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Feb 16, 2023 at 8:40 history edited Jukka Kohonen CC BY-SA 4.0
Even clearer this way?
Feb 16, 2023 at 8:34 history edited Jukka Kohonen CC BY-SA 4.0
Clarify the minimum
Feb 16, 2023 at 7:23 history edited domotorp
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Jul 16, 2020 at 3:09 review Close votes
Jul 16, 2020 at 13:15
Jul 9, 2020 at 12:02 answer added Chris timeline score: 10
Jul 9, 2020 at 8:42 answer added mike timeline score: 2
Jul 9, 2020 at 0:36 comment added Gerry Myerson Also posted to (but closed on) m.se, math.stackexchange.com/questions/3749357/…
Jul 8, 2020 at 20:02 comment added Brian Hopkins The last section of Problems and Snapshots from the World of Probability could also be relevant; $\S$17.6 "Palindromes" starting on p244. The book is by Blom, Holst, and Sandell, published by Springer in 1994.
Jul 8, 2020 at 19:43 comment added Brian Hopkins The unrestricted question was posed on Quora in 2016 and answers there suggest why the three flip minimum makes a more interesting question.
Jul 8, 2020 at 19:40 comment added user44143 I'd keep this open. There are some subtle dependencies, e.g.: If the first three flips are not a palindrome, and the first four flips are not a palindrome, then the first four are either HHTH, HHTT, HTTT, THHH, TTHH, TTHT. So the conditional probability of getting a palindrome at the fifth throw is 1/3.
Jul 8, 2020 at 19:34 comment added Brian Hopkins @DieterKadelka Well, you have to require some minimum length, else you'll never get past one flip.
Jul 8, 2020 at 19:16 comment added Dieter Kadelka 11 or 00 are no palindromes, but 101, 111, 000, 010 are palindromes? I'm irritated.
Jul 8, 2020 at 17:30 review Close votes
Jul 9, 2020 at 5:38
Jul 8, 2020 at 17:06 review First posts
Jul 8, 2020 at 17:13
Jul 8, 2020 at 16:59 history asked Joshua Haim Mamou CC BY-SA 4.0