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        riddles >> putnam exam (pure math) >> Choosing Two Reals in (0,1)
        
(Message started by: THUDandBLUNDER on Dec 20^th, 2004, 1:00am)

Title: Choosing Two Reals in (0,1)
Post by THUDandBLUNDER on Dec 20^th, 2004, 1:00am

Choose two real numbers x,y at random in (0,1) with a uniform distribution.
What is the probability that [x/y] is even, where [ ] denotes the nearest integer function?

Title: Re: Choosing Two Reals in (0,1)
Post by Barukh on Dec 21^st, 2004, 10:49am

Interesting! The attached drawing IMHO contains everything needed for the solution.

If I didn't make any mistakes, the answer is [hide](5-pi)/4[/hide].

Title: Re: Choosing Two Reals in (0,1)
Post by Sir Col on Dec 21^st, 2004, 12:22pm

I got the same answer, but I imagined it the other way round, and I'm afraid that I need to take things much more slowly...

::[hide]
If x/y < 1/2, then [x/y]=0. We can rearrange this to get y > 2x and as x,y < 1, this corresponds to the area above the line y=2x, with x and y limited to 1: when y=1=2x, x=1/2, so area = (1/2)/2 = 1/4.

Otherwise, [x/y] will become 2n when 2n-0.5 < x/y < 2n+0.5; that is, 4n-1 < 2x/y < 4n+1.

4n-1 < 2x/y leads to y < 2x/(4n-1)
2x/y < 4n+1 leads to 2x/(4n+1) < y

Therefore, 2x/(4n+1) < y < 2x/(4n-1)

I always like to substitute numbers in to get a feel for things at each stage, so... when n=1,2,3,4,..., we get: 2x/5 < y < 2x/3, 2x/9 < y < 2x/7, and so on.

That is, for the first interval we want the area below y = 2x/3 and above y = 2x/5. We can calculate this by subtracting the area of the triangles, noting that x=1 is the limit of this region; that is, (2/3)/2-(2/5)/2=1/3-1/5.

In general, the area of each interval will be given by (2/(4n-1))/2-(2/(4n+1))/2 = 1/(4n-1) - 1/(4n+1).

So we get the infinite series, S = (1/3-1/5)+(1/7-1/9)+(1/11-1/13)+...

And as pi/4 = 1-1/3+1/5-1/7+... is a well known identity, 1/3-1/5+1/7-... = 1-pi/4 = S

Hence the probability we seek is 1/4+1-pi/4 = (5-pi)/4.
[/hide]::

Title: Re: Choosing Two Reals in (0,1)
Post by Eigenray on Dec 21^st, 2004, 1:37pm

I got [hide](2-log 2)/2[/hide] before I realized it was the nearest integer function, not the greatest integer (floor). It's neat either way.

Title: Re: Choosing Two Reals in (0,1)
Post by Barukh on Dec 22^nd, 2004, 11:29pm

on 12/21/04 at 13:37:29, Eigenray wrote:

I got [hide](2-log 2)/2[/hide] before I realized it was the nearest integer function, not the greatest integer (floor). It's neat either way.

I made the same mistake ;D