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Topic: Rope Loops (Read 2511 times) |
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ThudnBlunder
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Rope Loops
« on: Jul 18th, 2007, 12:37pm » |
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Jill has six ropes. Randomly choosing two of the twelve loose ends (possibly of the same rope), she ties them together, leaving ten loose ends. Now she again randomly chooses two loose ends and joins them and continues like this until there are no loose ends. What is the expected number of loops Jill ends up with?
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SMQ
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Re: Rope Loops
« Reply #1 on: Jul 18th, 2007, 4:19pm » |
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Given n ropes:
After picking one loose end, there is a 1 in 2n-1 chance of picking the other end of the same rope and thus forming a loop.
After picking two loose ends, whether or not a loop was formed, there are now n-1 ropes remaining (where a "rope" may consist of multiple original ropes tied end-to-end.
So at every step there is a 1/(2n-1) chance of forming a loop. Since these events are independent, the expected number of loops is sum from 1 to n of 1/(2k-1).
I'm sure there's a closed form of that, but to answer the riddle as posed, for n=6, the expected number of loops is 19524/10395 or approximately 1.87821
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birbal
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Re: Rope Loops
« Reply #2 on: Mar 9th, 2011, 12:00am » |
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I was just wondering if we can say anything about the sizes of the loops.
Lets say to start with, we have n ropes and we keep on repeating the process until all the ends are vanished. What is the expected size of the maximum number of the loops?
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Noke Lieu
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Re: Rope Loops
« Reply #3 on: Apr 18th, 2011, 6:32pm » |
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Or, if the loops are linked (like a chain) or separate... 
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| « Last Edit: Apr 18th, 2011, 7:30pm by Noke Lieu » |
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ThudnBlunder
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Re: Rope Loops
« Reply #4 on: May 28th, 2011, 6:02am » |
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on Jul 18th, 2007, 4:19pm, SMQ wrote:
I'm sure there's a closed form of that, but to answer the riddle as posed, for n=6, the expected number of loops is 19524/10395 or approximately 1.87821
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Sumn = H2n - (1/2)*Hn (where Hn = 1 + 1/2 + 1/3 + 1/4 + ..... + 1/n)
= (1/2)*Hn + loge2
--> (1/2)*logen + 0.981755.... as n --> 
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| « Last Edit: May 28th, 2011, 8:18am by ThudnBlunder » |
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Hippo
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Re: Rope Loops
« Reply #5 on: Sep 12th, 2011, 11:36am » |
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on May 28th, 2011, 6:02am, ThudnBlunder wrote:
Sumn = H2n - (1/2)*Hn (where Hn = 1 + 1/2 + 1/3 + 1/4 + ..... + 1/n)
= (1/2)*Hn + loge2
--> (1/2)*logen + 0.981755.... as n -->
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I would use \approx ... instead of =
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