Cyclone Season

Solution

Let the five cyclone ratings be \[ 1,2,3,4,5 \] and suppose each day you generate one cyclone, with each rating equally likely: \[ P(\text{rating } i)=\frac15. \] You want to know two things: \[ \text{how long it takes on average to see all 5 ratings,} \] and \[ \text{the expected sum of the distinct ratings collected by the time this happens.} \] This is a standard coupon collector problem. \[ \text{Expected time to see all 5 ratings} \] At the beginning, you have seen none of the ratings. The expected time to get the first new rating is clearly \[ 1, \] because the very first cyclone always gives you a new rating. Once you have seen \(1\) distinct rating, there are \(4\) unseen ratings left. The probability that the next cyclone is a new one is \[ \frac45. \] So the expected number of days to get a new rating at that stage is \[ \frac{1}{4/5}=\frac54. \] Once you have seen \(2\) distinct ratings, there are \(3\) unseen ratings left, so the probability the next cyclone is new is \[ \frac35, \] and the expected waiting time is \[ \frac{1}{3/5}=\frac53. \] Similarly, after seeing \(3\) distinct ratings, the expected waiting time for a new one is \[ \frac{1}{2/5}=\frac52, \] and after seeing \(4\) distinct ratings, the expected waiting time for the final missing one is \[ \frac{1}{1/5}=5. \] So the total expected time is \[ 1+\frac54+\frac53+\frac52+5. \] Putting everything over a common denominator: \[ 1+\frac54+\frac53+\frac52+5 = \frac{60}{60}+\frac{75}{60}+\frac{100}{60}+\frac{150}{60}+\frac{300}{60} = \frac{685}{60}. \] Thus the expected number of days is \[ \frac{685}{60}=\frac{137}{12}\approx 11.42. \] Now we want the expected total \[ X_1+X_2+\cdots+X_T. \] The average rating on any single day is \[ \mathbb{E}[X_i] = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3. \] So each cyclone contributes \(3\) on average. Since you generate cyclones until time \(T\), the expected total sum is \[ \mathbb{E}\!\left[\sum_{i=1}^{T} X_i\right] = \mathbb{E}[T]\cdot \mathbb{E}[X_1]. \] Thus \[ \mathbb{E}\!\left[\sum_{i=1}^{T} X_i\right] = \frac{137}{12}\cdot 3 = \frac{137}{4}. \] So the expected sum of all cyclone ratings generated up until you have seen all five ratings is \[ \frac{137}{4}=34.25. \]