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Probability of getting a 2-drop

By Whebon

When you are the starting player of a match, you usually want to cast a 2-cost minion to potentially claim a mana orb in your next turn. You can play a solid body (2/3), a card for a specific tactic (such as kujata, manaforcer, etc.), a ranged minion or something like a sun wisp to make sure you keep the card advantage. Basically anything you want. This puts pressure on your opponent and forces them to play a removal spell instead of a stronger minion. In Duelyst, unlike in other TGC such as Hearthstone or MTG, not playing a card in your first turn can lead to an enormous disadvantage and a potential game loss.

Therefore, it is important to have a lot of 2-drops (or 1-drops) in your deck. Some people believe having a lot of 2-drops in their deck is a disadvantage for their end-game and they rather have an extra card in their hand instead of wasting it in turn 1. However, the replace mechanic helps players to easily get rid of early-game cards to search for their end-game bombs. Besides that, there are a lot of ways to refill your hand in a relatively cheap way. Passing the turn with unspent mana left seems like a huge waste to me.

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Whebon
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Languages: English

Posted

Updated

15 Apr, 2018 @ 6:40am

15 Apr, 2018 @ 7:51am

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Guide Index

Overview

Example

Generalizing the Problem

List of Probabilities

Comments

Example

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

Generalizing the Problem

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)!)

List of Probabilities

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

0.00 %

20.3 %

36.9 %

50.4 %

61.3 %

70.0 %

77.0 %

82.5 %

86.8 %

90.2 %

92.8 %

94.7 %

96.2 %

97.3 %

98.1 %

98.7 %

99.1 %

99.4 %

99.6 %

99.8 %

99.9 %

22+

~100 %

Keep in mind you'll have to mulligan two cards and replace a card that has two extra copies in your deck to achieve the assigned probability in the table.