tag:blogger.com,1999:blog-6288862798546085706.post4600531627611894310..comments2023-03-31T02:27:25.978-04:00Comments on Econometrics By Simulation: Flip a fair coin 4x. Probability of H following H is 40%???Francishttp://www.blogger.com/profile/16658586705916884436noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6288862798546085706.post-16500652611587419332015-10-15T06:32:10.971-04:002015-10-15T06:32:10.971-04:00The paper is not clear at all (on a first, quick r...The paper is not clear at all (on a first, quick read of the first 3 pages) about whether the authors show that people incorrectly compute probabilities or whether the authors believe that the incorrect method they derive is correct. A mess of a paper. Very good blog post. And great comment by Richard Moore.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6288862798546085706.post-85121361311917353442015-10-12T21:53:21.039-04:002015-10-12T21:53:21.039-04:00In calculating his 40.7% the author has calculated...In calculating his 40.7% the author has calculated the probability of the event within different sets of observations and then takes the average of these probabilities. This is effectively taking the average of averages. Bad Math!! <br /><br />The correct way to calculate the likelihood of p(HH|H) is to count out the number of instances of H, then see how many of them are followed by another H.<br /><br />Let’s examine the16 permutations of Heads and tails taken 4 at a time again. There are (16*3) / 2 or 24 Heads in the first 3 positions of the 16 sets of coin flips. Of those 24 Events of Heads that appear, 12 are followed by another H, or 50% probability of H following H. <br /><br />If the original article was trying to point out the issue of edge conditions then good for them. The calculated probability of a H following an H, counting H’s that are at the end of a set as a failure is 37.5%. This is easily calculated in the following way: <br /><br />1. 25% of The H’s that are observed in the strings of 4 observations occur at the end of the strings. These observations have a 0% chance of being followed by an H. <br />2. 75% of the H’s do not end a string and are followed by an H 50% of the time. <br />3. 25% * 0% + 75% * 50% = 37.5%. <br /><br />This edge effect attenuates as the strings get longer, because the edge condition H is a lower percentage of the total number of H’s observed. If the strings of observations were 100 observations long then the likelihood of an H followed by an H would be 49.5%: (1% * 0 + 99%* 50% = 49.5%)<br /><br />The message of the article should be that you need to take care that when you count initial observations. To calculate the probability of an event occurring, you should only count those observations that are observed at a time when they have the opportunity to should the tested for pattern. <br /><br />To bring this home, let’s say we were looking for the pattern of 4 H’s in a row. We expect this to happen 1 time in 8. But it happens only once in our 16 patterns and the count of H’s is 32. 75% of the H’s have a 0% chance of developing into 4H’s in a row due to the edge condition. If you do not take the edge condition into effect, you might think that the likelihood of an H being followed 3 more H’s is 1/32. (25% * 1/8 + 75% * 0%) <br /><br />-- Richard Moore (also as Anonymous because it was the only way to get it entered)<br />Anonymoushttps://www.blogger.com/profile/17413296405184274268noreply@blogger.comtag:blogger.com,1999:blog-6288862798546085706.post-52432706959722976432015-10-12T17:22:41.010-04:002015-10-12T17:22:41.010-04:00The problem here is the averaging. If you look at ...The problem here is the averaging. If you look at HHHH vs TTHH for example, both of them have P(H|H) = 1, but there are 3 successes in HHHH and only 1 in TTHH. Instead of averaging across samples, you could add 1 to a global counter if a head followed a head, and 0 otherwise. If you do it this way, you'll see 0.5.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6288862798546085706.post-20288971430452636332015-10-09T18:01:18.975-04:002015-10-09T18:01:18.975-04:00Your last table shows the 16 possible outcomes (al...Your last table shows the 16 possible outcomes (all of which presumably have equal probability).<br /><br />Of the 16, 8 have a sequence that includes “HH.” So there's a 50% chance of observing a “hot hand” in coin flipping, if by “hot hand” you mean a pair of heads in a row. The test seems not very stringent, roughly the same test as whether a basketballer shoots successful 3-pointers back to back. Is that a hot hand?<br /><br />But if there is a hot hand, there must be a similar period of cold hands, the likelihood of finding a “TT” sequence. Nobody will brag about cold hands, or pay much attention to them, but they are…uh, the flip side of the hot hands argument. Not surprisingly, there are 8 sequences that have a “TT” in them. The chance of a narrowly-defined cold hand is the same 50% as the chance of the narrowly-defined hot hand.<br /><br />If a hot hand can't exist without there being a corresponding cold hand, it's notable that only 2 of the sequences have neither HH or TT (THTH and HTHT). The weakness of looking for only a single pair of back-to-back lined-up results is shown by the fact that it's hard to NOT find a temperature effect by it.<br /><br />/ss WaltFrench. Anon is the only ID I can get to work; grr!Anonymousnoreply@blogger.com