Put more simply, if there is a 10(0.5)(0.6) calculation, why would you ever get a result other than 3 (in chess mode)? Does a Queen kill more pawns when she's surrounded?
Combat mechanics question
Moderator: Panzer Corps 2 Moderators
Re: Combat mechanics question
I follow the math, but not the logic. Chess mode should calculate the expected # of kills based on number of hits and return that every time. There should be no chance of 10 kills in chess mode, so why is that in the algorithm?
Put more simply, if there is a 10(0.5)(0.6) calculation, why would you ever get a result other than 3 (in chess mode)? Does a Queen kill more pawns when she's surrounded?
Put more simply, if there is a 10(0.5)(0.6) calculation, why would you ever get a result other than 3 (in chess mode)? Does a Queen kill more pawns when she's surrounded?
Re: Combat mechanics question
Hi Robo40,
I see what you mean. Yes there is a 10(0.5)(0.6) calculation but this ONLY DEFINES THE KILL CHANCE PER SHOT.
The simulation is more than that, it is a stochastic process. Every strength point takes a shot with a probability to hit defined by this calculation. If there is a hit a kill is registered and a strength point of the target is taken away. So the number of kills is a stochastic process defined in this way and depends on number of shots, probability to kill and number of targets. The logical error being made is that the EXPECTED NUMBER OF HITS (in our example 3) translates directly to an EXPECTED NUMBER OF THE SAME KILLS. We can not do this for stochastic processes. If X is a stochastic process (for example to hit) and f(X) is a derived stochastic process (for example to kill) than E[f(X)] =/= f(E[X]). (E meaning expectation), This is the logical error being made. In chess mode the outcome IS DEFINED as E[f(X)].
See it like this. You, me and 8 friends (10 of us) are going to a bar every night. There are always 3 people there, and we don't like them, no we don't. So we decide to kick their asses, but to make it more fair we each gone take a swing at them only once. We are not very good at bar fights, so we only swing and hit with a probability of 30%. When we hit someone is bruised of those 3 people.
Now the question of how many times did we bruise SOMEONE (number of hits in PzC) on average and how many people did we bruise on average (number of kills in PzC) is not the same! We can bruise 0, 1, 2, or 3 people, but we can hit 10 times!. Someone gets bruised multiple times of we hit more than 3 times. But because there is a finite probability of hitting less than 3 times the average number of bruised people is less than 3 (some times we miss a lot of times), like we calculated earlier it is 2.446. But how many times did we bruise someone on average? That would be 3 times exactly.
I see what you mean. Yes there is a 10(0.5)(0.6) calculation but this ONLY DEFINES THE KILL CHANCE PER SHOT.
The simulation is more than that, it is a stochastic process. Every strength point takes a shot with a probability to hit defined by this calculation. If there is a hit a kill is registered and a strength point of the target is taken away. So the number of kills is a stochastic process defined in this way and depends on number of shots, probability to kill and number of targets. The logical error being made is that the EXPECTED NUMBER OF HITS (in our example 3) translates directly to an EXPECTED NUMBER OF THE SAME KILLS. We can not do this for stochastic processes. If X is a stochastic process (for example to hit) and f(X) is a derived stochastic process (for example to kill) than E[f(X)] =/= f(E[X]). (E meaning expectation), This is the logical error being made. In chess mode the outcome IS DEFINED as E[f(X)].
See it like this. You, me and 8 friends (10 of us) are going to a bar every night. There are always 3 people there, and we don't like them, no we don't. So we decide to kick their asses, but to make it more fair we each gone take a swing at them only once. We are not very good at bar fights, so we only swing and hit with a probability of 30%. When we hit someone is bruised of those 3 people.
Now the question of how many times did we bruise SOMEONE (number of hits in PzC) on average and how many people did we bruise on average (number of kills in PzC) is not the same! We can bruise 0, 1, 2, or 3 people, but we can hit 10 times!. Someone gets bruised multiple times of we hit more than 3 times. But because there is a finite probability of hitting less than 3 times the average number of bruised people is less than 3 (some times we miss a lot of times), like we calculated earlier it is 2.446. But how many times did we bruise someone on average? That would be 3 times exactly.
Re: Combat mechanics question
Thank you for the explanation, esp the bar fight.
I get it, but I'm sure you can understand my issue with stochastic and chess mode being used in the same sentence. The derived process can be stochastic, but the basic probability is what chess mode should be based on.
But thank you, at least I understand why this seems counter-intuitive. Still a great game.
I get it, but I'm sure you can understand my issue with stochastic and chess mode being used in the same sentence. The derived process can be stochastic, but the basic probability is what chess mode should be based on.
But thank you, at least I understand why this seems counter-intuitive. Still a great game.