That’s true; despite having the same damage per minute, the red swords stats are harmed more by armour damage reduction (since if x<y, (x-a)y < x(y-a)).
It should be noted that the Armour Damage stat only affects a Sword’s Damage stat, while Dodge is global: Mitigated Damage per minute = (Sword Damage + Armour Damage) Speed (1-Dodge)
That’s true; despite having the same damage per minute, the red swords stats are harmed more by armour damage reduction
But the blue sword is harmed more by dodging, because it has fewer chances in a minute to do any damage at all. If we were looking at just dodge, and not damage absorption, the red one would do more damage in (100-dodge)% of a minute’s worth of attacks.
Does the absorption factor outweigh that, then? :)
But the blue sword is harmed more by dodging, because it has fewer chances in a minute to do any damage at all
No sword is harmed more or less by dodging than any other sword, under the stated assumption that hit points are large enough that the results are effectively always equal to their expectations.
That reasoning didn’t get the point across to me at all, but I sat down to do the math and see how it worked out for myself and you’re right. Never mind. :)
That’s true; despite having the same damage per minute, the red swords stats are harmed more by armour damage reduction
But the blue sword is harmed more by dodging, because it has fewer chances in a minute to do any damage at all. If we were looking at just dodge, and not damage absorption, the red one would do more damage in (100-dodge)% of a minute’s worth of attacks.
Does the absorption factor outweigh that, then? :)
(This and the one I made below can be seen properly at http://tinyurl.com/lwgttable , along with the ATT vs DEF tables I worked out the outcomes from)
Hmm. Unless this has gone wrong, the best combo is Sword 1 and Armour 4, with Sword 1/Armour 1 being close). But if you bank on people choosing 1⁄4, then 1⁄1 will beat them.
NB: Yes, I made a lot of mistakes and edits to get here, and probably have still made some...
You worked that out that yourself? (upvoted) Working out the best strategy is unfortunately a bit more complicated than just summing up each column, since by that logic everyone would go Green/Blue, and you’d then beat them with Red/Green or something else that beats Green/Blue. But I’m still impressed with your matricial mastery.
Yeah, it wasn’t that difficult once I worked out how to set it up. I used the table below of Sword vs Armour damages with an index function based on the numbers in the row/column headings. Here’s an example:
=INDEX($E$2:$H$5,RIGHT(O$14),RIGHT($A22))
E2:H5 is the Swords vs Armour table.
O14 is the s1 part of the a4|s1 column label.
A22 is the a2 part of the a2|s4 row label.
Thus, this works out the mitigated attack value of Sword 1 vs Armour 2. This table as a whole worked out the mitigated attacks for columns versus rows. A second table worked out rows vs columns, and the table shown above merely compared the two values.
I agree though, the strategy is complex and I think perhaps in these situations always comes down to how risky/analytical you think the other players going to be, and how you think they think they’re going to judge everyone else. And… well, how do you even start doing that? Especially since, most of the time, people will just… stop behaving rationally when faced with this kind of situation.
Cycle analysis is basically drawing out a graph (nodes and edges) of what beats what. For standard Rock-paper-scissors you get a graph that looks like this:
ROCK -------> SCISSORS -------> PAPER
^-----------------------------------------------|
In systems that aren’t balanced like the sword and armor problem here you can use it to decide what choice to make by giving each node a probability value based on how many people in game use that and then the best choice would be the node with the greatest (sum of probabilities on outgoing nodes) - (sum of probabilities on incoming nodes).
Nice chart. This one’s better, though; it clearly lists which sword and armor win, as well as listing number of losses and ties. I got it from the same thread as the one in the OP; I was just waiting until someone suggested doing something like this before I posted it.
The red sword is strictly worse than the blue sword. Beyond that I’m too lazy to figure it out.
That’s true; despite having the same damage per minute, the red swords stats are harmed more by armour damage reduction (since if x<y, (x-a)y < x(y-a)).
It should be noted that the Armour Damage stat only affects a Sword’s Damage stat, while Dodge is global: Mitigated Damage per minute = (Sword Damage + Armour Damage) Speed (1-Dodge)
But the blue sword is harmed more by dodging, because it has fewer chances in a minute to do any damage at all. If we were looking at just dodge, and not damage absorption, the red one would do more damage in (100-dodge)% of a minute’s worth of attacks.
Does the absorption factor outweigh that, then? :)
No sword is harmed more or less by dodging than any other sword, under the stated assumption that hit points are large enough that the results are effectively always equal to their expectations.
That reasoning didn’t get the point across to me at all, but I sat down to do the math and see how it worked out for myself and you’re right. Never mind. :)
I omitted my reasoning. I’m glad you found it for yourself :)
But the blue sword is harmed more by dodging, because it has fewer chances in a minute to do any damage at all. If we were looking at just dodge, and not damage absorption, the red one would do more damage in (100-dodge)% of a minute’s worth of attacks.
Does the absorption factor outweigh that, then? :)
This is the correct equation for any attack pair. Now just need to do a 16-16 table to see what wins over what. I’ll do it during lecture tomorrow.
You mean this table? :)
(This and the one I made below can be seen properly at http://tinyurl.com/lwgttable , along with the ATT vs DEF tables I worked out the outcomes from)
Hmm. Unless this has gone wrong, the best combo is Sword 1 and Armour 4, with Sword 1/Armour 1 being close). But if you bank on people choosing 1⁄4, then 1⁄1 will beat them.
NB: Yes, I made a lot of mistakes and edits to get here, and probably have still made some...
You worked that out that yourself? (upvoted) Working out the best strategy is unfortunately a bit more complicated than just summing up each column, since by that logic everyone would go Green/Blue, and you’d then beat them with Red/Green or something else that beats Green/Blue. But I’m still impressed with your matricial mastery.
Yeah, it wasn’t that difficult once I worked out how to set it up. I used the table below of Sword vs Armour damages with an index function based on the numbers in the row/column headings. Here’s an example:
=INDEX($E$2:$H$5,RIGHT(O$14),RIGHT($A22))
E2:H5 is the Swords vs Armour table.
O14 is the s1 part of the a4|s1 column label.
A22 is the a2 part of the a2|s4 row label.
Thus, this works out the mitigated attack value of Sword 1 vs Armour 2. This table as a whole worked out the mitigated attacks for columns versus rows. A second table worked out rows vs columns, and the table shown above merely compared the two values.
I agree though, the strategy is complex and I think perhaps in these situations always comes down to how risky/analytical you think the other players going to be, and how you think they think they’re going to judge everyone else. And… well, how do you even start doing that? Especially since, most of the time, people will just… stop behaving rationally when faced with this kind of situation.
That looks like what I got. It’d be interesting to do a cycle analysis like you do with Non-transitive Dice.
Thanks for the mini mind blow with the non-transitive dice link.
Can you elaborate on what cycle analysis is?
Cycle analysis is basically drawing out a graph (nodes and edges) of what beats what. For standard Rock-paper-scissors you get a graph that looks like this:
ROCK -------> SCISSORS -------> PAPER
^-----------------------------------------------|
In systems that aren’t balanced like the sword and armor problem here you can use it to decide what choice to make by giving each node a probability value based on how many people in game use that and then the best choice would be the node with the greatest (sum of probabilities on outgoing nodes) - (sum of probabilities on incoming nodes).
Nice chart. This one’s better, though; it clearly lists which sword and armor win, as well as listing number of losses and ties. I got it from the same thread as the one in the OP; I was just waiting until someone suggested doing something like this before I posted it.
Didn’t want to take away your fun, after all. ;)