# Brain Teaser: Making Ends Meet

A quick teas­er: Imag­ine you are one of 120 peo­ple in a room. Each per­son in the room is giv­en two lengths of rope and told to chose two of his or her four rope-ends at ran­dom and to tie them togeth­er. Then each per­son is told to tie the remain­ing two rope-ends together.

Then, we count up the loops of rope. How many should there be?

SOLUTION:

When each per­son pre­pares to choose his sec­ond rope-end, we note that one of the avail­able three rope-ends is the oth­er end of the rope he is hold­ing, and the oth­er two are from the oth­er length of rope.

He is equal­ly like­ly to pick any of these three rope-ends, so there is a one-in-three chance that he will cre­ate a loop at this time, and a two-in-three chance that he will instead sim­ply join two ropes into one. He’ll be left with one rope (plus pos­si­bly a loop). Either way, he’ll tie his final rope into a loop in the sec­ond step.

There­fore, we expect one third of the peo­ple (40) to have made two loops, and the remain­ing two-thirds (80) to have made one, for a total of (40 x 2 + 80 = ) 160 loops.

– Wes Car­roll is the head of Do The Math pri­vate tutor­ing ser­vices, Puz­zle Mas­ter for the Ask A Sci­en­tist lec­ture series, and much more. Find out more at wescarroll.com.

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1. B on March 26, 2008 at 9:11

He needs to define what a loop is first.

2. zenkat on April 4, 2008 at 7:05

Fun lit­tle prob­lem, but I have one minor quib­ble. Since the ini­tial selec­tion of an open or closed end is ran­dom, the answer is a sta­tis­ti­cal dis­tri­b­u­tion, with 160 loops as the expect­ed mean.

Bonus ques­tion: what is the stan­dard devi­a­tion of the distribution?

3. Paul on April 11, 2008 at 12:59

You can not base math­e­mat­i­cal log­ic on a human con­di­tion — Assump­tion. You’re argu­ment is valid, pro­vid­ed your assump­tion of the peo­ple will do, what you expect.

4. Govardhan on April 30, 2008 at 1:35

i dont agree with, what Wes is saying…This ques­tion can have mul­ti­ple answers…

5. Sue on May 1, 2008 at 8:38

Yeah..yeah
Try this..2 loops for each knot tied and 1 loop that the ties create.

6. C on May 11, 2008 at 6:27

B is right. What is a loop for the author?
If I take two ropes and tie them one end to one end, I get a longer rope wit two ends free, Then I tie those two free ends and I get a big loop. Ecah per­son gets one big loop, which gives a total of 120 loops. What is wrong with this log­ic ver­sus the for­mu­la­tion of the problem?

7. CB on May 15, 2008 at 4:38

I agree com­plete­ly with C. I don’t see the log­ic in attempt­ing to com­pli­cate a rather sim­ple solution.

8. J on May 15, 2008 at 5:02

I agree with both B and C’s replies; the ques­tion was both unclear and com­pli­cat­ed. How­ev­er, I did like fig­ur­ing it out and came up with 120 loops, also.

9. Ali on May 17, 2008 at 1:31

The answer giv­en does not address the prob­lem as ini­tial­ly set out, which was to start by tak­ing a ran­dom selec­tion of any two ends at which point you have a 50:50 chance of select­ing ends from two dif­fer­ent ropes and you end up with an aver­age of 180 loops. The giv­en answer only works if hav­ing already select­ed one of 4 avail­able ends you are asked to tie that to one of the reamin­ing 3.

10. Ali on May 17, 2008 at 5:49

Cor­rec­tion, your solu­tion is the right one since the answer is the same whichev­er of the 4 ends is select­ed first.

11. paula on May 21, 2008 at 11:24

i don’t think you can say just like that how many peo­ple will fin­ish tying the ropes…asumpting that evry­one man­ages to tie the first ends of the ropes and than the oth­er two,we get 2 loops for evry per­son ..wich means 240 loops…i mean we can only say wich is the max­i­mum num­ber of loops that the 120 per­sons can cre­ate and not how many of them will man­age to do what they’re asked

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