I was surprised the search came up with nothing. I roll a d6. Is the result a binary evens/odds or is it a numerical value that is assigned the evens/odds qualifier? For example, I roll a 3. Should this be read as "odds" or "3 is an odd number?" Assuming the d6 is a check, I ask because of blessing the Ivory Dice. Calling odds on these two d6s, would you need two "odds" binaries (a 3 and a 5) or two "x is an odd number" qualifiers (a 3 and a 4) to succeed? Even if it isn't a check, it's an interesting thought experiment.
Bottom line do wathever suits you as long as you get 50% chance.
Since the rules generally do not require you to remember which die came from what source, I would say you're going to want to use the total of the check. That being said, I don't have the cards in front of me, so I don't remember how Ivory Die is worded.
What Alex319 said, it isn't a check the say way the 1d4 from Cure isn't a check. And I think that means that the answer to your question doesn't have any impact on the way it is play, since with only 1 die no matter how you understood it the results do the same.
When you would fail a check reveal this card and choose odd or even. Then roll 1d6. If the roll matches your choice, recharge this card to succeed at the check. If the roll does not match, bury 3 cards from you hand.
If you fail a check and reveal it and choose even then roll 1d6, on a 2, 4, or 6 you succeed. On a 1, 3, or 5 you bury 3 cards from your hand.
What Alex319 said, it isn't a check the say way the 1d4 from Cure isn't a check. And I think that means that the answer to your question doesn't have any impact on the way it is play, since with only 1 die no matter how you understood it the results do the same.
When you would fail a check reveal this card and choose odd or even. Then roll 1d6. If the roll matches your choice, recharge this card to succeed at the check. If the roll does not match, bury 3 cards from you hand.If you fail a check and reveal it and choose even then roll 1d6, on a 2, 4, or 6 you succeed. On a 1, 3, or 5 you bury 3 cards from your hand.
Ah, gotcha. Sorry for the misinformation, I thought it was added to a check, and something happened based on the result (similar to Mogmurch for example).
This said, the math teacher in me would argue that as long as you add up standard Pacg dice, the occurence of an odd or even sum is always 50/50. Wouldn't be true though obviously if you add a d3 or other strange dice.
The issue lies in that the 'roll' is the total of all dice modifiers, whereas, the die/dice are just the dice.
I think what the OP is asking is whether it's the individual dice that count towards it, as in 2d6 needing both dice to be odd numbers to be considered 'odd' or if the total of the roll needs to be 'odd'. Getting both numbers to be odd is harder (1/2 * 1/2 = 1/4 of the time) as opposed to just the total (1/2 of the time).
...is whether it's the individual dice that count towards it, as in 2d6 needing both dice to be odd numbers...
Well actually if indeed that was the intent, that somehow would be a fun twist (the less dice you roll, the max chance you have). However that doesn't make any sense since, as stated earlier, it is not the dice used for the check that are taken into account but rather a different roll that you use only to check for the ivory dice power. So there is absolutely no incentive to multiply dice on this one.
It's always fun to play this card when there's nothing left in your hand to bury.
