In Magic, at the start of the game, you draw 7 cards. How would you calculate the likelihood of drawing a specific card in your opening hand?

For example, let's say I have a 60 card deck, and I'm running 4 Birds of Paradise. What is the percent chance that I will have at least one Bird in my opening hand?


3 Answers 3


The calculation you are looking for is called a Hypergeometric Distribution. This calculated your chances of drawing a particular number of "successes" from a population, without replacement.

  • Population Size: 60 cards
  • Successes in Population: 4 Birds of Paradise
  • Sample Size: 7 cards
  • Successes in Sample: 1 (the minimum number we want to draw)

  • Results: 39% chance of drawing at least 1 Birds of Paradise.

In the Hypergeometric Distribution calculator linked above, that result is represented in the Cumulative Probability: P(X ≥ 1) field: the chance of drawing greater than or equal to 1.

The online calculator will also give you the odds of drawing greater than that many successes in the sample (6%, the P(X > 1) result), and exactly that number (33%, the Hypergeometric Probability: P(X = 1) result).

You can see the calculation on the Wikipedia page, or searching math.stackexchange.com for Hypergeometric Distribution. Unfortunately, this site doesn't support math formatting. (Note: You will also need to know how to calculate binomial coefficients (and factorials).

  • Interesting program, but i was looking for the formula so that i could figure it out on my own.
    – DForck42
    Commented Jul 12, 2012 at 18:06
  • 4
    @DForck42 why would you want to do it by hand? :) Anyway, I think the full answer would fall out of scope for this stack exchange. The answer to your question: stats.stackexchange.com/questions/24211/…
    – ghoppe
    Commented Jul 12, 2012 at 18:10
  • 7
    Why WOULDN'T you want to do it by hand!? :P
    – Johno
    Commented Jul 12, 2012 at 18:15
  • 1
    @DrunkCynic, yes my answer is for exactly a certain number of successes, but using that, you can calculate it for exactly 1, exactly 2, exactly 3, etc. successes and then add them all together.
    – user1873
    Commented Feb 22, 2016 at 23:26
  • 1
    @DrunkCynic Isn't that what user1873 said in the sentence directly following his bullet points?
    – ghoppe
    Commented Feb 23, 2016 at 15:30

The odds of drawing a particular card in a 60-card deck are obviously 1/60. If there are four such cards, the odds are 4/60. The odds of NOT drawing one of those cards in the first draw is 1 - 4/60 = 56/60.

To calculate the odds of the entire first hand, we can do it backwards:

The odds of not having any of the four cards in the first card is 56/60 (as I said above). The second card has odds of 55/59 (i.e. one of the remaining non-Bird cards after a non-Bird card was drawn to start), and then 54/58 and so on:

  • Card 1: 56/60 chance of not being the card you targeted
  • Card 2: 55/59
  • Card 3: 54/58
  • Card 4: 53/57
  • Card 5: 52/56
  • Card 6: 51/55
  • Card 7: 50/54

The odds of ALL of these happening (i.e. none of the four cards being in your hand) is the result of multiplying all these odds together:

(56*55*54*53*52*51*50) / (60*59*58*57*56*55*54) = ~0.6005 or ~60%

To calculate the odds of at least one of these cards being the one you're looking for, you can subtract this result from 1 (or 100%) to get a ~40% chance that (at least) one of your four cards will occur in a 7-card draw from a 60-card deck.

  • 4
    user1873 had the correct answer, but I've written out the working to answer the comments to that first answer.
    – Johno
    Commented Jul 12, 2012 at 18:16

Magic Workstation besides many other tools for collection management, deck building, and online play has a very powerful probability calculator. It will go beyond opening hand and will let you see by what turn are you likely to have drawn the combo that you need.

  • Magic Workstation appears to be offline/gone, but I found this alternative (which seems pretty easy to use): mtgnexus.com/tools/drawodds
    – Venryx
    Commented Jul 7, 2022 at 11:38

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