Probability of getting a 2-drop

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In order to be able to play a 2-drop in turn 1, you’ll need a high chance of having at least one 2-drop in your opening hand. Let’s say I have a deck with full playsets of 2 different 2-drops. That makes a total of 6 2-drops in my deck (or 39-6 = 33 non-general, non-2-drops cards). Let’s begin by calculating the probability that the first card of the deck is NOT a 2-drop: P(1st card is non-2-drop) = 33 (non-2-drop) / 39 (total cards) The same logic applies to all the cards in your first starting hand (remember: the amount of cards in your deck is decreasing) P(1st card is non-2-drop) = 33 / 39 P(2nd card is non-2-drop) = 32 / 38 P(3rd card is non-2-drop) = 31 / 37 P(4th card is non-2-drop) = 30 / 36 P(5th card is non-2-drop) = 29 / 35 Combining the probabilities gives: P(starting hand contains 5 non-2-drops) = 33 / 39 * 32 / 38 * 31 / 37 * 30 / 36 * 29 / 35 = (33!/28!) / (39!/34!) = (33!34!) / (39!28!) = approx. 0.41 If we still don’t have a single 2-drop, we can mulligan twice. The probability of getting a non-2-drop for the 5th time is equal to the probability of getting a non-2-drop from a mulligan. Because the mulligan at the beginning of every match doesn't count as replace, so it is possible to get the same cards back after the mulligan. P(mulligan gives two non-2-drops | starting hand contains only non-2-drops) = (29 / 35)^2 P(starting hand after mulligan contains 5 non-2-drops) = (33!34!) / (39!28!) * (29 / 35)^2 = (29^2*33!34!)/ (35^2*39!28!) = approx. 0.28 If you still don’t have a single 2-drop, the last chance of getting a 2-drop in your first turn is to replace a card during your first turn. According to the Duelyst wiki “you never get a copy of the card or the same card you replaced”. For the sake of simplicity, let’s say we don’t run a deck with the Mythron Wanderer and we have 3 copies of every card in our deck. Keep in mind that the following calculation also assumes that you are replacing a card for which you still have 2 copies left in your deck. This means that the calculation is invalid if you have 2 duplicates of card type A and 3 duplicates of card type B in your hand. P(replace gives a non-2-drop) = (34(deck)-2(extra copies)-6(2-drops)) / (34-2) = 26/32 Combining everything gives: P(final opening hand only contains non-2-drops) = (29^2*33!34!)/ (35^2*39!28!) * 26/32 = (26*29^2*33!34!) / (32*35^2*39!28!) = approx. 0.23 P(opening hand contains at least one 2-drop) = 1-(26*29^2*33!34!) / (32*35^2*39!28!) = approx. 0.77 Alright, now we know we have a 77% chance to get our desired 2-drop. But what if our deck contains x 2-drops instead of 6 2-drops? Let’s generalize the probability: P(starting hand contains 5 non-2-drops) = (39-x)!*34! / (39!(34-x)!) P(starting hand after mulligan contains 5 non-2-drops) = ((35-x)^2*(39-x)!*34!) / (35^2*39!(34-x)!) y = amount of extra copies in your deck of the card you’re replacing P(final opening hand contains 5 non-2-drops) = ((35-x)^2*(39-x)!*34!) / (35^2*39!(34-x)!) * (34-y-x)/(34-y) For the sake of simplicity, let’s say y=2: P(final opening hand contains 5 non-2-drops) = ((35-x)^2*(39-x)!*34!) / (35^2*39!(34-x)!) * (32-x)/(32) = ((32-x)*(35-x)^2*(39-x)!*34!) / (32*35^2*39!(34-x)!) P(opening hand contains at least one 2-drop) = 1 - ((32-x)*(35-x)^2*(39-x)!*34!) / (32*35^2*39!*(34-x)!) In this table you can check out the probability of getting at least one 2-drop, based on the total amount of 2-drops in your deck: Amount of 2-drops in your deck Probability of getting at least one 2-drop --- Источник: https://steamcommunity.com/sharedfiles/filedetails/?id=1362097403