The coast guard station has received a distress call from the S.S. Minnow, sinking near one of two islands in opposite directions from the station. The call was interrupted by radio failure on the Minnow before the tourboat could specify which of the two islands it was near.
The station chief knows from experience with that tour operator that there is a 20% chance the boat is near island A and an 80% chance it's near island B. The station has 13 rescue boats, and, again from experience, it is known that each rescue boat has, independently of the other search boats, a 20% probability of finding a distressed boat if indeed a distressed boat is present, effecting a rescue.
How should the 13 boats be split between the two islands to maximize the probability that the people aboard the Minnow will be rescued? What is the probability that they will in fact be rescued if that optimal strategy is followed?
Part 2:
Suppose the coast guard station has 40 boats available but each one has only a 5% probability, independently, of finding a ship in trouble (given there is one in that location to be found). And further, there's only a 10% probability the boat is near island A, 90% of being near island B.
I started with a spreadsheet to calculate which island the boats should go to in a sequential order. Simply comparing the probability that island A or B may still have a survivor on it after the previous boats were sent is enough to determine where the next boat went.
Boat Send A B Saved
0 X 0.200000 0.800000 0.000000
1 B 0.200000 0.640000 0.160000
2 B 0.200000 0.512000 0.288000
3 B 0.200000 0.409600 0.390400
4 B 0.200000 0.327680 0.472320
5 B 0.200000 0.262144 0.537856
6 B 0.200000 0.209715 0.590285
7 B 0.200000 0.167772 0.632228
8 A 0.160000 0.167772 0.672228
9 B 0.160000 0.134218 0.705782
10 A 0.128000 0.134218 0.737782
11 B 0.128000 0.107374 0.764626
12 A 0.102400 0.107374 0.790226
13 B 0.102400 0.085899 0.811701
This resulted in the same 3/10 split that others reported with a final rescue chance of 81.1701%. But viewing the spreadsheet I can make some observations. Initially, rescue boats are all sent to the more likely island as long as the ratio between the two islands is larger than the chance of a rescue. Then the boats alternate between islands.
So a shorter calculation method presents itself. Step 1 find the smallest integer N such that the chance of rescuing from the more likely island times the rescue rate becomes less than the chance of rescue from the less likely island; Step 2 divide the remaining boats equally.
In the first problem island B is more likely and the equation becomes 0.8*(0.8)^N > 0.2, which makes N > log(0.2/0.8)/log 0.8 = 6.21257. So the first N=7 are sent to island B. Then the remaining (13-7)/2=6 are split between A and B for a total of 10 to island B and 3 to island A.
Then for the second problem: 0.9*(0.95)^N>0.1 implies N>log(0.1/0.9)/log 0.95 = 42.836. So the first 43 boats should head to island B but with only 40 boats available then every boat available is sent to island B. We would need 44 boats to start sending to island A.
Solution
