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18.? ORE VEIN CONTENT AND PROBABILITIES
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As far as I can tell, here is how the game proceeds with ore veins:
First the game determines the total number of items in the vein.
Then the game rolls each item to determine whether it is an ore, a gem of the
first type (gem1) or a gem of the second type (gem2).
If we call:
N the number of items in the vein
N_O the number of ore in the vein
N_G1 the number of gem1 in the vein
N_G2 the number of gem2 in the vein
P_N the probability of getting N items in the vein
P_O the probability of rolling an ore
P_G1 the probability of rolling a gem1
P_G2 the probability of rolling a gem2
We have:
N = N_O + N_G1 + N_G2
P_O + P_G1 + P_G2 = 1
The probability of getting N items with N_O ores, N_G1 gem1s and N_G2 gem2 is:
P_N x Power(P_O,N_O) x Power(P_G1,N_G1) x Power(P_G2,N_G2)
x Combination(N_G1+N_G2,N_G1) x Combination(N,N_G1+N_G2)
Power(P_O,N_O) corresponds to the probability of getting N_O ores
Power(P_G1,N_G1) corresponds to the probability of getting N_G1 gem1s
Power(P_G1,N_G2) corresponds to the probability of getting N_G2 gem2s
The Combinations correspond to the count of the various scenarii leading to the
same result. For exemple, getting one ore then one gem1 gives the same result
as getting one gem1 then one ore.
So the probability of getting 5 ores and no gem (5-0-0) is
P_N x Power(P_O,5) x Power (P_G1,0) x Power (P_G2,0) x Combination(0,0)
x Combination(5,0) = P_N x Power(P_O,5)
The probability of getting 4 ore, 4 gem1 and no gem2 (4-4-0) is
P_N x Power(P_O,4) x Power(P_G1,4) x Power(P_G2,0) x Combination(4,4)
x Combination(8,4) = P_N x Power(P_O,4) x Power (P_G1,4) x 70
Intensive testing gave the following results:
N varies between 5 (no less) and 8 (no more) and P_N = 0.25
That means there is an equal chance of getting 5, 6, 7 or 8 items in a vein.
N_O, N_G1 and N_G2 all vary between 0 and N
That means we can potentially find 8 malachite gems in a vein!
Sure, the odds are extremly low as you can imagine. We'll see that later.
For information the max I have got so far is 5 malachite gems in a single vein.
P_O = 0.818 (approximately)
P_G1 = P_G2 = 0.091 (approximately)
So, the previous examples translate into the following probability:
Probability of 5-0-0 (5 malachite ores, no gem)
0.25 x Power(0.818,5) = 0.0915 = 9.15%
Probability of 4-4-0 (4 malachite ores, 4 malachite gems, no aquamarine)
0.25 x Power(0.818,4) x Power(0.091,4) x 70 = 0.0005 = 0.05%
That 0.05% is not the probability of getting 4 malachite gems.
There are other results which lead to 4 malachite gems.
More precisely we have:
0-4-1, 0-4-2, 0-4-3, 0-4-4, 1-4-0, 1-4-1, 1-4-2, 1-4-3, 2-4-0, 2-4-1, 2-4-2,
3-4-0, 3-4-1 and our previous 4-4-0.
The probability of getting 4 malachite gems is then the sum of the probabilities
of those results.
I am not going to list every possible results (there are exactly 130 different
results, such as 0-0-8 or 8-0-0) and its corresponding probabilities.
I will rather list what you're probably looking for.
Malachite Gems (MG) in a vein:
MG Probability Attempts
0 54.09% 2
1 34.55% 3
2 9.64% 10
3 1.54% 65
4 0.16% 641
5 0.01% 9,732
6 0.0004% 230,955
7 0.00001% 9,357,497
8 0.0000001% 850,606,805
So, getting 8 malachite gems is possible, yes, but well, I'm sure you're not
going to reload 850 million times, right?
And then we have the expected question: "Wait a minute, you say it is possible
to get no ore at all? I have played the game many times, have reloaded for
malachite veins hundreds, or even thousands of times and I have always got some
ore. So, what are you talking about?"
The fact is that the probability to roll an ore is very high. And there are at
least 5 items to roll. So it is very unlikely you're going to get no ore at all.
Malachite Ores (MO) in a vein:
MO Probability
0 0.006%
1 0.14%
2 1.37%
3 6.86%
4 18.97%
5 28.61%
6 23.98%
7 15.05%
8 5.01%
So, you will need around 16,000 attempts on average to get no ore at all in a
vein.
There is one place in the game where two malachite veins are so close together
that exploring the content of the first triggers the content of the other.
Taking that into account, here are the probabilities:
MG Probability Attempts
0 29.26% 3
1 37.38% 3
2 22.37% 4
3 8.33% 12
4 2.17% 46
5 0.42% 240
6 0.06% 1,626
7 0.007% 14,065
8 0.0007% 153,435
9 0.00005% 2,101,805
10 0.000003% 36,234,928
11 0.0000001% 793,869,559
12 gems and above have a totally insignificant probability and are not listed.
As you can see, getting 6 gems in total from those two veins (1,626 attempts) is
much more difficult than getting 3 gems from each of two distant veins (130
attempts, 65 + 65).
There is a workaround however. The fact is that those two veins are not exactly
at the same place. That means it is possible to trigger one and not the other.
The issue is that you cannot see the content of the first vein whitout
triggering the other. So here is one method to deal with that:
1. Make a save (save01) far enough from the two veins.
2. Quicksave.
3. Approach from the right angle and stop at the right distance in order to
trigger the closest vein (vein1) but not the other (vein2). Make a different
save (save02) - That step can be tricky until you find a mark.
4. Explore the two veins and keep note of the content of vein1.
If you're happy with the total result, well, save and go on, lucky one.
If not, if you're happy with the content of vein1, reload the save02, don't
move, quicksave and proceed with vein2 as usual.
Finally, if you're not happy with the total result and the content of vein1,
reload the quicksave and proceed with step 3 again (you can overwrite save02).
That way you can deal with those two veins (almost) like two regular ones.