Have you ever wondered what the actual probability is to discover a resource deposit in one of your mines? What is the actual chance of getting such an event?
According to the game XML, the chance of discovering a certain resource (which must be known to you) is 1/10000 (see e.g. http://forums.civfanatics.com/showthread.php?t=140111) per known resource. So the chances are higher if you know more (e.g. bronze, iron, coal etc. in addition to the precious metals which are known from the start).
Based on this little tidbit, I tried to make an analysis which shows the probability of discovering a mineral deposit as a function of turns elapsed, assuming that you know of five resources (e.g. copper, iron, gold, gems, silver) and that you have a fixed number of mines (yes, this is a bit silly).
The results are presented in two ways,:
A. Percentage chance to find a resource as a function of turns elapsed:
The horizontal-axis is the number of turns elapsed
The vertical-axis is the probability of discovering at least one deposit (so 1.0 corresponds to 100%)
So e.g. after a hundred turns, with 10 mines, you have roughly 40% chance of finding a mineral resource.
B. Turns elapsed to get a certain percentage chance, as a function of mines worked
The horizontal-axis is the number of mines worked
The vertical-axis is the turns elapsed to get a certain probability of discovery
So e.g. with 20 mines worked, you'd have to wait roughly 160 turns to have an 80% of discovering a resource.
This is based on the following calculations (there might be faults, if so post here and I'll fix the graphs!)
Part A
P=1/10000 (base probability of discovery)
(1-P) is the probability of not finding the first resource
(1-P)^2 is the probability of finding neither the first or second resource
etc.
So (1-P)^5 is the probability of not finding a mineral in a mine, assuming you know of 5 types.
Now if you have M mines, the probability of not finding a mineral in any of them is (1-P)^(5*M), and if you wait T turns the probability for not having found any mineral in all your mines is (1-P)^(5*M*T), where T is the number of turns elapsed.
and since we are interested in the chance of having found something, the graph is of
f(T) = 1-(1-P)^(5*M*T)
Part B
Fixing f to a certain value, one has
(1-P)^(5*M*T)=1-f
so that, taking logarithms:
ln(1-f)=ln(1-P)*5*M*T, leading to
T=1/(5*M)*ln(1-f)/ln(1-P)
According to the game XML, the chance of discovering a certain resource (which must be known to you) is 1/10000 (see e.g. http://forums.civfanatics.com/showthread.php?t=140111) per known resource. So the chances are higher if you know more (e.g. bronze, iron, coal etc. in addition to the precious metals which are known from the start).
Based on this little tidbit, I tried to make an analysis which shows the probability of discovering a mineral deposit as a function of turns elapsed, assuming that you know of five resources (e.g. copper, iron, gold, gems, silver) and that you have a fixed number of mines (yes, this is a bit silly).
The results are presented in two ways,:
A. Percentage chance to find a resource as a function of turns elapsed:
The horizontal-axis is the number of turns elapsed
The vertical-axis is the probability of discovering at least one deposit (so 1.0 corresponds to 100%)
So e.g. after a hundred turns, with 10 mines, you have roughly 40% chance of finding a mineral resource.
B. Turns elapsed to get a certain percentage chance, as a function of mines worked
The horizontal-axis is the number of mines worked
The vertical-axis is the turns elapsed to get a certain probability of discovery
So e.g. with 20 mines worked, you'd have to wait roughly 160 turns to have an 80% of discovering a resource.
This is based on the following calculations (there might be faults, if so post here and I'll fix the graphs!)
Part A
P=1/10000 (base probability of discovery)
(1-P) is the probability of not finding the first resource
(1-P)^2 is the probability of finding neither the first or second resource
etc.
So (1-P)^5 is the probability of not finding a mineral in a mine, assuming you know of 5 types.
Now if you have M mines, the probability of not finding a mineral in any of them is (1-P)^(5*M), and if you wait T turns the probability for not having found any mineral in all your mines is (1-P)^(5*M*T), where T is the number of turns elapsed.
and since we are interested in the chance of having found something, the graph is of
f(T) = 1-(1-P)^(5*M*T)
Part B
Fixing f to a certain value, one has
(1-P)^(5*M*T)=1-f
so that, taking logarithms:
ln(1-f)=ln(1-P)*5*M*T, leading to
T=1/(5*M)*ln(1-f)/ln(1-P)