# How do you calculate the likelihood of drawing certain cards in your opening hand?

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?

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. Jul 12 '12 at 18:06
• @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/… Jul 12 '12 at 18:10
• Why WOULDN'T you want to do it by hand!? :P Jul 12 '12 at 18:15
• @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. Feb 22 '16 at 23:26
• @DrunkCynic Isn't that what user1873 said in the sentence directly following his bullet points? Feb 23 '16 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.