Game Coding Q&A

[furballdn]

Is there a cap on bonewalls?

[Absol]

Is there a cap on bonewalls?

It has been addressed in the other thread, but no one seems to reach the cap.
I myself have reached 4500+, and that's not the cap. Speculatively, it could be 9999.
But i think to accurately answer that, we have to look into the Bone Wall coding.

Xeno, do you have access to the game code?

[OldTrees]

Xeno, do you have access to the game code?

Yes. Yes he does.

[furballdn]

Which is why I'm asking here.

[Xenocidius]

Is there a cap on bonewalls?

There is no hard cap. Technically, however, the stack couldn't exceed the maximum size of an integer in Actionscript (approximately 1.79769313486232 x 10³⁰⁸). That's a pretty damn big stack.

[Sevs]

What is the AI priority in discarding?

[Tirear]

From earlier in this thread (6th link in the original post)

What's the AI's priority for discarding cards?

Basically, it goes through all the cards in its hand and assigns them a score based on this formula:

Score = (cost of card)/10 + suspicious

In general, suspicious is 1 if that creature was Nightmared last turn.

However, if the card is Ghost of the Past:

Score = 0

For each card, if its score > (random number between 0 and 1), it will discard it.

If once it has run through all its cards it has not found one to discard, it discards a random card.

[Sevs]

From earlier in this thread (6th link in the original post)

What's the AI's priority for discarding cards?

Basically, it goes through all the cards in its hand and assigns them a score based on this formula:

Score = (cost of card)/10 + suspicious

In general, suspicious is 1 if that creature was Nightmared last turn.

However, if the card is Ghost of the Past:

Score = 0

For each card, if its score > (random number between 0 and 1), it will discard it.

If once it has run through all its cards it has not found one to discard, it discards a random card.

thanks

[teffy]

Ok, another question I always wanted to know:

How high is the chance to get a rare card in the rare spin (Arena) ? If you don´t know it, could you perform a test ?
If an exact chance is too complicated - could you make a chance estimation (higher/lower than x%) ? I have an idea for a "higher than" estimation - but would need help.

Edit:
Is it right, that the spinner chooses 4 cards (not 5) ?

[OldTrees]

Ok, another question I always wanted to know:

How high is the chance to get a rare card in the rare spin (Arena) ? If you don´t know it, could you perform a test ?

Assuming the goal is to win a rare card regardless of the rare, follow these rules
1) Always try to match a fixed slot (0 spins remaining)
2) Avoid breaking pairs
3) If there are no pairs then spin the slot with the most spins remaining

See here for raw probabilities.
http://elementscommunity.org/forum/index.php/topic,28014.msg358556.html#msg358556

P(A3A3A3) = (1/r)^3
P(R|A3A3A3) = 1

From A3A3B3 [P = (1-r)(1/r)^3]
Chance of winning from A3A3B3
P(R|A3A3B3)=(1/r)^9 * (- r^9 + 5 r^8 - 11 r^7  + 19 r^6 - 20 r^5 + 13 r^4  - 5 r^3 + r^2)
P(A3A3A2|A3A3B3) = (1/r)
P(A3A3B2|A3A3B3) = (1-r)(1/r)
P(A3A3A1|A3A3B3) = (1-r)(1/r)^2
P(A3A3B1|A3A3B3) = (1-r)^2(1/r)^2
P(A3A3A0|A3A3B3) = (1-r)^2(1/r)^3
P(A3A3B0|A3A3B3) = (1-r)^3(1/r)^3

I will be back.

PS: If Xeno would confirm the number of rares in the arena spin that would be useful.

P(A3B2B0|A3A3B0) = (1/r)
P(A3A2B0|A3A3B0) = (1-r)(1/r)
P(A3B1B0|A3A3B0) = (1-r)(1/r)^2
P(A3A1B0|A3A3B0) = (1-r)^2(1/r)^2
P(A3B0B0|A3A3B0) = (1-r)^2(1/r)^3

P(B2B0B0|A3B0B0) = (1/r)
P(A2B0B0|A3B0B0) = (1-r)(1/r)
P(B1B0B0|A3B0B0) = (1-r)(1/r)^2
P(A1B0B0|A3B0B0) = (1-r)^2(1/r)^2
P(B1B0B0|A3B0B0) = (1-r)^2(1/r)^3

P(BBB|A3B3B0) = (  (1/r) + (1-r)(1/r)^2 + (1-r)^2(1/r)^3  ) * (  (1/r) + (1-r)(1/r)^2 + (1-r)^2(1/r)^3 )
= (  (1/r) + (1-r)(1/r)^2 + (1-r)^2(1/r)^3  ) ^ 2
= (1/r)^6 * (  r^3 + (1-r)r^2 + (1-r)^2*r  ) ^ 2
= (1/r)^6 * (  r^3 + r^2 - r^3 + r - 2r^2 + r^3  ) ^ 2
= (1/r)^6 * (  r^3 - r^2 +r  ) ^ 2
= (1/r)^6 * (r^6 - 2 r^5 + 2 r^4 +r^4 - 2 r^3 + r^2)
= (1/r)^6 * (r^6 - 2 r^5 + 3 r^4  - 2 r^3 + r^2)

P(BBB|A3A3B3) = (1-r)^3(1/r)^9 * (r^6 - 2 r^5 + 3 r^4  - 2 r^3 + r^2)
=(1/r)^9 * (1 - 3 r + 3 r^2 - r^3) * (r^6 - 2 r^5 + 3 r^4  - 2 r^3 + r^2)
=(1/r)^9 * (r^6 - 2 r^5 + 3 r^4  - 2 r^3 + r^2)
+(1/r)^9 * (- 3 r^7 + 6 r^6 - 9 r^5  + 6 r^4 - 3 r^3)
+(1/r)^9 * (3 r^8 - 6 r^7 + 9 r^6  - 6 r^5 + 3 r^4)
+(1/r)^9 * (- r^9 + 2 r^8 - 3 r^7  + 2 r^6 - r^5)
=(1/r)^9 * (- r^9 + 5 r^8 - 12 r^7  + 18 r^6 - 18 r^5 + 12 r^4  - 5 r^3 + r^2)
P(AAA|A3A3B3) = (1/r) + (1-r)(1/r)^2 + (1-r)^2(1/r)^3
P(R|A3A3B3) = (1/r)^9 * (  r^8 + r^7 - r^8 + r^6 - 2 r^5 + r^4 )    +   (1/r)^9 * (- r^9 + 5 r^8 - 12 r^7  + 18 r^6 - 18 r^5 + 12 r^4  - 5 r^3 + r^2)
=(1/r)^9 * (- r^9 + 5 r^8 - 11 r^7  + 19 r^6 - 20 r^5 + 13 r^4  - 5 r^3 + r^2)

Summary:
P(R|A3A3A3)= 1
P(R|A3A3B3)=(1/r)^9 * (- r^9 + 5 r^8 - 11 r^7  + 19 r^6 - 20 r^5 + 13 r^4  - 5 r^3 + r^2)
P(R|A3B3C3)=??

[Rutarete]

Can permanents have passives?

[eaglgenes101]

Short/long bow and the eyes have the ranged passive.