tag:blogger.com,1999:blog-1055932257464975902.post2303722052119710415..comments2022-07-23T17:34:20.401-05:00Comments on Stupid Motivational Tricks / Bemsha Swing: OddsJonathanhttp://www.blogger.com/profile/09371893596402673898noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-1055932257464975902.post-72946931199540846572013-03-28T10:16:53.105-05:002013-03-28T10:16:53.105-05:00Yes. You are right. It is not two to the power of...Yes. You are right. It is not two to the power of whatever round it is (counting backwardds from the end), but rather a sequence of 2, 4, 16, 256... <br /><br />(not 2, 4, 8, 16, 32, 64)...<br /><br />So it is more exponential than I thought. Jonathanhttps://www.blogger.com/profile/09371893596402673898noreply@blogger.comtag:blogger.com,1999:blog-1055932257464975902.post-62450929872570332892013-03-28T00:35:45.814-05:002013-03-28T00:35:45.814-05:00Your conjecture was already disproven in round 3, ...Your conjecture was already disproven in round 3, when you multiplied by 1/16 (16 is 2^4 not 2^3). I'd guess you multiply each time by 1/[2^(2^(y-1))]...<br /><br />Proof:<br /><br />let y be the round number<br /><br />this has 2^y teams<br />hence 2^y / 2 = 2^(y-1) games<br /><br />so chance of guessing that round entirely correctly is 1/2^(2^(y-1)), which you have to multiply onto the Jameshttps://www.blogger.com/profile/05168447723700552378noreply@blogger.comtag:blogger.com,1999:blog-1055932257464975902.post-91756616832441580252013-03-28T00:34:44.606-05:002013-03-28T00:34:44.606-05:00This comment has been removed by the author.Jameshttps://www.blogger.com/profile/05168447723700552378noreply@blogger.comtag:blogger.com,1999:blog-1055932257464975902.post-80473720945848772962013-03-25T15:47:28.589-05:002013-03-25T15:47:28.589-05:00I feel the need to use my junior high school math ...I feel the need to use my junior high school math once in a while. Jonathanhttps://www.blogger.com/profile/09371893596402673898noreply@blogger.comtag:blogger.com,1999:blog-1055932257464975902.post-22561340708697884772013-03-25T13:42:10.055-05:002013-03-25T13:42:10.055-05:00You're right. Assuming N, the number of teams,...You're right. Assuming N, the number of teams, is a power of two, the number of games played is N - 1. (Three games in the case of four teams, etc.) Assuming a 50% chance of picking each game right, your chance of getting it right overall is 1 / (2 ^ (N - 1)), or 2 / (2 ^ N). And of course the relationship of the number of (complete) rounds to the number of games is exponential already, so Vance Maverickhttps://www.blogger.com/profile/07477306994564623348noreply@blogger.com