<p>The calculation you are looking for is called a <a href="http://stattrek.com/online-calculator/hypergeometric.aspx" rel="noreferrer">Hypergeometric Distribution</a>. This calculated your chances of drawing a particular number of "successes" from a population, without replacement.</p>

<ul>
<li>Population Size: 60 cards</li>
<li>Successes in Population: 4 Birds of Paradise</li>
<li>Sample Size: 7 cards</li>
<li><p>Successes in Sample: 1 (the minimum number we want to draw)</p></li>
<li><p>Results: 39% chance of drawing at least 1 Birds of Paradise.</p></li>
</ul>

<p>In the Hypergeometric Distribution calculator linked above, that result is represented in the <em>Cumulative Probability: P(X &ge; 1)</em> field: the chance of drawing greater than or equal to 1.</p>

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

<p>You can see the calculation on the <a href="http://en.wikipedia.org/wiki/Hypergeometric_distribution#section_1" rel="noreferrer">Wikipedia page</a>, 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 <a href="http://en.wikipedia.org/wiki/Binomial_coefficient#Computing_the_value_of_binomial_coefficients" rel="noreferrer">binomial coefficients</a> (and factorials).</p>

CC BY-SA 3.0
2012-07-12T18:06:37.703Z
Interesting program, but i was looking for the formula so that i could figure it out on my own.
DForck42
CC BY-SA 3.0
2012-07-12T18:08:23.303Z
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
CC BY-SA 3.0
2012-07-12T18:10:36.030Z
@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: http://stats.stackexchange.com/questions/24211/what-is-the-probability-of-n-people-from-a-list-of-m-people-being-in-a-random-se
ghoppe
CC BY-SA 3.0
2012-07-12T18:15:30.197Z
Why WOULDN'T you want to do it by hand!? :P
Johno
CC BY-SA 3.0
2016-02-22T17:14:07.333Z
Question is looking for "At least one," yet answer provides "exactly one."
Drunk Cynic
CC BY-SA 3.0
2016-02-22T23:26:28.953Z
@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
CC BY-SA 3.0
2016-02-23T00:09:26.470Z
@user1873 I understand the inter-workings of the hypergeometric distribution, and took no other issue with the answer. Using the calculator you linked, the at least one is just the "X >= 1" result, ~40%.
Drunk Cynic
CC BY-SA 3.0
2016-02-23T15:30:31.197Z
@DrunkCynic Isn't that what user1873 said in the sentence directly following his bullet points?
ghoppe
CC BY-SA 3.0
2016-02-23T23:28:59.003Z
@ghoppe, well bollocks for tunnel vision.
Drunk Cynic
CC BY-SA 4.0
2012-07-12T17:17:54.403Z
2018-05-05T00:43:46.563Z
2018-05-05T00:43:46.563Z
user1873
<p>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.</p>

<p>To calculate the odds of the entire first hand, we can do it backwards:</p>

<p>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:</p>

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

<p>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:</p>

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

<p>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 <strong>~40% chance that (at least) one of your four cards will occur in a 7-card draw from a 60-card deck</strong>.</p>

CC BY-SA 3.0
2012-07-12T18:16:11.930Z
user1873 had the correct answer, but I've written out the working to answer the comments to that first answer.
Johno
CC BY-SA 3.0
2012-07-12T18:14:55.317Z
2012-11-29T12:52:12.413Z
2012-11-29T12:52:12.413Z
Johno
<p><a href="http://www.magicworkstation.com/" rel="noreferrer">Magic Workstation</a> 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.</p>

CC BY-SA 4.0
2022-07-07T11:38:43.030Z
Magic Workstation appears to be offline/gone, but I found this alternative (which seems pretty easy to use): https://www.mtgnexus.com/tools/drawodds
Venryx
CC BY-SA 3.0
2012-07-12T22:03:45.340Z
2012-07-12T22:03:45.340Z
Andrey
<p>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?</p>
<p>For example, let's say I have a 60 card deck, and I'm running 4 <a class="mtg-autocard" target="_blank" rel="noreferrer" href="https://gatherer.wizards.com/Pages/Card/Details.aspx?name=Birds%20of%20Paradise">Birds of Paradise</a>. What is the percent chance that I will have at least one Bird in my opening hand?</p>

<magic-the-gathering><probability>
How do you calculate the likelihood of drawing certain cards in your opening hand?
DForck42
# How do you calculate the likelihood of drawing certain cards in your opening hand?

(Asked by: username_0 on 2012-07-12)

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](https://gatherer.wizards.com/Pages/Card/Details.aspx?name=Birds%20of%20Paradise). What is the percent chance that I will have at least one Bird in my opening hand?

**Tags:** magic-the-gathering, probability


---

## 3 answers

### Answer by username_1 on 2012-07-12. Score: 57

The calculation you are looking for is called a [Hypergeometric Distribution](http://stattrek.com/online-calculator/hypergeometric.aspx). 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](http://en.wikipedia.org/wiki/Hypergeometric_distribution#section_1), 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](http://en.wikipedia.org/wiki/Binomial_coefficient#Computing_the_value_of_binomial_coefficients) (and factorials).

> **Comments:**
> **username_0**: Interesting program, but i was looking for the formula so that i could figure it out on my own.  
> **username_2**: 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%.  
> **username_2**: @username_0 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: http://stats.stackexchange.com/questions/24211/what-is-the-probability-of-n-people-from-a-list-of-m-people-being-in-a-random-se  
> **username_3**: Why WOULDN'T you want to do it by hand!? :P  
> **username_4**: Question is looking for "At least one," yet answer provides "exactly one."  
> **username_1**: @username_4, 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.  
> **username_4**: @username_1 I understand the inter-workings of the hypergeometric distribution, and took no other issue with the answer. Using the calculator you linked, the at least one is just the "X >= 1" result, ~40%.  
> **username_2**: @username_4 Isn't that what username_1 said in the sentence directly following his bullet points?  
> **username_4**: @username_2, well bollocks for tunnel vision.  

---

### Answer by username_3 on 2012-07-12. Score: 34

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**.

> **Comments:**
> **username_3**: username_1 had the correct answer, but I've written out the working to answer the comments to that first answer.  

---

### Answer by username_5 on 2012-07-12. Score: 8

[Magic Workstation](http://www.magicworkstation.com/) 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.

> **Comments:**
> **username_6**: Magic Workstation appears to be offline/gone, but I found this alternative (which seems pretty easy to use): https://www.mtgnexus.com/tools/drawodds  

---

