tag:blogger.com,1999:blog-4528393111359731672.post327460161124544909..comments2024-03-24T00:19:53.054+00:00Comments on Green All Over: Flaw in the Fifth ColumnCassinihttp://www.blogger.com/profile/05879449876804295094noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-4528393111359731672.post-58607141836944065342020-05-30T15:45:46.482+01:002020-05-30T15:45:46.482+01:00I agree it’s a badly written section in an otherwi...I agree it’s a badly written section in an otherwise fine book.<br /><br />What I believe the writer is trying to get across is that even though the coin toss is always 50%, the chance of a streak continuing in a finite sequence of tosses appears to be less than 50%.<br /><br />Even that only holds if you measure streaks in a flawed way “mean of each player’s percentage of heads after heads”, rather than the logical “percentage heads”.RedBallDatahttps://www.blogger.com/profile/00262008267537096647noreply@blogger.comtag:blogger.com,1999:blog-4528393111359731672.post-1164357573799740732020-05-15T10:41:20.055+01:002020-05-15T10:41:20.055+01:00Yeah, that's totally invalid. Their error is t...Yeah, that's totally invalid. Their error is to take a simple average of the percentages, thus giving equal weight to each of the 6 cases. They needed to take a weighted average of the percentages taking into account what proportion of the 'after heads' situations occurred in that case. So of 8 total 'after heads' situations, 1 (1/8 of the total) occurred in four of the cases and 2 (1/4 of the total) occurred in the last 2 cases. So the weighted average is <br /><br />[1/8 x (0% + 0% + 100% + 0%)] + [1/4 x (50% + 100%)] = 12.5% + 37.5% = 50%Anonymoushttps://www.blogger.com/profile/15720479083984981176noreply@blogger.com