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I am playing a game of Commander and I have a deck of 99 cards. In this deck are 37 lands. Using a hypergeometric calculator it returned a 29.12% chance of drawing exactly 3 Lands on a 7 card draw using the probability density function. My question is how do I calculate the probability for drawing 3 lands with one or more mulligans?

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    Which mulligan rule are you using?
    – LeppyR64
    Jun 1 at 20:27
  • 2
    @LeppyR64 I think we should assume the default Commander rules as specified in the CR unless a poster specifies otherwise. Jun 1 at 21:29

2 Answers 2

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The probability to draw exactly 3 lands after n attempts (including your first non-mulligan draw) is 1-(1-0.2912)^n, with n<6. With the maximum of 4 mulligans, you have a ~82.1% chance to get a 3-lander.

On each mulligan, you draw a new hand of 7 cards and then put as many cards on the bottom of the library as you have taken mulligans.

103.4. Each player draws a number of cards equal to their starting hand size, which is normally seven. (Some effects can modify a player’s starting hand size.) A player who is dissatisfied with their initial hand may take a mulligan. First, the starting player declares whether they will take a mulligan. Then each other player in turn order does the same. Once each player has made a declaration, all players who decided to take mulligans do so at the same time. To take a mulligan, a player shuffles the cards in their hand back into their library, draws a new hand of cards equal to their starting hand size, then puts a number of those cards equal to the number of times that player has taken a mulligan on the bottom of their library in any order. [..]

This limits you to a maximum of 4 mulligans because if you take 5 or more, you will have fewer than 3 cards left in your starting hand and you can never have a 3-lander.

Also, since you shuffle the library between mulligans, the attempts are independent of each other.

If you have an event A with probability P(A), the probability of it failing is its inverse, or

1-P(A)

If you want to calculate the odds of n independent instances of A having at least 1 success, you first calculate the odds of all of them failing. Since the attempts are independent of each other in your case, you can simply multiply their probabilities with each other to get the result:

(1-P(A))^n

The inverse of that is the probability that at least one of them does not fail:

1-(1-P(A))^n

That means that if you have seen n=5 hands (after 4 mulligans), the probability of having at least one of them being a 3-lander is

1 - (1-0.2912)^5 = 1 - 0.7088^5 = 1 - 0.1789 = 0.8211, or 82.1%

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  • Does this analysis include the fact that the first mulligan in Commander is free? ("103.4c In a multiplayer game and in any Brawl game, the first mulligan a player takes doesn’t count toward the number of cards that player will put on the bottom of their library or the number of mulligans that player may take.") Jun 2 at 8:04
  • No. Only Brawl has the free mulligan, and the question makes no mention of Brawl.
    – Hackworth
    Jun 2 at 13:12
  • Commander has a free mulligan "In a multiplayer game"; the Brawl reference is there because Brawl is two player. Jun 2 at 17:07
  • The question makes no mention of multiplayer either.
    – Hackworth
    Jun 3 at 5:57
  • 99% of Commander games are a 4 player pod. Yes, 1v1 Commander exists but it is very much not what people mean when they say "Commander". Jun 3 at 6:15
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37% times the 7 cards you initially draw gives you an average of 2.59 lands per opening hand for 40 lands you get 40% * 7 which gives you 2.8 lands on average. It seems you would need a 43% of lands in your deck to get a 3 lands opener on average. That means 26 lands for a 60 card deck and 37 for a commander deck.

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