Windows 7: Today's Puzzle

Correct, John. That was too easy .

X = 2 + ½X = 4
3 * 4 = 12

7 and a half kilos?

I thought it was a trick question...

NExt one, a bit more difficult but still doable. As this is widely used in teaching statistics and probabilities, the answer can be easily found around the interwebs but try to figure this out by yourself before searching the answer.

A group of people. How big must the group be to be absolutely 100% sure at least two persons in the group have the same birthday, are born on the same day and month for instance one in April 14th 1970 and another the same April 14th 1967?

How big must the group be that probability to find two people with same birthday is 99%?

How big must the group be that probability to find two people with same birthday is 50%?

I know it's 23 people for 50% and I seem to recall that 99% is achieved at around the 50 mark but my calculator can't handle calculations larger than 365^39 so I haven't got a hope in hell of working that one out!

Well it's got to be 367 for 100% certainty (in a leap year)
The other two I will have a go at later after taking my dog for a walk (Never know might meet Carl)

Correct, Tom and John.

In probability theory this is known as the Birthday problem.

Due leap years, 100% probability needs 367 people, but surprisingly the 99% probability only needs 57 people, and 50% probability only 23 people.

It is an interesting example of probability theory, how statistics work.

For those of you who remember the Flame malware that came to light last year, this is the main theory behing the MD5 collision attack that gave Flame a, what seemed to be, legitimate digital signature:


http://www.trailofbits.com/resources/flame-md5.pdf

This is also an interesting read: Birthday attack - Wikipedia, the free encyclopedia

I'm still trying to figure out the bricks.