<p>In the situation you describe, each player reveals their card independently, and you get mana equal to the number of non-land cards revealed. This is called a <a href="https://en.wikipedia.org/wiki/Binomial_distribution" rel="nofollow noreferrer">Binomial Distribution</a>, with <code>N = 4</code> and <code>P = 2/3</code> in your particular case. So, the probability of getting exactly <code>k</code> mana is <code>(4 choose k) * (2/3) ^ k * (1/3) ^ (n - k)</code>. You are specifically looking for the probability of getting at least a certain amount of mana, so you are looking for the Cumulative Binomial Distribution. To get the probability of getting <em>at least</em> a certain amount of mana, you simply add up the probabilities for each amount that is at least that number. So, to calculate that for your particular situation, you would calculate</p>

<p><a href="https://i.sstatic.net/HfEvu.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/HfEvu.gif" alt="Sum from k = 2 to 4 of (4 choose k) * (2 / 3) ^ k * (1 / 3) ^ (4 - k)"></a></p>

<p>The fully general formula for this, if you want to get at least M mana is</p>

<p><a href="https://i.sstatic.net/coiQa.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/coiQa.gif" alt="Sum from k = M to N of (N choose k) * (P) ^ k * (1 - P) ^ (N - k)"></a></p>

<p>In the particular case of <code>k = 2</code>, we can instead simply consider the cases where you <em>don't</em> get at least 2 mana, and then subtract that total probability from 1. For each player, the probability of only that player flipping a non-land is 2/3 * (1/3) ^ (N - 1). The probability of every player flipping a land is (1/3) ^ N. So, the total probability of getting at least two mana is</p>

<p><a href="https://i.sstatic.net/8A5EK.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/8A5EK.gif" alt="1 - (N * 2/3 * (1/3) ^ (N - 1) + (1/3) ^ N)"></a></p>

<p>For 4 players, this probability is 8/9.</p>

CC BY-SA 3.0
2016-04-21T19:01:59.073Z
Thanks this is what I was looking for also, how would you use this to figure out the probability of generating x mana
Peter Kwiatek
CC BY-SA 3.0
2016-04-21T19:46:57.493Z
I added a fully general formula
murgatroid99
CC BY-SA 3.0
2016-04-22T00:28:06.360Z
What does ^ mean?
Zinma
CC BY-SA 3.0
2016-04-22T00:57:30.560Z
That's exponentiation. We don't have MathJax on this site, so I don't really have a better way to represent it.
murgatroid99
CC BY-SA 3.0
2016-04-24T06:55:29.140Z
You could use a Latex generator to create a static image: http://arachnoid.com/latex/?equ=%5Ctextrm%7BSum%20from%20%7D%20k%20%3D%20M%20%5Ctextrm%7B%20to%20%7D%20N%20%5Ctextrm%7B%20of%20%7D%20(N%20%5Ctextrm%7B%20choose%20%7D%20k)%20%5Ctimes%20P%5E%7Bk%7D%20%5Ctimes%20(1%20-%20P)%5E%7B(N%20-%20k)%7D for middle equation, http://arachnoid.com/latex/?equ=1%20-%20(N%20%5Ctimes%20%5Cfrac%7B2%7D%7B3%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7B3%7D%5E%7B(N%20-%201)%7D%20%2B%20%5Cfrac%7B1%7D%7B3%7D%5EN) for the last one.
doppelgreener
CC BY-SA 3.0
2016-04-24T17:22:54.897Z
OK, I've made the large equations into LaTeX images.
murgatroid99
CC BY-SA 3.0
2016-04-21T18:44:51.120Z
2016-04-24T17:20:59.597Z
2016-04-24T17:20:59.597Z
murgatroid99
<p>I am trying to figure out what are the chance if getting 2 or more mana after parleying <a class="mtg-autocard" target="_blank" rel="noreferrer" href="https://gatherer.wizards.com/Pages/Card/Details.aspx?name=Selvala%2C%20Explorer%20Returned">Selvala, Explorer Returned</a>, assuming that using commander/EDH one third of the deck is land and that at least four people are playing. Mainly curious after my friends mentioned way to go infinite if I could reliably generate mana.</p>

CC BY-SA 3.0
2016-04-21T18:20:22.573Z
I'm not sure how "infinite" it would be, since everyone will be drawing cards.  At some point people will run out of cards and lose.
Becuzz
CC BY-SA 3.0
2016-04-21T18:21:20.427Z
The answer to this question also depends on how many people you are playing with.
Becuzz
CC BY-SA 3.0
2016-04-21T18:24:20.667Z
Lets assume its a four player game, and I know it won't be infinite per say I just wanted to know the probability
Peter Kwiatek
CC BY-SA 3.0
2016-04-21T18:24:43.713Z
I think that assuming that only 1/4 of a deck is land in EDH is unrealistic. Going below 1/3 is already risky, and 2/5 is common.
murgatroid99
CC BY-SA 3.0
2016-04-21T18:30:56.920Z
I said assuming a one land to 3 nonland ratio, not one to four
Peter Kwiatek
CC BY-SA 3.0
2016-04-21T18:32:37.233Z
@PeterKwiatek 1 land to 3 nonland ratio means that 1/4 of the deck is land, and 3/4 of the deck is nonland.
Rainbolt
CC BY-SA 3.0
2016-04-21T18:34:56.490Z
@Rainbolt My bad, i got confused i meant a third of the deck is land
Peter Kwiatek
<magic-the-gathering><mtg-commander>
What's the odds of Selvala's parley granting 2 or more mana?
Peter Kwiatek
# What's the odds of Selvala's parley granting 2 or more mana?

(Asked by: username_0 on 2016-04-21)

I am trying to figure out what are the chance if getting 2 or more mana after parleying [Selvala, Explorer Returned](https://gatherer.wizards.com/Pages/Card/Details.aspx?name=Selvala%2C%20Explorer%20Returned), assuming that using commander/EDH one third of the deck is land and that at least four people are playing. Mainly curious after my friends mentioned way to go infinite if I could reliably generate mana.

**Tags:** magic-the-gathering, mtg-commander

> **Comments:**
> **username_1**: I'm not sure how "infinite" it would be, since everyone will be drawing cards.  At some point people will run out of cards and lose.  
> **username_1**: The answer to this question also depends on how many people you are playing with.  
> **username_0**: Lets assume its a four player game, and I know it won't be infinite per say I just wanted to know the probability  
> **username_2**: I think that assuming that only 1/4 of a deck is land in EDH is unrealistic. Going below 1/3 is already risky, and 2/5 is common.  
> **username_0**: I said assuming a one land to 3 nonland ratio, not one to four  
> **username_3**: @username_0 1 land to 3 nonland ratio means that 1/4 of the deck is land, and 3/4 of the deck is nonland.  
> **username_0**: @username_3 My bad, i got confused i meant a third of the deck is land  

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## 1 answers

### Answer by username_2 on 2016-04-21. Score: 5 | Accepted answer

In the situation you describe, each player reveals their card independently, and you get mana equal to the number of non-land cards revealed. This is called a [Binomial Distribution](https://en.wikipedia.org/wiki/Binomial_distribution), with `N = 4` and `P = 2/3` in your particular case. So, the probability of getting exactly `k` mana is `(4 choose k) * (2/3) ^ k * (1/3) ^ (n - k)`. You are specifically looking for the probability of getting at least a certain amount of mana, so you are looking for the Cumulative Binomial Distribution. To get the probability of getting *at least* a certain amount of mana, you simply add up the probabilities for each amount that is at least that number. So, to calculate that for your particular situation, you would calculate

[![Sum from k = 2 to 4 of (4 choose k) * (2 / 3) ^ k * (1 / 3) ^ (4 - k)](https://i.sstatic.net/HfEvu.gif)](https://i.sstatic.net/HfEvu.gif)

The fully general formula for this, if you want to get at least M mana is

[![Sum from k = M to N of (N choose k) * (P) ^ k * (1 - P) ^ (N - k)](https://i.sstatic.net/coiQa.gif)](https://i.sstatic.net/coiQa.gif)

In the particular case of `k = 2`, we can instead simply consider the cases where you *don't* get at least 2 mana, and then subtract that total probability from 1. For each player, the probability of only that player flipping a non-land is 2/3 \* (1/3) ^ (N - 1). The probability of every player flipping a land is (1/3) ^ N. So, the total probability of getting at least two mana is

[![1 - (N * 2/3 * (1/3) ^ (N - 1) + (1/3) ^ N)](https://i.sstatic.net/8A5EK.gif)](https://i.sstatic.net/8A5EK.gif)

For 4 players, this probability is 8/9.

> **Comments:**
> **username_0**: Thanks this is what I was looking for also, how would you use this to figure out the probability of generating x mana  
> **username_2**: I added a fully general formula  
> **username_4**: What does ^ mean?  
> **username_2**: That's exponentiation. We don't have MathJax on this site, so I don't really have a better way to represent it.  
> **username_5**: You could use a Latex generator to create a static image: http://arachnoid.com/latex/?equ=%5Ctextrm%7BSum%20from%20%7D%20k%20%3D%20M%20%5Ctextrm%7B%20to%20%7D%20N%20%5Ctextrm%7B%20of%20%7D%20(N%20%5Ctextrm%7B%20choose%20%7D%20k)%20%5Ctimes%20P%5E%7Bk%7D%20%5Ctimes%20(1%20-%20P)%5E%7B(N%20-%20k)%7D for middle equation, http://arachnoid.com/latex/?equ=1%20-%20(N%20%5Ctimes%20%5Cfrac%7B2%7D%7B3%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7B3%7D%5E%7B(N%20-%201)%7D%20%2B%20%5Cfrac%7B1%7D%7B3%7D%5EN) for the last one.  
> **username_2**: OK, I've made the large equations into LaTeX images.  

---

