I found some interesting statistics about Travel Bugs (each probability applies to a one-day period, only one incident per day, i.e. a travel bug can't be retrieved and get lost during the same day):
If a travel bug is in a cache in Gibraltar, the probability that it is retrieved is 0.11 and if it is retrieved from a cache in Gibraltar, the probability of it getting lost is 0.004 and getting dropped back to a cache is 0.333, otherwise it stays in the pocket of a cacher. If it is dropped back to a cache, the probability that it is dropped back to a cache in Gibraltar is 0.603, to a cache in Spain 0.287 and otherwise to a cache in another country.
If a travel bug is in a cache in Spain, the probability that it is retrieved is 0.05 and if it is retrieved from a cache in Spain, the probability of it getting lost is 0.009 and getting dropped back to a cache is 0.1, otherwise it stays in the pocket of a cacher. If it is dropped back to a cache, the probability that it is dropped back to a cache in Gibraltar is 0.105, to a cache in Spain 0.84 and otherwise to a cache in another country.
If a travel bug is in a cache in another country, the probability that it is retrieved is 0.075 and if it is retrieved from a cache in another country, the probability of it getting lost is 0.011 and getting dropped back to a cache is 0.25, otherwise it stays in the pocket of a cacher. If it is dropped back to a cache, the probability that it is dropped back to a cache in Gibraltar is 0.0008, to a cache in Spain 0.12 and otherwise to a cache in another country.
I want to drop my travel bug in Gibraltar (day 0) and want to know the following probabilities after one year (365 days):
a) probability of my travel bug being lost
b) probability of my travel bug being in a cache is Gibraltar
c) probability that my travel bug is in the pocket of any cacher (after retrieved from a cache in Spain)
d) after how many days the probability of my travel bug being lost is > 0.9
When you have the probabilities (a-c), round first each probability using 6 decimal places and count the sum of the digits. Add the number of days (d) to the sum and give the result to the checker to get the final coordinates.
e.g. if a = 0.1234567, b = 0.0012341 and c = 0.0123456, the rounded probabilities are 0.123457, 0.001234 and 0.012346 and the sum of the digits is 137037. If d) is 12345 days, the final result = 149382.
You can validate your puzzle solution with certitude.