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EdisonCruise
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Joined: September 15th, 2012, 4:22 am

tie ropes to form a ring

August 1st, 2013, 4:14 am

there are 100 ropes ? if you tie 100 ends together, what would be the expected number of ropes that would have tied to their own end and be in a ring?
 
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Vanubis1
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Joined: February 21st, 2011, 7:41 am

tie ropes to form a ring

August 1st, 2013, 7:33 am

If the choice of ends is randomly, i find sum(p=1 to 100) 1/(2*p-1)=3.2843
 
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And2
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Joined: January 29th, 2007, 5:24 pm

tie ropes to form a ring

August 1st, 2013, 5:18 pm

A quick simulation (if i did not screw it up - was very quick coding endeed) indicates that it should be around 0.5.
 
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crmorcom
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Joined: April 9th, 2008, 4:09 pm

tie ropes to form a ring

August 1st, 2013, 6:57 pm

Would the number you're after be rings formed of one single rope? Or do you want total rings including rings made up of several ropes tied together?
 
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Vanubis1
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Joined: February 21st, 2011, 7:41 am

tie ropes to form a ring

August 2nd, 2013, 7:08 am

My answer is for rings composed of any number of ropes.
 
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Vanubis1
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Joined: February 21st, 2011, 7:41 am

tie ropes to form a ring

August 2nd, 2013, 12:08 pm

For rings formed of one single rope, the answer is 100/199 so close to 0.5
 
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crmorcom
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Joined: April 9th, 2008, 4:09 pm

tie ropes to form a ring

August 2nd, 2013, 1:25 pm

So, if I have n non-ring ropes or composite ropes, pick 1 end. There are 2n-1 other ends. If you pick the other end of the same rope (I assume equial probabilities of picking any end) you get a ring, otherwise you just get n-1 composite ropes. If E(n) is the expected number of rings from n pieces,E(n) = p(pick same rope end).1 + E(n-1) = 1/(2n-1) + E(n-1).E(1) is 1, soE(n) = sum_s=1^n 1/(2n-1)For n=100 I get 3.284342 from Excel.
 
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And2
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Joined: January 29th, 2007, 5:24 pm

tie ropes to form a ring

August 5th, 2013, 1:27 pm

Crmorcom,if you pick wrong end (not the other end of the same rope), you tie together (and thus disqualify) two ropes, so you can't use E(n-1)
Last edited by And2 on August 4th, 2013, 10:00 pm, edited 1 time in total.
 
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Vanubis1
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Joined: February 21st, 2011, 7:41 am

tie ropes to form a ring

August 6th, 2013, 6:25 am

When you tie 2 ropes, you have one bigger rope.That' s the way we understand the problem.
 
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Traden4Alpha
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Joined: September 20th, 2002, 8:30 pm

tie ropes to form a ring

August 6th, 2013, 12:13 pm

Each time you perform a tie, it either: converts two ropes into one longer rope or converts one rope into one ring. Both cases reduce the number of ropes available for tying in the next round. That is, the number of remaining ropes (of varying lengths) that are NOT in rings drops by 1.Given 100 ropes and 100 tie operations, ALL ropes will become tied into rings of varying circumference.If "expected number of ropes that would have tied to their own end" includes long multi-segment ropes tied to their own ends, then the answer is all 100 ropes will be rings
Last edited by Traden4Alpha on August 5th, 2013, 10:00 pm, edited 1 time in total.
 
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vinpat
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Joined: December 16th, 2013, 7:36 pm

tie ropes to form a ring

January 23rd, 2014, 1:21 am

Actually a better question is, how many rings (of any circumference) do you get in expectation?I think the answer to that is the harmonic number.