tag:blogger.com,1999:blog-6841654461773705283.post5525167300707624449..comments2008-09-28T07:34:39.944+03:00Uri Kalishhttp://www.blogger.com/profile/08561692601016236573noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6841654461773705283.post-78582017317581057072008-09-28T07:34:00.000+03:002008-09-28T07:34:00.000+03:00

On principle, everyone should read Fernat's Equasion. <BR/><BR/>That way we can be suitably down to reality. Particularly when it comes to tossing coins.<BR/><BR/>I love your blog. Keep up the good work.

Dano MacNamarrahhttp://www.blogger.com/profile/02718969376533836939noreply@blogger.comtag:blogger.com,1999:blog-6841654461773705283.post-67231991304244367022008-05-18T03:22:00.000+03:002008-05-18T03:22:00.000+03:00

Hey Yair,<BR/><BR/>1. You must agree that consecutive tosses are much easier to spot when looking at a result.<BR/><BR/>2. Not all x-letter-words have the same probabilities for appearing in a result. For example, the odds for having 'HHHHH' are not equal to the odds for having 'HHTHH'. In this case our intuition telling us that the word 'HHHHH' is not as common as the word 'HHTHH' is correct,

Uri Kalishhttp://www.blogger.com/profile/08561692601016236573noreply@blogger.comtag:blogger.com,1999:blog-6841654461773705283.post-16275535430564012782008-05-18T00:21:00.000+03:002008-05-18T00:21:00.000+03:00

Hi Uri,<BR/><BR/>4-heads is only a private case of any other 4-tuple (e.g. HTHT, HHHT) and therefore not having at-least one of the 2^n n-tuples is equally strange. <BR/><BR/>Moreover, what about non-consecutive occurences? for example, having H?H?H?H (? stands of whatever) is expected to occur as many times as H???H???H???H and so on.

yair h.noreply@blogger.comtag:blogger.com,1999:blog-6841654461773705283.post-33095502092065778932008-05-08T11:01:00.000+03:002008-05-08T11:01:00.000+03:00

I just love stuff, and observations, like this! :)

Markushttp://bloggingthent.blogspot.comnoreply@blogger.com