Section 4.2: What are my chances of smelting Valorite (or other high level ores)?
At GM Mining you will fail 48% (!) of all Valorite smelts (ie. 52% success rate). Each failure leaves you with 50% of the ore, though. Smelting what remains gives you another 52% chance to succeed and so on.
The basic rule is this: At the exact level where you start mining an ore color you have exactly 50% chance to smelt it successfully. Each additional 0.1 in mining skill adds 0.2% to the success rate. Thus at 99 mining skill your chance to smelt Valorite is 50% and at 100 skill its 52% (and at 105 skill it is therefore 62%)
At 100 Mining Skill, your smelting chances are as follows:
100% Iron
100% Dull Copper
100% Shadow Iron
100% Copper
90% Bronze
80% Golden
70% Agapite
60% Verite
52% Valorite
In the end the OVERALL gain rate from Valorite ore to ingots will be about 64.5% (at GM Mining).
52%*2+(48%*52%*1) = average ingot payout per large ore, smelted individually.
1.04 + 0.2496 = 1.2896 ingots.
In other words smelting 1000 valorite ore pieces individually will on average give some 1290 valorite ingots. The numbers change slightly as your piles get larger.
4 ore example:
(52%*8)+(48%*52%*4)+(48%*48%*52%*2)+(48%*48*48%*52*1) = average payout for 4 ore smelting.
4.16+0.9984+0.239616+0.05750784 = 5.45552384, or an average of 1.36388096 ingots per ore.
HOWEVER, this is misleading to an extent. If one fails 4 times in a row, smelting 4 ore, 1 at a time, one loses 2 ore completely, but still has a chance of getting 1-4 ingots from the remaining two ore. If one fails 4 times in a row on a pile of 4 ore, smelted as a unit, one is left with ZERO ingots. Given that the UO Random number generator is excessively prone to streaks (granted, ones that can be either to your detriment or benefit), it all comes down to a form of loss prevention decision.
Either you can smelt high-end ore in smaller numbers, sacrificing the chance for maximum gain, to get something closer to the expected average ingots returned,
OR....
You can go all at once, hoping for a 100% return, but risking an automatic loss of 50% on one failure, 75% on 2 failures, 87.5% on 3 failures, etc.
One should EXPECT the number of successes and failures to balance out in the end, but one has to consider one does not have the same number of ore, each smelt. Failing once on a 1000 ore smelt, statistically, is the still the same as failing on a 1 ore smelt, despite the fact one has a loss of 1000 ingots potential, and the other, only 1. Choosing a set amount to smelt at one time, you can limit your short-term losses, by forcing your results to more closely approach the mean (theoretical average).
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Real in-game Example:
It is near certain that one will get over 1000 ingots from smelting 1000 Valorite ore, in groups of 4. In fact, the average amount one will get back is about 1364 (1363.88).
However, smelting the pile as a whole, gives you a 52% chance of getting 2000 ingots, BUT a 48% chance of getting 1000 or less! In fact, there is about an 11% chance that you will get 250 ingots or less (a little under the odds of flipping 3 heads in a row on a coin, which is 12.5%).
If one is needing to fill a 10-count Exceptional Valorite Plate LBOD for a Verite Hammer, one will need at least 1000 ingots. Which is the method most likely to give you the ingots you need, without having to go back and mine more?
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Even though you know that it will even out with future smelts it will still frustrate you to see all that ore being lost, and having a short-term big failure is not helped in the slightest by a future offset - a future offset that is statistically likely, but not in anyway predictable as to WHEN it could happen.
The best bet is to do the following, if you are in immediate need of the ore you are to smelt:
A. For an ore that is your worst chance of smelting, pick a single-digit power of 2 (1, 2, 4 or 8). This will be your base number to smelt. Powers of 2 are best, because you will not lose ingots from rounding errors (for example, 3 small ore are reduced to 0 on a failure, and other rounding issues along the way from odd numbers can lose the equivalent of a small ore piece). Use this number for that range, whenever you need the best average payout.
B. For the next lowest ore, double this number. For example, if you choose to smelt Valorite in piles of 4 or less, smelt Verite in piles of 8 or less.
C. Continue doubling as you progress to easier ores. Ores at 100% can be done as one pile without worries.
Now, if the need for the ingots is NOT pressing, it is your choice to stick to the routine, or make the gamble for a higher return, as you have less riding on success or failure. Still, it would probably be best to go for piles in the 3-digit powers of 2 (128, 256, 512) or maybe 1024 if you are working with piles in the 10k to 60k range (like looted from a house fall). As great as it might be to succeed on the first try on a pile of 10,000 Valorite ore, do you really want to live with the memory of failing 3 times, getting only 2500 ingots from it? 10 groups of 1000 would get you closer to 13,000 ore, and very unlikely to be under 10,000 (as you will average 5+ first-chance successes over 10 tries).