A question for Mathematicians/Statisticians

[PhilipJFry]

What is the proper way to calculate the odds of, when you draw Mox Opal, drawing 2 additional artifacts.  I assume that the important pieces of information are the number of cards in your deck, and the number of non-Mox Opal artifacts in your deck, and I guess the number of cards you draw.  But I am not gifted with numbers, and I was hoping that someone would provide the appropriate equations.

Likewise, what is the equation to calculate the odds of drawing 2 Mox-Opals in a given hand?

Thanks for nay help that can be provided.

[meadbert]

For simplicities sake assume an infinite sized deck with A being the probability that a card is an artifact.

The probability of getting 0 artifacts in 6 draws is (1-A)^6
The probability of getting 1 artifact in 6 draws is 6*A*(1-A)^5

Thus the probability of getting 2 or more artifacts in 6 draws is 1 - prob(0) - prob(1) = 1 - (1-A)^6 - 6*A*(1-A)^5


Probability of getting one of a 4of in your opening hand is roughly 40%.
Prob of getting duplicate of a 4of in hand is roughly 10%.  That drops to 5% is you only run 3 and only 1.7% or something like that if you run 2.

[DubDub]

Hypergeomdist
The 'hypergeomdist' function (in Excel) is very useful for solving questions like these.

The syntax for that function is:
=hypergeomdist(sample_successes, sample_size, population_successes, population_size)

If we are talking about opening hands, sample_size will be 7 (cards in hand at the beginning of the game), and population_size will be 60 (cards in deck).

Sample_successes is how many Mox Opals, or artifacts you want in the opening hand.
Population successes is how many Mox Opals, or qualifying artifacts there are in your deck (excluding things like Inkwell Leviathan, which don't really help).

You're asking about a situation that's a combination of a) having Mox Opal, and b) turning it on.  Let's take each in turn.

Having Mox Opal
Our function will look like (assuming you're playing just a singleton Mox Opal)
=hypergeomdist(1, 7, 1, 60)

Which is equivalent to 7/60, or ~ 11.7%.

Turning Mox Opal on
The key variable here is how many qualifying artifacts you plan to play.  I'll assume, full normal moxen, Lotus/Petal/Crypt/Sol Ring/Voltaic Key/Sensei's Divining Top, which is 11.

Something to note is that, if we want to find the chance of having two or more of the above in our opening hand, we a) only have six cards remaining to fill in, since we're taking cases where Mox Opal is in our opening hand, and the population is 59, since Mox Opal is in our opening hand.

Another note, the chance of having exactly two + exactly three + ... + exactly six of these 11 is equivalent to (1- chance of having 0 and chance of having 1).  So we have:
=(1- (hypergeomdist(0, 6, 11, 59)+(hypergeomdist(1, 6, 11, 59)))
Which returns ~30.9%.

The composite probability then is 11.7% times 30.9%, which is only roughly 3.6%..  Not a very good chance of having Mox Opal active on one's first turn, although this doesn't factor in being on the draw, seeing more cards quickly (with Brainstorm, Ancestral, etc), mulliganing hands of seven that don't contain the acceleration you want, or other factors.

[Delha]

Hypergeomdist
The 'hypergeomdist' function (in Excel) is very useful for solving questions like these.

That's awesome. I work with Excel a lot, but I'm almost entirely self trained, and never heard of that function.

[Harlequin]

DubDub is right.  But to bring that a bit further (and make a math lesson out of it, we need to look at Laws of Conditional Probability!

Pr(Event A) = Pr(A | B) x Pr(B) + Pr(A | Not-B) x Pr(not-B)
Where "|" is our "Conditional on" or "...Given..."

Quick example:  Supose we're at seaworld and we have time for 1 more attraction.  We're going to flip a coin to either swim in the shark tank or the dolphin pool.  And we also know that 4 out of every 10 people who swim in the shark tank are eaten.  What is our probability of getting eaten by sharks?  Intuatively we may be able to derive that its going to be 20%, but using our law:

Event A ~> Eaten by a shark
Event B ~> we are swiming in the shark tank.

Prob(Eaten by a shark) =
Prob( Getting eaten by a shark GIVEN we are swimming in the shark tank ) x prob( ending up in the shark tank)
+ Prob( Getting eaten by a shark GIVEN we are swimming in the doplhin pool ) x prob( we decide to swim with dolphins)

Which ends up being = 40% x 50% + 0% x 50% = 20%


So with our exmaple:
Event A ~> A 7 card hand that looks like
 - Exactly 1 Mox Opal
 - At least 2 of the remaining 6 cards are Artifacts (that arn't other mox Opals, because its legendary)

For this, lets say that...
Event B ~> Drawing Exactly 1 Mox Opal

Then we can say that:
Prob( Event A ) = Prob( drawing 2+ artifacts in 6 cards given we've already drawn our Mox Opal ) x Prob( drawing 1 Mox Opal in 7 cards ) + 0% *
* because our probability of encountering a shark disguised as a dolphin is negligibly small
** wait no, I mean that we can't satifiy Event A without actually drawing a Mox Opal.

So to type it out in long hand: 
Prob( drawing 1 Mox Opal in 7 cards ) =
=HYPGEOMDIST(1, [Drawn], [Opals] ,60)
 ~ Where [Opals] is the Address of the Cell containg the number of Opals you are Running
 ~ And [Drawn] is in this case 7.  But if youre going do this in excel, might as well make a variable out of it.

We also need: Prob( drawing 2+ artifacts in 6 cards given we've already draw a mox opal ) =
=1-(HYPGEOMDIST(0,[Drawn]-1,[Artifacts],59) + HYPGEOMDIST(1,[Drawn]-1,[Artifacts],59))
 ~ where [Artifacts] is the number of other artifacts you're running.

 *Note that this formula is actually saying "The probability of Not drawing 0 or 1 artifacts" Which is simpiler than typing out prob(*2*) + prob(*3*) + prob(*4*) + prob(*5*) + prob(*6*)
 * Also note that this time its actually out of 59 cards, because we have to account for our "GIVEN we've already draw a mox opal" statement.  Meaning there are only 59 cards left in the deck we could draw from.

Now simply Multiply Event A|B and event B and you're good to go.
So if we plug in 1 Opal, 11 Other Artifacts, 7 Cards drawn => about 3% like DubDub said.

Now we can plug in all sorts of combinations.  Like running 3 Opals we end up with about 13%.  If we instead are on the draw (meaing we draw 8 cards) we end up with almost 18%.