Did the Carlos player perchance play a card first and have Ambre? in that case it would be (7+3)*3=30 against 6*5=30 => random 50:50 chance. About the only scenario I can imagine within 5 minutes of thinking in which the above seems a possible correct outcome.
I did count it. I only interpret "with 3 pillz" as actually 3 pillz (including the free one) actually attached to the card - it was not said "with 3 extra pillz played". But in this case it does, for once, not matter. I did your above calculation and a few more as well. (With 3 total pillz on the card and with 3 +1 free incidentally. But terminology gets fuzzy here )
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