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

Aditya promised the chocolate, I didn’t.

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.

Also, I believe the path you follow while switching doors matters, and that your analysis of strategy four is imprecise (though you are correct that by the last door, it is inferior to strategy two).

Intuitively, strategy two is the best if you know the game will proceed until the last door: you choose a random door with P of C 1/n. While losing doors are revealed, this probability will never change so long as it is your choice. P of C for the other closed doors will rise each time a losing door is opened. When you make the switch to the final door, it has P of C (n-1)/n … but only if you stayed with your initial door for the rest of the game. Do you see why this casts some doubt on your analysis of strategy four?

Image 1: Logo of The Rolling Stones, of which Sir Mick Jagger was a member

Image 2: Cricinfo, with which Mick Jagger collaborated to form Jagged Internetworks and bought online rights for watching a cricket tournament

Image 3: Adam Hollioake, the captain of the England team that participated in the tournament whose online rights Mick Jagger bought

Bonus:

Image 4: Johnny Depp — He plays Capt. Jack Sparrow in the Pirates of the Caribbean series of movies. In the third movie of this series, the role of Jack Sparrow’s father, Capt. Teague, has been portrayed by Keith Richards, a member of The Rolling Stones. Also, Jack Sparrow’s character was partly modelled after Keith Richards.

(1) and (4): The role of Jack Sparrow played by Johny Depp is allegedly inspired by Keith Richards. And he also featured in one of the movies.