tag:blogger.com,1999:blog-29763791.post6882173139316473861..comments2023-10-31T05:07:19.353-04:00Comments on Delenda est Carthago: A Conspiracy Theory, ExaminedDr. Φhttp://www.blogger.com/profile/14086783503820477029noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-29763791.post-37478361714034723222020-06-03T12:58:28.369-04:002020-06-03T12:58:28.369-04:00I stumbled across this old post. I don't know ...I stumbled across this old post. I don't know if you are paying attention to comments, but I thought you might be intrigued by this tidbit on proximity to killings. Robert Todd Lincoln was in town visiting his family when his father was shot at Ford's Theater. He became a well respected lawyer and politically astute enough to be Secretary of War in 1881, and at a train station about to leave with President Garfield when Garfield was assassinated. Years later he was in Rochester, New York, to meet President McKinley when McKinley was shot. He lived into the 1920's, but refused to ever be near a President again. What are the odds?Rod Jeaanhttps://www.blogger.com/profile/05669244047978746963noreply@blogger.comtag:blogger.com,1999:blog-29763791.post-13559105727824157652019-10-16T14:22:37.700-04:002019-10-16T14:22:37.700-04:00Perhaps wasted, as this post is really old, but he...Perhaps wasted, as this post is really old, but here is my comment:<br /><br />If I use the intuitive back of the envelope way, I come up with 3 people at both events.<br /><br />For simplification, assume intersection of both populations does not get changed by scaling one by 0.5 and the other by 2, furthermore, simplify US population to 300 million.<br /><br />80K -> 40K<br />20K -> 40K<br /><br />Now assume 3/4 of audience present at time of shooting.<br /><br />40K -> 30K<br /><br />Assume independent probabilities, that presence at one event does not affect probability of being at other event.<br /><br />By belonging to the population of "Americans", the probability of being at either event is 1/10K, or without the scaling, 1/20K and 1/5K.<br /><br />Independent probabilities: p12 = p1*p2 = 1/100 million or 10^-8.<br /><br />The naive answer is that the expected number present at both events is US population times p12 = 3*10^8 * 10^8 = 3.<br /><br />I understand there is a binomial expansion that shows the relative probabilities of 0 to 10 or whatever cutoff you want, however here the intuitive way of calculating is not far off, the problem is that the average Vox or mainstream media writer is simply innumerate.<br /><br />If taking it one step further, assuming that people at the Las Vegas shooting are less likely to go to future big public events, that is a negative cross correlation, however the geographic proximity between LA and LV is a positive cross correlation, and in this case one could conclude that they somewhat cancelled out.Vikinghttps://www.blogger.com/profile/09347653720514068701noreply@blogger.comtag:blogger.com,1999:blog-29763791.post-22012984152137236742019-10-16T14:02:51.153-04:002019-10-16T14:02:51.153-04:00In this case, the naive method to estimate works f...In this case, the naive method to estimate works fine.<br /><br />For simplicity, assume no significant change in odds by doubling 20K to 40K, and reducing the 80K to 40K.<br /><br />Then, to simplify further, Use 300 million instead of 320 million for US population.<br /><br />Furthermore, assume 75% present at shooting for each event.<br /><br />Now 30K/300 Million = 1/10K.<br /><br />Multiply the 2 probabilities, and you get 1/100 million.<br /><br />Probability x US Population =<br />10^-8 x 3 x 10^8 = 3 people expected to be present for both events.<br /><br />This is of course coincidence, but I expect not too different from the expectancy value from a Monte Carlo simulation.Vikinghttps://www.blogger.com/profile/09347653720514068701noreply@blogger.comtag:blogger.com,1999:blog-29763791.post-50295503972476160772019-08-31T11:34:36.104-04:002019-08-31T11:34:36.104-04:00Thanks! Yes, I had that same birthday math trick ...Thanks! Yes, I had that same birthday math trick presented in a seminar by <a href="https://academywatch.blogspot.com/2007/06/but-what-if-youre-idiot.html" rel="nofollow">this guy</a> back in '99 I think.<br /><br />I nearly gave up on this problem a couple of times trying to find a solution that didn't require iteration. If one exists, I don't know it.<br /><br />I've had another math post in my "draft" list for a year. I'm going to try to finish it up.Dr. Φhttps://www.blogger.com/profile/14086783503820477029noreply@blogger.comtag:blogger.com,1999:blog-29763791.post-30017091138775335292019-08-17T10:35:20.015-04:002019-08-17T10:35:20.015-04:00I have a degree in Aerospace Engineering. I am a ...I have a degree in Aerospace Engineering. I am a high school math teacher teaching as high as AP Calculus. I love math and it makes perfect sense to me most of the time.<br /><br />GOD do I hate probability. It is so confusing. <br /><br />That being said, your explanation jibes better with what I do remember than either Barbie's or Vox's.<br /><br />I will add that the answer to this problem is pretty counter-intuitive and why we shouldn't trust our feelings when it comes to probability. Sort of like the problem where you calculate the probability that two people in a room have the same birthday. Without rehashing the math it works out that you only need about 23 people to exceed 50%. <br /><br />PS I enjoy your annual posts. Thanks.heresolonghttps://www.blogger.com/profile/00461382067580153600noreply@blogger.com