Tying the Knot
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
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.
HintThere are three possible combinations of loop sizes that the strings can form.
The possibility that the strings will form one large loop is 8 in 15 times, or 8/15.
You determined the probability of each situation by finding out the probability of many events. You only had to deal with the bottom knots, because if you left out the top knots, you will just be simplifying the problem a little bit. Then you figure out all the possibilities for the first knot (of the bottom) that could be tied. Then, using those, you found out the three possibilities for each basic diagram, having only the first knot. That gives you 45 possible knot combinations, which is every combination for the bottom half. Then you simply count out the different groups (1 loop, 2 loops or 3 loops) and simplified it. There are 24/45 for 1 loop, 18/45 for 2 loops, and 3/45 for 3 loops. Simplified that is:
**1 Loop: 8/15**
2 Loops: 6/15
3 Loops: 1/15
Oct 13, 2006
|Cool, that was fun. And I'm the first to say so! wheee!|
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:
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 --
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.
Oct 13, 2006
|Thank you for doing that! I myself had a hard time explaining it |
Oct 14, 2006
|That was really confusing.|
Mar 01, 2007
|I did the exact same thing as Qrystal. The original explanation was kind of confusing. Good teaser though, made me think |
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:
We can lose 7 of these which have ab,cd, or ef, leaving 8 out of 15.
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
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.
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|
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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