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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?

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Only weakly related, but good deck-building software (including the one I usually used when I still played MtG a lot, at Essential Magic) will give you sample opening hands and some statistical information. – Aesin Jul 12 '12 at 19:57
take the odds that you won't get any birds of paradise, and invert it – Sam I am Apr 9 '13 at 21:39

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: exactly 1

  • Results: 33%

The online calculator will also give you the odds of drawing greater than (6%) the exact number of successes in the sample, and "at least" as many successes (greater than or equal to 1 = 40%).

You can see the calculation on the Wikipedia page, or searching 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).

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Interesting program, but i was looking for the formula so that i could figure it out on my own. – DForck42 Jul 12 '12 at 18:06
That program calculates the odds of getting at least one BoP at 40% if you are going first. Remember, since you draw when you go second the odds rise to 44%. – ghoppe Jul 12 '12 at 18:08
@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:… – ghoppe Jul 12 '12 at 18:10
Why WOULDN'T you want to do it by hand!? :P – Johno Jul 12 '12 at 18:15
Question is looking for "At least one," yet answer provides "exactly one." – Drunk Cynic Feb 22 at 17:14

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.

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user1873 had the correct answer, but I've written out the working to answer the comments to that first answer. – Johno Jul 12 '12 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.

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