Magic Math: Stats on Drawing Cards

By: Meta_Psychosis

What is This?

This is a small write-up on some of the math for finding the probability of drawing certain cards. The context will be framed around the Pioneer Breakfast Bears deck (also known as Pioneer Lumra or Simic Scapeshift Mill). For the average game of magic, this won’t matter too much. Most decks just draw their card for the turn and maybe an extra card for some card advantage. This will primarily apply to combo decks that rely on some amount of randomness, such as the Pioneer Breakfast Bears deck.

Basic Probabilities

In magic, the typical deck is 60 cards. For the purpose of simplicity, ignore Yorion decks or decks that play 61 cards. These principles will still apply, but there is no need to discuss every deck size possible. When drawing a card from a deck of 60 cards, there is a 1.67% chance of drawing any single card from the deck, assuming all 60 cards are different. This is found by taking the number of “hits” and dividing that by the population. In this instance, since all 60 cards are different and you want to know the chance of drawing one specific card from the top, it would be 1/60, then multiply that by 100 to get the percent chance. Now, say there are four copies of a card in the deck, and you want the chance of drawing one of those four cards from the top of the 60-card deck. That would be 4/60 * 100, which is 6.67%. Lastly, for the basics here, say the deck is at 50 cards, and you have 3 copies of a particular card left in your deck. The chances that the top card of the deck is one of those three copies of the card you want is 3/50 * 100, which is 6%. 

The next step in this is to find the probability of a failure. To do this, firstly, you need to know your population size as well as the number of successes. So, for example, say I have a 60-card deck, and I have 4 successes. What are the chances of not drawing one of those four cards? Firstly, take your population of 60 and subtract 4 to get 56. This is the number of failures in the population. Then divide that by the total population. This comes out to 93.3%. That means that out of 60 cards, there is a 93.3% chance that the top card of the deck is not one of the four cards you want. 

Let's say you wanted the probability of drawing four cards and all four of the cards are not the one you wanted. For this example, assume we still have a 60-card deck, 4 successes, and we will draw 4 cards. Firstly, you need to know the chances of the top card in the 60-card deck not being that card. Earlier, this was already found to be 93.3%. Now, assuming that the card is not replaced back to the top, whether by going to the graveyard or it stays in your hand, you next need to find the chance that the next card is not the one you wanted. In this case, the population size has been reduced by 1, making it 59 total. So you would take this new population of 59, subtract the number of successes, which is 4, then divide by the population. This comes to 93.2%. Continuing this, we get 93.1% and 93%. Keep in mind, this is the probability of NOT drawing one of the four cards you wanted. Now, to combine all of this, you would multiply all four percentages together. This gives us a 75.3 percent chance of not drawing the card we want. 

So we know the chance that one of the top 4 cards isn't the one we want, but what if we wanted to know the chance that one of the top 4 cards is the one we want? To do this, firstly, you need to find the chance that it isn't the card you wanted. In this example, we said we would draw 4 cards from a 60-card deck, and there are 4 successes in a 60-card deck. It was found that there was a 75.3 percent chance that we did not draw the card we wanted. To find the chances that it is you would have to take one minus the chance of not drawing the card. So for our case, 1-75.3% gives us 24.7%. So we have a 24.7% chance that of the 4 cards we drew, one of them is a success. 

This process can be long, and it is prone to rounding errors, but it is the baseline for how the hypergeometric distribution works. One downside of doing this the long way like this is that we can only ever find the chances of finding a single success, whereas a hypergeometric distribution can find the chance of getting multiple successes. A bonus of doing this the long way is that it is much simpler to think about and do by hand. 

Hypergeometric Equations

Before beginning, there are online tools to automatically calculate these probabilities. So if you're not interested in learning how to do this by hand, I would recommend using https://aetherhub.com/Apps/HyperGeometric to find the probabilities quickly.

To start with, when it comes to hypergeometric distributions, the first thing to recognize is that there are only two "categories". These categories are success and failure. If, for some reason, a third category is needed, then this hypergeometric distribution will not work. When it comes to magic, these categories are easy enough to identify, but outside of Magic, there may be some uses for having more than one category. Another important aspect is that the hypergeometric distribution does not replace back into the population. So if a success is removed, then the probability of drawing a second success is changed. This can be powerful in that the hypergeometric distribution can give the probability of drawing more than one success. This may not come up very often in Magic: The Gathering, but it's important to know how to understand the equation itself.

The equation for the hypergeometric distribution is below:

y=Number of successes in the number of trials n 

N=Total Population

r=Successes in the population

N-r=Number of failures in the population

n=Number of trials

n-y=number of failures in n trials

So firstly, let’s break down what these variables even mean. For our purposes, the trials will refer to the number of cards drawn. So y is the number of successes we want in the number of cards drawn. N will be the deck size when we are finding the probability of drawing any particular card. R is the number of cards in the deck we would consider a success. So, using the previous example, we will have a 60-card deck, drawing 4 cards, and 4 of the cards will be successes. Let's say we only want to draw one of those 4 cards.

y=1(we only need one copy of the card)

N=60(the deck size at the time of finding the probability)

r=4(there are 4 copies of the card we want to draw)

N-r=56(56 of the cards in the deck are not what we want to draw)

n=4(we are drawing 4 cards total)

n-y= 3(of the 4 cards drawn, 3 of them will be cards we do not want to draw)

Now for the actual equation,

​

After plugging in the numbers, we get the following:

Which is equal to:

So now the question is, what does each of the parts actually mean? 

This is going to be the probability that there is one success within the sample size.

This is the number of ways the single card can be chosen from the 4 cards drawn. This is spoken as “4 choose 1”. 

Similar to the 4 choose 1, this is how many combinations of 3 cards are within the 56-card total population.

Finally, 60 choose 4 represents the number of combinations of 4 cards within the 60-card deck.

Now, to calculate each of these sections, you will need the binomial coefficient equation. 

In this case, n is equal to the top number; it is not equal to the number from the hypergeometric distribution. For this lets plug in the 56 choose 3. This gives us the equation

Now, to calculate the individual pieces of this equation. Firstly, 56! is also called 56 factorial. What this represents is the multiplication of all previous numbers. For example, 3! = 1*2*3. So 56! Is 1*2*3*4…*56. For 56! We get an extremely large number that I'm not even going to try to type out here. For actually calculating this equation, https://www.desmos.com/scientific is a great online calculator that has the factorial function. Clicking the Func option will show the ! to calculate this by hand if needed. As for calculating the 56 choose 3, you should get 27,720. For the 4 choose 1, you will get the answer of 4, and the 56 60 choose 4 will give the answer of 487,635.

Now, back to the hypergeometric equation, we can plug all of this back in to get

This calculates to 22.7%. Something major to note here is that this is not the same 24.7% we got earlier. This is because with this hypergeometric calculation, we found the probability of drawing exactly 1 copy, whereas the other calculation found the probability of drawing 1 or more copies. To find the probability of drawing one or more successes, you need to calculate things a bit differently. You would find the probability of not drawing a single copy of the card, then take one minus that probability to find the chance of drawing one or more copies. I will work through the example, but not as in-depth. 

This will equal 75.3%, which is the chance of not drawing any copies of the card wanted. Taking 1-75.3% = 24.7%. This is the chance of drawing one or more copies of the card.

Events in Series

Now the question may be “what if I draw 4 cards, then my number of successes changes after drawing another 4 cards”. Well, when we manually found the probability of not drawing a card by changing the population size, that same principle applies here. So let's continue with our example. We will still start with the 60-card deck, drawing 4 with 4 successes. We found the chances of not drawing one of the four cards was 75.3%. So let's say we continue and draw another 4 cards, but this time there are only two successes. This means our new population size is starting at 56. So, plugging in these numbers, we find that there is a 86.1% chance that in the second draw 4 we do not draw one of the two cards we wanted to draw. 

This information can be useful, but how do we find the total story? Similar to how we calculated each failure manually, then multiplied them together to find the total failure rate, we can do the same thing here. So, to start from the top. We start with a 60-card deck, and draw 4 cards. Within the 60-card deck are 4 successes. After the initial draw of 4, we draw an additional 4 (8 cards total), and within the second set of 56 cards, there are 2 successes. For event one, we found a 75.3% failure rate. For event two, we found a 86.1% failure rate. The combined failure rate is 75.3%*86.1%=64.9%. So between the two events, there is a 64.9% chance that both events fail. If we then take 1 minus this, we get 1-64.9%=35.1%. So we have a total percent chance of success of 35.1%.

A Few Use Cases for this Math

So that covers the math itself, but where is this useful? If you want to find the chances of drawing a number of lands in your opening hand, you would plug in the number of lands you have and the number of lands you want to draw, and you will get that probability. If you wanted to find the chance of milling a Lumra with a specific line, you would plug in the number of Lumra and the number of cards milled, and that would give the chance of milling one or more Lumra. There are many other situations that could be useful; those are just a few quick ones off the top of my head.