Current Known Bugs list

[Marek14]

3. When Ai play second oblivion ring the creature removed with the first one didn't return to play.

Why should it return? Or did it exile the first Ring?

[Ichigo1100]

Lichenthrope doesn't trigger right it prevents damage and doesn't put the -1/-1 counters on it.

[freestorageaccount]

I used the pump ability of Umezawa's Jitte on a double-striking creature after first-strike combat damage was dealt and before regular combat damage was assigned (so its power had increased by this time), but the old power was used rather than the new.

EDIT: Now I understand no one will believe the boy who called Kithkin Rabble, but I've counted this one super carefully and these Conqueror's Pledge tokens should be 8/7: http://tinyurl.com/mtgjihad. (I did make absolutely sure to write down "Soldier" and neither "soldier" nor "Soldiers" for the Shared Triumph dialog; it unfortunately seems to treat the choice made as case-sensitive.)

[Ichigo1100]

An error has occured. You can copy/paste this message or save it to a file.
Please report this, plus what you tried to do, to:
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Index: 0, Size: 0


Version:
Forge -- official beta: $Date: 2010-02-22 10:48:25 -0600 (Mon, 22 Feb 2010) $, SVN revision: $Revision: 415 $

Detailed error trace:
java.lang.IndexOutOfBoundsException: Index: 0, Size: 0
at java.util.ArrayList.RangeCheck(Unknown Source)
at java.util.ArrayList.remove(Unknown Source)
at forge.MagicStack.pop(MagicStack.java:151)
at forge.Input_StackNotEmpty.selectButtonOK(Input_StackNotEmpty.java:23)
at forge.GuiInput.selectButtonOK(GuiInput.java:36)
at forge.GuiDisplay3.okButtonActionPerformed(GuiDisplay3.java:975)
at forge.GuiDisplay3.access$3(GuiDisplay3.java:974)
at forge.GuiDisplay3$26.actionPerformed(GuiDisplay3.java:767)
at javax.swing.AbstractButton.fireActionPerformed(Unknown Source)
at javax.swing.AbstractButton$Handler.actionPerformed(Unknown Source)
at javax.swing.DefaultButtonModel.fireActionPerformed(Unknown Source)
at javax.swing.DefaultButtonModel.setPressed(Unknown Source)
at javax.swing.AbstractButton.doClick(Unknown Source)
at javax.swing.AbstractButton.doClick(Unknown Source)
at forge.GuiDisplay3$27.keyPressed(GuiDisplay3.java:786)
at java.awt.Component.processKeyEvent(Unknown Source)
at javax.swing.JComponent.processKeyEvent(Unknown Source)
at java.awt.Component.processEvent(Unknown Source)
at java.awt.Container.processEvent(Unknown Source)
at java.awt.Component.dispatchEventImpl(Unknown Source)
at java.awt.Container.dispatchEventImpl(Unknown Source)
at java.awt.Component.dispatchEvent(Unknown Source)
at java.awt.KeyboardFocusManager.redispatchEvent(Unknown Source)
at java.awt.DefaultKeyboardFocusManager.dispatchKeyEvent(Unknown Source)
at java.awt.DefaultKeyboardFocusManager.preDispatchKeyEvent(Unknown Source)
at java.awt.DefaultKeyboardFocusManager.typeAheadAssertions(Unknown Source)
at java.awt.DefaultKeyboardFocusManager.dispatchEvent(Unknown Source)
at java.awt.Component.dispatchEventImpl(Unknown Source)
at java.awt.Container.dispatchEventImpl(Unknown Source)
at java.awt.Window.dispatchEventImpl(Unknown Source)
at java.awt.Component.dispatchEvent(Unknown Source)
at java.awt.EventQueue.dispatchEvent(Unknown Source)
at java.awt.EventDispatchThread.pumpOneEventForFilters(Unknown Source)
at java.awt.EventDispatchThread.pumpEventsForFilter(Unknown Source)
at java.awt.EventDispatchThread.pumpEventsForHierarchy(Unknown Source)
at java.awt.EventDispatchThread.pumpEvents(Unknown Source)
at java.awt.EventDispatchThread.pumpEvents(Unknown Source)
at java.awt.EventDispatchThread.run(Unknown Source)

[freestorageaccount]

Sometimes the computer becomes very reluctant to play anything after being partially milled, and occasionally Mind Funeral hits it for nearly 40 cards. This land-stacking feature sure is a great asset to the AI!

Bonus question: For a library containing an infinite number of cards of which 1/3 are lands, the expected number of cards milled by one Mind Funeral is 12. Instead of that, what's the probability of 40 or more cards being milled? Assume the library was just perfectly shuffled and "Stack AI land" is off. The first correct ( ...) answer will win a cookie (virtual) and a nominal but non-zero prize!

[DennisBergkamp]

I used the pump ability of Umezawa's Jitte on a double-striking creature after first-strike combat damage was dealt and before regular combat damage was assigned (so its power had increased by this time), but the old power was used rather than the new.

EDIT: Now I understand no one will believe the boy who called Kithkin Rabble, but I've counted this one super carefully and these Conqueror's Pledge tokens should be 8/7: http://tinyurl.com/mtgjihad. (I did make absolutely sure to write down "Soldier" and neither "soldier" nor "Soldiers" for the Shared Triumph dialog; it unfortunately seems to treat the choice made as case-sensitive.)

Ah, what this is, is a bug in the card display of Forge. Notice how they're 8/7 in the combat panel? This seems to happen sometimes with cards that don't have pictures associated with them. In this case, Forge will actually treat them as 8/7s, not 7/6s.

[Marek14]

Sometimes the computer becomes very reluctant to play anything after being partially milled, and occasionally Mind Funeral hits it for nearly 40 cards. This land-stacking feature sure is a great asset to the AI!

Bonus question: For a library containing an infinite number of cards of which 1/3 are lands, the expected number of cards milled by one Mind Funeral is 12. Instead of that, what's the probability of 40 or more cards being milled? Assume the library was just perfectly shuffled and "Stack AI land" is off. The first correct ( ...) answer will win a cookie (virtual) and a nominal but non-zero prize!

Hmm. For simplicity, I am going to assume that each card can be a land, a card A or a card B, with equal probabilities (this allows me to compute discretely).

The question is equivalent to "3 or less lands in the first 39 cards".

Since there are 3 states for a card, the total number of possibilities is 3^39: 4 052 555 153 018 976 267.

Now, if there is NO land, we get 2^39 possibilities, which is 549 755 813 888.

With ONE land, we get 39*2^38 possibilities, which is 10 720 238 370 816.

With TWO lands, we get (39*38*2^37)/2 possibilities, which is 101 842 264 522 752.

With THREE lands (maximum allowed), we get (39*38*37*2^36)/6 possibilities, which is 628 027 297 890 304.

The total number of favored cases is then 741 139 556 597 760.
Dividing by 3^39, we get 0,018288%.

I.e, the chance is roughly 1 in 5468.

[freestorageaccount]

Sometimes the computer becomes very reluctant to play anything after being partially milled, and occasionally Mind Funeral hits it for nearly 40 cards. This land-stacking feature sure is a great asset to the AI!

Bonus question: For a library containing an infinite number of cards of which 1/3 are lands, the expected number of cards milled by one Mind Funeral is 12. Instead of that, what's the probability of 40 or more cards being milled? Assume the library was just perfectly shuffled and "Stack AI land" is off. The first correct ( ...) answer will win a cookie (virtual) and a nominal but non-zero prize!

Hmm. For simplicity, I am going to assume that each card can be a land, a card A or a card B, with equal probabilities (this allows me to compute discretely).

The question is equivalent to "3 or less lands in the first 39 cards".

Since there are 3 states for a card, the total number of possibilities is 3^39: 4 052 555 153 018 976 267.

Now, if there is NO land, we get 2^39 possibilities, which is 549 755 813 888.

With ONE land, we get 39*2^38 possibilities, which is 10 720 238 370 816.

With TWO lands, we get (39*38*2^37)/2 possibilities, which is 101 842 264 522 752.

With THREE lands (maximum allowed), we get (39*38*37*2^36)/6 possibilities, which is 628 027 297 890 304.

The total number of favored cases is then 741 139 556 597 760.
Dividing by 3^39, we get 0,018288%.

I.e, the chance is roughly 1 in 5468.

Correct! (I was actually expecting someone to cheat and use the negative-binomial distribution. How silly of me.) Here is your cookie. Unfortunately, I didn't think of a suitable award at the time I posed the question -- I really should have done so! Any ideas (keeping in mind the exigencies of likely being located thousands of miles away)?

[Almost_Clever]

I don't see how you can choose to not sacrifice at least one creature when you play a creature with devour, such as Gluttonous Slime.

[DennisBergkamp]

I don't see how you can choose to not sacrifice at least one creature when you play a creature with devour, such as Gluttonous Slime.

Whoa, true, I never noticed that... I'll see if I can fix this.

[lm01]

AI related:
AI uses Force of Will without paying it's mana cost and without removing a blue card from hand.

[apthaven]

A very interesting thing Triadasoul pointed out to me in the forge decks forum, is that Ali from Cairo does not work as it should.

Ali from Cairo should only consider damage when activating its effect from cards like Lightning Bolt but not from cards like Dark Confidant (or other cards that cause you to lose life but not damage).

So if for example, I don't block the AI's creatures I don't lose the game because Ali from Cairo works. However if I'm currently stuck at 1 life and I play a Dark Confidant, or I have a Dark Confidant in play already, once its effect triggers the next time, I must lose the game. However this is not the case.

The way Ali from Cairo is handled in Forge now is that it considers both damage or loss of life from an effect or cost when its "Damage that would reduce your life total to less than 1 reduces it to 1 instead" ability triggers.

[silly freak]

I don't see how you can choose to not sacrifice at least one creature when you play a creature with devour, such as Gluttonous Slime.

Whoa, true, I never noticed that... I'll see if I can fix this.

sorry, my fault, i fixed it. i called getChoices instead of getChoicesOptional

[Almost_Clever]

It looks like you cannot cast Lull. The only option available is to cycle it.

[lm01]

AI casts Imperiosaur tapping Basic Forests and creatures that produce mana.