tag:blogger.com,1999:blog-8890204.post7973315237240827651..comments2023-01-24T11:41:07.927-05:00Comments on My Biased Coin: Probability Assignments Using SetMichael Mitzenmacherhttp://www.blogger.com/profile/06738274256402616703noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-8890204.post-37548160040982488552012-01-21T18:03:00.342-05:002012-01-21T18:03:00.342-05:00Through simulations similar to Peter Norvig’s, I f...Through simulations similar to Peter Norvig’s, I found that the odds against there being no set in 12 cards when playing a game of Set start off at 30:1 for the first round. Then they quickly fall, and after about the 4:th round they are 14:1 and for the next 20 rounds they slowly fall towards 13:1. So for most of the rounds played, the odds are between 14:1 and 13:1.<br /><br />Also, it turns out that for the probabilities it doesn’t matter much if a “most similar” set is picked, or if a random set is picked. See <a href="http://henrikwarne.com/2011/09/30/set-probabilities-revisited/" rel="nofollow"> Set Probabilities Revisited</a> for more details.Henrik Warnehttp://henrikwarne.comnoreply@blogger.comtag:blogger.com,1999:blog-8890204.post-62921416296320375982011-11-09T08:18:35.990-05:002011-11-09T08:18:35.990-05:00Can I say that the degree of achieving a condition...Can I say that the degree of achieving a conditional statement (if x <br /> then y) is equivalent to a <a href="http://calculate-conditional-and-distributio.blogspot.com/2011/10/online-help-in-solving-probability.html" rel="nofollow">conditional probability</a> p(y|x)?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8890204.post-25471414815478800902011-06-26T04:00:01.882-04:002011-06-26T04:00:01.882-04:00"Apparently the largest number of cards witho...<i>"Apparently the largest number of cards without a set is 20. I wonder if there's a short proof of that without just doing exhaustive search."</i><br /><br />Yes: Davis and Maclagan. <a href="http://www.warwick.ac.uk/staff/D.Maclagan/papers/set.pdf" rel="nofollow">The card game Set.</a> The Mathematical Intelligencer, 25, No. 3, 2003, 33-40.<br /><br />One might also think about the <a href="http://www.cs.illinois.edu/~pbg/papers/set.pdf" rel="nofollow">computational complexity of set</a>.Anonymoushttps://www.blogger.com/profile/13177925339061636642noreply@blogger.comtag:blogger.com,1999:blog-8890204.post-86437711332186917602011-06-24T22:31:51.549-04:002011-06-24T22:31:51.549-04:00Hey Andy -- stop giving away parts of the assignme...Hey Andy -- stop giving away parts of the assignment they're supposed to think about! <br /><br />Though to the first order my model would be when multiple sets can be taken a random one is selected. Maybe a better model is to sort so that sets that are similar in more features have priority over sets that are different in more features. I don't think I care about that level of detail for the assignment, except to the extent they should be thinking about the underlying model assumptions and questioning them.Michael Mitzenmacherhttps://www.blogger.com/profile/02161161032642563814noreply@blogger.comtag:blogger.com,1999:blog-8890204.post-66388299114191501652011-06-24T22:10:21.013-04:002011-06-24T22:10:21.013-04:00What's your model of which sets are removed by...What's your model of which sets are removed by players? It wouldn't be a random set from those present, since some are harder to see. Does the kind of set you grab affect the conditional probabilities you mention?<br /><br />For a given set of players, you could probably build a decent probability model for which sets they'll find; just gather data by having them play the computer version.Andy Dhttps://www.blogger.com/profile/03897281159810085972noreply@blogger.comtag:blogger.com,1999:blog-8890204.post-73449075781975559172011-06-24T16:44:37.176-04:002011-06-24T16:44:37.176-04:00One exercise that might be useful as an example to...One exercise that might be useful as an example to drive home your "cards left on the table not random" point:<br /><br />Show that if 26 sets have been removed, and that there are 3 cards left on the table, those remaining 3 cards *always* form a set.Anonymousnoreply@blogger.com