tag:blogger.com,1999:blog-5811124440838283502.post2866945396558511570..comments2024-03-02T07:28:06.384+00:00Comments on Tony's Maths Blog: Mathematics and TennisTonyhttp://www.blogger.com/profile/08832715837375830128noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-5811124440838283502.post-19613458931962780092013-05-28T01:31:02.230+01:002013-05-28T01:31:02.230+01:00That makes perfect sense. Thank you so much. I am ...That makes perfect sense. Thank you so much. I am in the process of setting up a tennis blog answering interesting tennis questions using regression techniques, and I would love to hear your thoughts/suggestions. I will leave a link here once it's up and running (:Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5811124440838283502.post-72840182922381817972013-05-27T08:03:11.318+01:002013-05-27T08:03:11.318+01:00The draw divides the players into two equal halves...The draw divides the players into two equal halves, with 2^(n-1) in each. (Sorry, typing maths as text without using exponents does make it a lot more difficult to follow!) So A's half of the draw contains 2^(n-1)-1 players other than A, and the other half contains 2^(n-1): the probability that B is in the former is [2^(n-1)-1)]/[2^(n-1)-1 + 2^(n-1)].<br />For example, if there are 8 players, three are in the same half of the draw as A and four aren't. So the probability that B is in the same half of the draw as A is 3/7.Tonyhttps://www.blogger.com/profile/08832715837375830128noreply@blogger.comtag:blogger.com,1999:blog-5811124440838283502.post-84909914816204177632013-05-27T06:06:40.284+01:002013-05-27T06:06:40.284+01:00I understand that there are (2^n)-1 players who ar...I understand that there are (2^n)-1 players who are not A, but why is it that 2^(n-1)-1 are in the same half of the draw as A? In particular, why is the exponent (n-1)? Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5811124440838283502.post-89987705479357219412013-05-25T15:07:07.051+01:002013-05-25T15:07:07.051+01:00Probability that B is in the same half of the draw...Probability that B is in the same half of the draw as A if there are 2^n players:<br />of the 2^n -1 players who are not A, 2^(n-1)-1 are in the same half of the draw as A, and 2^(n-1) are in the other half. So the probability that B is in the first group is [2^(n-1) - 1]/[2^n -1].<br />Hope that explains it!<br /><br />TonyTonyhttps://www.blogger.com/profile/08832715837375830128noreply@blogger.comtag:blogger.com,1999:blog-5811124440838283502.post-44436633319684878722013-05-25T14:49:51.045+01:002013-05-25T14:49:51.045+01:00"If they are in the same half of the draw, th..."If they are in the same half of the draw, they meet earlier and, in a random draw with 2^n players, the probability of that is (2^(n-1)-1)/((2^n)-1) which is only just under 1/2." <br />How did you come up with that equation?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5811124440838283502.post-35282036344902574482012-07-09T08:49:28.917+01:002012-07-09T08:49:28.917+01:00Thanks. That's very nice!Thanks. That's very nice!Tonyhttps://www.blogger.com/profile/08832715837375830128noreply@blogger.comtag:blogger.com,1999:blog-5811124440838283502.post-27578554652524791082012-07-09T07:58:37.591+01:002012-07-09T07:58:37.591+01:00You might enjoy a couple of the posts at http://ed...You might enjoy a couple of the posts at http://education.theage.com.au/content/maths/gjj0s8Anonymousnoreply@blogger.com