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More ways to get Braingle...

Tying the Knot

Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

 

Puzzle ID:#33450
Fun:*** (2.35)
Difficulty:*** (2.83)
Category:Probability
Submitted By:medster99*

 

 

 



There are 6 strings clustered together. One end of each string is at point B (the top), and the other at point A (the bottom). First, two of the ends at point A (randomly) are tied together. Then the two more are tied together, and then the last two. Next, two ends at point B are tied together. Then the two more are tied together, and then the last two.
What is the possibility that all the strings will be tied together in one large loop?

Note: Simplify the answer.





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Comments

QrystalAca*
Oct 13, 2006

Cool, that was fun. And I'm the first to say so! wheee!
QrystalAca*
Oct 13, 2006

What I actually wanted to say is how I figured this out ... your explanation kinda confused me.

First, I simplified it by starting to solve the problem after the three bottom knots were already tied. So, all that needed figuring out is: what are the odds of making a big loop, given that the ropes already look like this:
|_| |_| |_|
We can pretend now that there are just three ropes, bent into loops.

Take one of the top ends at random (1 of the 6, marked by the exclamation mark):
|_| |_! |_|
Grab another of the ends at random (1 of the remaining 5). What are the odds that your second choice doesn't screw up the chance of making a big loop? 4/5.

Now the situation looks kinda like this:
|_| |_|^|_|
Grab one free end randomly. There are two situations possible:
Case 1:
You grabbed an end of one of the ropes which is already tied to another rope:
|_| !_|^|_|
Thus, to avoid screwing your chance of getting a big loop, you need to grab one of the ends of the rope that isn't already tied up top:
|_! !_|^|_|
Odds of doing that at random: 2/3.
-- OR --
Case 2.
You grabbed one of the ends of the rope that is not already tied up top:
|_! |_|^|_|
The only way to screw this up is to grab the other end of the same piece of rope. Odds of not doing that: 2/3.

If you have successfully gotten this far:
|_|^|_|^|_|
the two remaining ties WILL make a big loop, 100% probability (1/1).

Thus, the overall probability of making a big loop, given those random choices, is the product of the probabilities of each of the steps: 4/5 * 2/3 * 1/1 = 8/15.
medster99*
Oct 13, 2006

Thank you for doing that! I myself had a hard time explaining it
blondebookwormAus*
Oct 14, 2006

That was really confusing.
KellingtonAca*
Mar 01, 2007

I did the exact same thing as Qrystal. The original explanation was kind of confusing. Good teaser though, made me think
cms271828
Apr 22, 2007

I got 8/15, but I didn't do it that way.
First imagine the ends at A are all tied.
Then for the ends at B:
Choosing one of the ends, this has 5 options of where it joins to.
So joining this and choosing the next end, we see it has 3 options, giving a total of 3*5=15 possible ways to connect ends at B.

If we label the ends at B as a,b,c,d,e,f, such that (a,b),(c,d),(e,f) are joined at A, then the 15 options are:
ab,cd,ef
ab,ce,df
ab,cf,de
ac,bd,ef
ac,be,df
ac,bf,de
ad,bc,ef
ad,be,cf
ad,bf,ce
ae,bc,df
ae,bd,cf
ae,bf,cd
af,bc,de
af,bd,ce
af,be,cd

We can lose 7 of these which have ab,cd, or ef, leaving 8 out of 15.
Eshootzi_scrs
Apr 24, 2007

Cool. The answer was confusing. 45 possible knot combinations?
I could only come up with 15 possible correct solutions tha would make one big loop.
Did anyone else find it amazing that with the given choices it would be a loop over 50% of the time.
Im gonna try it on ten people and see if it works. Nice one
javaguru*us*
Dec 09, 2008

I also did it the same as Qrystal except that I didn't bother to separate the selection of the second pair into two cases. There are two sets of two ends for the second pair and you need one end from each set, so the two cases are really the same case.

Very cool puzzle though. My first intuition was that the problem would be more complicated than the simple solution and that the odds would be lower than they turned out to be.
rnd2101
Dec 27, 2010

There is a very simple way to do this. Say that you first of all do all of the preliminary "A" ties, leaving only three independent strings (B_a, B_b, B_c) where B_a has sides B_1 and B_2,; B
rnd2101
Dec 27, 2010

There is a very simple way to do this. Say that you first of all do all of the preliminary "A" ties, leaving only three independent strings: B_a, B_b, B_c, where B_a has sides B_1 and B_2, B_b has sides B_3 and B_4, and B_c has sides B_5 and B_6.

When we go to tie our first "B-knot" there are 5 possible knots we can tie, and thus a 4/5 chance that we don't tie the string on to itself. Assuming we have not tied the string on to itself, we now have two strings, and we thus have a 2/3 chance of not tying either string onto itself. After we have done this second "B" knot, there is only one knot left to tie (completing the loop).

Therefore, the probability of tying the strings into one large loop is

P = 4/5 * 2/3 = 8/15.



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