Jane Street Interview Question: Say I take a rubber band and ... | Glassdoor

Interview Question

Summer Trading Intern Interview(Student Candidate) New York, NY

Say I take a rubber band and randomly cut it into three

  pieces. What's the probability that one of the pieces has length greater than 1/2 of the original circumference of the rubber band.
Answer

Interview Answer

9 Answers

6

3/4

Interview Candidate on Oct 4, 2011
0

Suppose you have two cuts on the rubber band placed randomly.
The probability of having one segment greater than half the circumference is the probability that the third cut will be inside the combined range of 90* to either side of the cuts.
Since the average distance between the first two cuts is also 90*, the combined range is 270*, or 3/4 of the circle.

Anonymous on Oct 13, 2011
3

You need 3 cuts to end up with 3 pieces. The first cut doesn't matter. The second cut can also be anywhere and the largest piece will still be at least half the circumference. What matters is the third cut, which should lie in the same half as the second cut. So the probability is actually 1/2.

Anonymous on Oct 19, 2011
5

The correct answer is 3/4, as this problem is equivalent to the famous 3-points-on-semicircle problem.

Why? If one of the pieces has length greater than 1/2 the circumference, then the three points of cutting must lie in the same semicircle. On other hand, if the three points of cutting lie on the same semicircle, then the longest piece must be at least 1/2 of the circumference.

Ed on Oct 25, 2011
3

For reference to the 3-points-on-same-semicircle problem, see e.g., http://godplaysdice.blogspot.com/2007/10/probabilities-on-circle.html

Ed on Oct 25, 2011
0

1/4

1 -3/4

Anonymous on Nov 27, 2011
3

suppose I have two points whose minor arc distance is t <= 1/2. Then the range of semicircles covering both points gives an arc length of (1/2+1/2)-t = 1-t.

say we fix the first point, tracing the second point around gives minor arc lengths from 0 to 1/2 and then 1/2 to 0. Therefore the answer is

2*integral (1-t) from 0 to 1/2, which is 2(1/2-1/8) = 3/4

pothead on Jan 7, 2012
0

It's 3/4. Cut it into 1 piece make a line. Cut it close. Pretend the length is 100. If you cut the first at x=1, as long as it isn't between x=50-51, it will have a length greater than 50% so there's 99% chance. You can imagine that if the cut was infinitely close to the end it would be about 100%. Now cut at x=2 you can't do between x=50-52. For x=3 it's 50-53 etc. So when you get to right to infinitely close to 50 it is pretty much between x=50-100 so there is a 50% chance you hit your spot. (obviously 50-50 is 100%, but since this length is continuous there's little chance it lands on that point). Obviously since this is symmetrical you can see this pattern going from 50% to 100% at the other end. Since each point on the continuous line has the same probability of happening the answer is clearly 75%.

KG on Jan 21, 2013
1

This problem is also equivalent to the probability that, if you have a line segment from 0 to 1 and you make 2 random cuts on that line segment, what is the probability that the three resulting pieces do NOT make a triangle?

anon on Feb 20, 2014

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