tag:blogger.com,1999:blog-3829599.post109616016247565860..comments2022-12-23T08:17:48.984-08:00Comments on Agoraphilia: The Three-Sided Coin: AnswersUnknownnoreply@blogger.comBlogger9125tag:blogger.com,1999:blog-3829599.post-88655561464257880652007-03-21T22:57:00.000-07:002007-03-21T22:57:00.000-07:00No. There is only one way to get three heads (a s...No. There is only one way to get three heads (a sum of three), but there are three ways each to get one head or one tail.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1096841744069753332004-10-03T15:15:00.000-07:002004-10-03T15:15:00.000-07:00another take that is not mentioned: label head and...another take that is not mentioned: label head and tail 0 and 1; list your options as 1 2 3; now throw the coin 3 times; add up the result; if its 0 start over; is this correct?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1096228987518322532004-09-26T13:03:00.000-07:002004-09-26T13:03:00.000-07:00To see that Mike's and Lauren's approaches are equ...To see that Mike's and Lauren's approaches are equivalent, modify Mike's approach to answer "Yes" in case TH and "No" in cases HH,HT. Then what Mike is doing is just flipping coins two at a time until a head comes up, and answering "Yes" if and only if that first head comes up on an even-numbered flip.<br /><br />This problem could be a starting point for a nice philosophy-of-math discussion about measure theory, probability, and certainty. The Mike-Lauren approach will give an answer in finite time with probability 1. However, this does not quite mean it will *always* give an answer-- the sequence of flips that results in all tails forever is one possible sequence of flips; it just has zero probability since there are infinitely many (indeed uncountably many) sequences.<br /><br />It is not hard to prove, using the fact (noted by a previous commenter) that 2^n is not divisible by 3, that this is the best you can do, in the sense that no event in a probability space generated by a bounded sequence of coin flips-- a sequence with a finite maximum time to give an answer-- can have probability exactly 1/3.<br /><br />Nick WeiningerAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1096214855191517882004-09-26T09:07:00.000-07:002004-09-26T09:07:00.000-07:00I wolk up thinking about Glen's solution and I thi...I wolk up thinking about Glen's solution and I think he was too hard on himself. His solution can be improved on! Suppose A flips with B and wins. Next A flips with C and wins. Game over! No need for the third flip unless you want to know where to go to after going to A. So I suspect that all three methods are as efficient as each other. A true elimination process!<br /><br />TrumpitAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1096210359217340972004-09-26T07:52:00.000-07:002004-09-26T07:52:00.000-07:00Correction to the end of second paragraph: then 2^...Correction to the end of second paragraph: then 2^(n-2) is divisible by 3 and so on...hchttps://www.blogger.com/profile/04434414106884235127noreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1096210264562440872004-09-26T07:51:00.000-07:002004-09-26T07:51:00.000-07:00Seems like there are no powers of 2 divisible by t...Seems like there are no powers of 2 divisible by three. This just occured to me: every power of 2 is divisible by 2. <br /><br />Let us assume 2^n is divisible by 3. Then it is also divisible by 6. 6 is 2*3, so we can deduce 2^n/2 = 2^(n-1) is also divisible by 3, and consequently it is divisible by 6, so 2^(n-1) is divisible by 3 too, and so on...<br /><br />This might be trivial to you, but I'm not a mathematician and it was a great fun to come out with this while reading your posting.hchttps://www.blogger.com/profile/04434414106884235127noreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1096185355379039882004-09-26T00:55:00.000-07:002004-09-26T00:55:00.000-07:00I believe my answer took me longer than most (abou...I believe my answer took me longer than most (about 30/40 mins). That might just break the tie in Mike's favor. Not to mention I came up with a very wrong answer before this one. <br /><br />I thought through the problem in the same way that it is illustrated in the original post. However, there are other ways to "see" it. <br /><br />The binary expansion of 1/3 is .010101010..... <br />If you treat a head as zero and a tail as one it becomes easier to see. Thus, if what you have flipped, when the first head appears, is less than than 1/3 then you "win" (with a 1/3 probability). You could end up in the situation where you flip forever. . .but as Professor Whitman points out, 2 flips are expected.<br /><br />I like the problem! Thanks.<br />Lauren<br />LFisher03@mckenna.eduAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1096166975863380792004-09-25T19:49:00.000-07:002004-09-25T19:49:00.000-07:00To help remedy the selection bias, I thought about...To help remedy the selection bias, I thought about it for around 15 minutes and did not come up with a solution (although what I did come up with was closest to Glen's -- I just didn't quite get to the repeatability part).Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1096165012993646982004-09-25T19:16:00.000-07:002004-09-25T19:16:00.000-07:00Just because you can find a solution to a problem ...Just because you can find a solution to a problem quickly doesn't mean the problem isn't worth posing. The fact that there are different solutions is insightful in itself. I think Lauren's answer is the most elegant and I would like to know how it occurred to him/her. To "see" something that requires an infinite series to prove is rather remarkable to me. Next, I liked Glen's "knockout" solution which although a tad longer seems very practical requiring almost no understanding of probability. Sorry, Mike, if I got the same solution as you did in about a minute, it means we both think rather linearly (here we go from one toss to two.) I'm not sure what to make of those that gave "weird" non-mathematical answers. Maybe they are voting for Bush!<br /><br />TrumpitAnonymousnoreply@blogger.com