tag:blogger.com,1999:blog-15628310.post5694121090199730675..comments2021-04-15T02:41:46.574-04:00Comments on Question of the day: Sliding CoinsAnonymoushttp://www.blogger.com/profile/18153935609499338685noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-15628310.post-4622967966023110772010-11-26T12:17:48.955-05:002010-11-26T12:17:48.955-05:001 coin1 coinAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-15628310.post-22282885004147825382010-11-15T12:40:02.190-05:002010-11-15T12:40:02.190-05:00Hmmm.... I think it depends how you interpret &quo...Hmmm.... I think it depends how you interpret "...the chances are...". <br /><br />For example, with seven roles there is a 39% chance (i.e (7/8)^7) that no coins would make and, therefore, a 61% (1-0.39) that at least one coin would make it. So the chances are that at least one coin would make it.<br /><br />Similarly, for 6 coins there is a 45% that no coins will make it and a 55% chance that at least one will. So, agian, the chances are that at least one coin would make it.Dave UKnoreply@blogger.comtag:blogger.com,1999:blog-15628310.post-25515413237072032182010-11-12T16:30:17.200-05:002010-11-12T16:30:17.200-05:00Chances for a coin to get past the first hole are ...Chances for a coin to get past the first hole are 1:2, chances to pass the second hole are then 1:2 x 1:2 = 1:4, and finally chances to pass the third hole are 1:4 x 1:2 = 1:8.<br /><br />We should release 8 coins and theoretically one of them should make it all the way down.<br /><br />Why do I get a feeling it can't be that simple?Antehttps://www.blogger.com/profile/06616417014707665130noreply@blogger.com