Bloomberg LP Interview Question
Country: United States
Interview Type: Phone Interview
This question did not mention the speed of each of the 3 person, and if they will stop running once they reach a corner. The probability could be 1 under some conditions...
Each person can go in one of two directions.
So there are 2*2*2 = 8 possible 'group moves' for these three guys.
From these 8 moves only 2 in which no one meets:
1 - when all three go clockwise
2 - when all three go counter clockwise
So P (no one meets another) = 2/8 = 1/4.
And then:
P (at least two persons meet) = 1 - P (no one meets another) = 1-1/4 = 3/4.
It is 3/4.
My approach was that,
Probability the n'th person goes left: p[n,l]=1/2
Probability the n'th person goes right: p[n,r]=1/2
Since each probability is independent of the other then the overall probability of an outcome is the products of each probability.
So, the probability of them all (here n=1,2,3 for 3 people) turning left is: p[1,l]*p[2,l]*p[3,l]=1/8. Also the probability of them all turning right is: p[1,r]*p[2,r]*p[3,r]=1/8.
The only way that at least one collision occurs is that they don't all turn right or all don't turn left: 1-(1/8+1/8)=1-1/4=3/4.
Total 2*2*2 = 8 possible configurations after the three person move randomly from the triangle vertices. Out of these, only two configurations lead to people not meeting with anybody - purely clockwise / anti-clockwise motion. So, Probability of meeting = (8-2)/8 = 0.75
- Sharma February 13, 2012PS: I assume probability that two person meet each other doesn't exclude the probability of one person meeting both the other person (one pair on the edge, other upon reaching the vertex).