Mar 7, 2008
Brain Teaser: Making Ends Meet
By: Alvaro Fernandez
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.
He needs to define what a loop is first.
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?
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.
i dont agree with, what Wes is saying…This question can have multiple answers…
Yeah..yeah
Try this..2 loops for each knot tied and 1 loop that the ties create.
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?
I agree completely with C. I don’t see the logic in attempting to complicate a rather simple solution.
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.
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.
Correction, your solution is the right one since the answer is the same whichever of the 4 ends is selected first.
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