Azothi Posts: 437
21 days ago

Tsar Koschei wrote:
However, I do wonder if those models people have built take into account the cost of having to counter rivals. It happens every now and then that on the 10th and final action, as you get to the goal, your rival also reaches 10 progress at the same time, and taking precedence, has to be paid off. Brings down the profits a fair bit. For the simulations I've looked over, confrontations are accounted for.
Simulation, Python (EPA range: ~2.24  2.28): This fails to account for the actions saved from multiple signs (which would increase the EPA slightly, as it takes fewer actions), as well as failing to account for the Nightmares gain from the confrontation (which would decrease the EPA slightly, as 1 CP of healing is worth 1/6 of an action). The strategy employed also accounts for confrontations by always beginning with at least 40 Supplies. It, however, fails to account for the opportunity cost of obtaining Absinthe.
Simulation, Java (EPA range: ~2.27): This accurately represents Nightmares gain, though it assumes the player will always play the optimal Nightmares reduction action (the social action). It also fails to account for actions saved from multiple signs, as well as the opportunity cost for obtaining Absinthe.
These aren't perfect, but they have taken most variables into account. The complication of obtaining Absinthe in a costeffective way likely reduces the EPA for the whole system, but I believe it's reasonable to treat players as having sufficient Absinthe to treat it as an Echo cost in the short term. Trying to account for obtaining Absinthe is a spanner in the works, and it's likely to massively overcomplicate things for relatively little effect.

Mathematically, we can evaluate the effect of confrontation:
1 action +5 CP Nightmares 10 Absinthe 10 Supplies +35 Progress
The Nightmares will take 5/6 of an action to heal via social action. On average, 4 Progress is gained, equivalent to a sign, which means 6 Supplies are lost on average during the confrontation. Supplies are valued at 2.7 Echoes per supply, so this is 16.2 Echoes lost. Add in the 5 Echoes of Absinthe lost, and we get a loss of ~2 actions and 21.2 Echoes from the confrontation.
Comparatively, a sign gives the equivalent of 10.8 Echoes in supplies, while two signs give 21.6 Echoes while saving 1 action. Three signs give 32.4 Echoes and 1 saved action. The echoes gained from two signs balance out the echoes lost from one confrontation, at the cost of 1 action. Now we can consider the probabilities of these events.
Using the binomial formula for confrontations (which require 9 out of 10 steps to increase Rival's Progress), we have (10 C 9)*(0.5)^9*(0.5)^1 = ~0.00977 probability, or about 1%. The simulations reflect this figure.
For two signs, we have (10 C 2)*(0.05)^2*(0.95)^8 = ~0.0746 probability, or about 7.5%. In other words, one's almost ten times as likely to find two signs compared to confronting rivals. For three signs, we have (10 C 3)*(0.05)^3*(0.95)^7 = ~0.0105 probability, or about 1%. For every confrontation, one is just as likely to find three signs in one expedition.
It's like there are two opposing forces: the confrontations with rivals and the appearance of multiple signs. Both are very rare, with the appearance of multiple signs being slightly more likely to occur, but they pull against each other, leaving the final EPA at around the simplified calculation value taking neither into account, in the 2.2  2.3 EPA range.

And while it's outside the scope of your original question a few months ago ("What are considered the best readily repeatable money grinds these days, anyway?"), the ability to use Dock favours massively increases the EPA of the expedition, and for moneymaking purposes, the ability to stay in London and potentially draw highvalue cards also increases the EPA.
The caprices of the RNG fall away as the number of trials increases, so there clearly can be rough patches and smooth patches in terms of the actual expeditions, but the math seems to suggest an EPA in the range of 2.2 to 2.3.
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dov Posts: 2531
21 days ago

Azothi wrote:
For two signs, we have (10 C 2)*(0.05)^2*(0.95)^8 = ~0.0746 probability, or about 7.5%. In other words, one's almost ten times as likely to find two signs compared to confronting rivals. For three signs, we have (10 C 3)*(0.05)^3*(0.95)^7 = ~0.0105 probability, or about 1%. For every confrontation, one is just as likely to find three signs in one expedition. Note that encountering A Sign twice means that you can complete the expedition in 9 actions instead of 10, which further boosts profitability and reduces the chance of a confrontation (unless that last A Sign is on that tenth action).
After many hundreds of expeditions myself, I've only ever had a confrontation with a rival once, and I've had several runs in which I've has A Sign appear two, or even (rarely) three times.
At the time this was discussed on reddit, I've modified the Java simulation to add the cost of dealing with a confrontation (should it occur) and the effect on the total EPA was completely negligible.
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Skinnyman Posts: 1779
20 days ago

Azothi wrote:
The Nightmares will take 5/6 of an action to heal via social action. On average, 4 Progress is gained, equivalent to a sign, which means 6 Supplies are lost on average during the confrontation. Supplies are valued at 2.7 Echoes per supply, so this is 16.2 Echoes lost. Add in the 5 Echoes of Absinthe lost, and we get a loss of ~2 actions and 21.2 Echoes from the confrontation.
Comparatively, a sign gives the equivalent of 10.8 Echoes in supplies, while two signs give 21.6 Echoes while saving 1 action. Three signs give 32.4 Echoes and 1 saved action. The echoes gained from two signs balance out the echoes lost from one confrontation, at the cost of 1 action. Now we can consider the probabilities of these events.
Using the binomial formula for confrontations (which require 9 out of 10 steps to increase Rival's Progress), we have (10 C 9)*(0.5)^9*(0.5)^1 = ~0.00977 probability, or about 1%. The simulations reflect this figure. My yesterday's napkin math thought that you finish the expedition only after 9 uses of progress. Let's see if I get it right with your info from your post:
You may not gain any progress because the rival can reach 10 progress when you already have 28 or 30 progress; on average, only 1 progress is gained. But the chance for this to happen means you must get A Sign at most.
If you never get A Sign, I'll use your 0.977% rate and multiply it with (0.96)^9 for a 0.676% chance. However, if you get A Sign, I won't even take into account when this happens and how many rolls mustn't be over 96 because only the chance for the rival to always get a progress is (0.5)^9 which is 0.19%. Even with the lower progress gain, this boosts the EPA a little more as it has a lower chance of occurrence than getting two Signs during an expeditions.
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