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Topic: Choosing Two Reals in (0,1) (Read 524 times) |
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ThudnBlunder
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Choosing Two Reals in (0,1)
« on: Dec 20th, 2004, 1:00am » |
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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?
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| « Last Edit: Dec 20th, 2004, 8:25am by ThudnBlunder » |
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Barukh
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Interesting! The attached drawing IMHO contains everything needed for the solution.
If I didn't make any mistakes, the answer is (5-pi)/4.
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Sir Col
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Re: Choosing Two Reals in (0,1)
« Reply #2 on: Dec 21st, 2004, 12:22pm » |
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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...
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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.
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| « Last Edit: Dec 21st, 2004, 12:29pm by Sir Col » |
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Eigenray
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Re: Choosing Two Reals in (0,1)
« Reply #3 on: Dec 21st, 2004, 1:37pm » |
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I got (2-log 2)/2 before I realized it was the nearest integer function, not the greatest integer (floor). It's neat either way.
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| « Last Edit: Dec 21st, 2004, 1:40pm by Eigenray » |
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Barukh
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Re: Choosing Two Reals in (0,1)
« Reply #4 on: Dec 22nd, 2004, 11:29pm » |
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on Dec 21st, 2004, 1:37pm, Eigenray wrote:| I got (2-log 2)/2 before I realized it was the nearest integer function, not the greatest integer (floor). It's neat either way. |
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I made the same mistake 
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