Just thought I should re-post this. One of the most beautiful poems I’ve come across.

]]>I think that has been the struggle throughout history. However, if there is any hope it is in America. Although, never a perfect nation, the ideals that were placed in our governing documents, such as our constitution empower us to live in a country where governing on a whole is not just about vested interests. However, Americans better wake up fast and corporately become more active to clean things up….. at some point it will be too late and I’m afraid that day is coming quickly.

]]>No problem, dude. All the best with the experiments.

]]>Aditya promised the chocolate, I didn’t. ðŸ˜‰ ðŸ˜€

By the way, a discussion on Aditya’s work has been going on at the Oktave Forum. You may find it interesting: http://forum.oktave.in/index.php?PHPSESSID=e054896eef4374174db8d47172f4e475&topic=78.0

Thanks!

]]>It was an interesting one. By comparing a large set of answers we can get a good idea of what are the terms that people commonly associate with something while which others are more eclectic …

]]>P(one person having a birthday): (365/365)

P( two people having different birthdays):(365/365)(364)(365)

P(3 people having diff bdays):(365/365)(364/365)(363/365)

etc…

P(23 diff bdays):(365/365)(364/365)…(343/365)=.4927…

This means that the prob of 23 people having different birthdays is 49.27%. OR that the probablility of there being at least one bday the same is 50.73%

]]>This contradiction occurs because the birthdays of the pairs of people are not independent events. They’re related. That is, if I’m in the group of people, I don’t get a new birthday each time I’m paired with someone to compare our birthdays.

One person has a birthday. P of having a pair of people with the same birthday (with only one person) is 0, since there are no pairs. P1 = 0.

The second person has a 1/365 chance of matching birthdays with the first person. P2 = 1/365.

The third person may not be needed for a birthday matching pair to be present. If needed, there is a 2/365 chance of a match. P3 = 1/365 + (364/365 * 2/365).

P1 = 0, For n>1

Pn = P(n-1) + (1-P(n-1) * (n-1)/365.

This should mean that P365 = 1, so P366 = 1 + (1-1 * (365/365)) = 1 + (0 * 1), and likewise for n > 365.

]]>{2,1} () 0

{3,1} (2) 1

{3,2} (1) 1

{4,1} (3,2) 2

{4,2} () 0

{4,3} (1,2) 2

{5,1} (4,3,2) 3

{5,2} (3,1,4) 3

{5,3} (2,1,4) 3

{5,4} (1,3,2) 3

{6,1} (5,4,3,2) 4

{6,2} (4) 1

{6,3} () 0

{6,4} (2) 1

{6,5} (1,4,3,2) 4

moves in game: max(m,n)/gcd(m,n) – 2

If the number of moves is even, make the other player go first.

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