)/n directly.<br /><br />What I find also instructive for students here is that the counting of permutations is equivalent to determining the probability of each separate draw using dependent events, then applying the multiplication principle. This equivalency is subtle for learners.Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-12638026258666337972009-06-30T21:00:09.488-04:002009-06-30T21:00:09.488-04:00The other way I was thinking of it was, like you, ...The other way I was thinking of it was, like you, to assume the black cards are distinguishable, and think about all the permutations of orders in which all N cards could be chosen. There are N! of these. How many of them have the red card last? (N-1)! permutations of the non-red cards in the first N-1 spaces. So probability of "red card last" is ((N-1)! / N!) = 1/Nmathmomhttps://www.blogger.com/profile/05869925405540832241noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-63145949421577618292009-06-30T06:44:28.127-04:002009-06-30T06:44:28.127-04:00Nice way to explain it, Mathmom!
I do feel it is s...Nice way to explain it, Mathmom!<br />I do feel it is still an elusive idea. One can think of it as N different colors, so it's no more or less likely that the red card (or any other color) is drawn each time. It still seems somewhat paradoxical to me since we are dealing with dependent events (cards are not replaced). Very interesting...Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-5238161318839350792009-06-28T17:52:28.968-04:002009-06-28T17:52:28.968-04:00Here's what I think is an easier way to think ...Here's what I think is an easier way to think of it, without the difficulty (I think) of having to prove the possibilities equally likely.<br /><br />Imagine that the game does not end when the red card is drawn, and that all N cards are drawn. Dorothy wins as long as the red card is not drawn last. If all N cards are drawn, the red card is equally likely to be drawn first, second, third, ... , Nth. So it is drawn last with probability 1/N, in which case Dorothy loses. Otherwise she wins, probability 1 - 1/N. <br /><br />Middle schoolers could *follow* this, but even given a fair bit of experience with this type of problem, I don't think they'd *come up with* it. Still a great problem to make them think. :)mathmomhttps://www.blogger.com/profile/05869925405540832241noreply@blogger.com