Probability, part 3
So, now we’ve defined all these wonderfully confusing terms. How do they help?
Potentially quite a bit, but let’s start by getting some cold, hard numbers. We’ll start with our original 35/25 deck, and figure out a few expectancies. We’ll start with the expected turn by which we’ll have drawn 4 lands.
The probability that we’ll have drawn exactly four lands on turn one, i.e. after having drawn seven cards, is, as we saw last week,
L(1) = (C(m,k)*C(N-m,n-k))/C(N,n)=(C(25,4)*C(35,3)/C(60,7)≈0.214
So the first term of the sum is 1*0.214. For the second term, we draw one more card, and therefore one more spell, for a probability of
L(2) = (C(m,k)*C(N-m,n-k))/C(N,n)=(C(25,4)*C(35,4)/C(60,8)≈0.258
and a term of 2*0.258 = 0.516. Adding these two terms together gets us 0.214+0.516=0.730. We go on creating these terms, with the probabilities of
L(n) = (C(25,4)*C(35,n+2)/C(60,n+6)
Which we can multiply by n and then add all up together. We do this for all values of n… okay, no, not really. Once we’ve drawn our 39th card, even if we started out by drawing all 35 non-lands, the next four cards will have to be lands, so any further draws will put us over the exact number of lands we are looking for, namely four. That means we can stop adding up terms after n=33, which happens to coincide with the last number for which C(35,n+2) is defined. Using our handy-dandy calculator (WolframAlpha is your friend!), we arrive at an expectancy of
E(L) = Σ n * (C(25,4)*C(35,n+2)/C(60,n+6) ≈ 10.647.
In other words, we expect to have drawn exactly 4 lands on average between turns 10 and 11. If that number seems high, well, it is, and there’s a few reasons for that, the biggest being that we only have used positive numbers for n. If we accept negative values of n (in other words, we take into account the cards drawn before the start of the game), we arrive at the value 3.894 – a much more reasonable (and realistic) value. So why use the first value at all?
Well, firstly it’s just a mathematical construct that will help us in making mulligan decisions, not an actual representation of what turn we’ll draw four lands by, and secondly shut up! I goofed, all mathematicians are entitled to a false start or two!
Ahem, anyway, there actually is a good reason for using the first expectancy, and that is because the expectancy I calculated actually doesn’t assume we know anything about our opening hand. What we need to do is look at how the expectancy changes based on the contents of our opening hand. To do this, we have to use something called the conditional expectation.
Let’s say the opening hand has three lands in it already. So what is the probability then that we draw our fourth land in the second turn? Well, the library still contains 22 lands and 31 spells, so the probability is a straight 22/53 ≈ 0.423. The probability that we have four lands in the next turn is the probability that you draw first a land and then a spell plus the probability you draw a spell and then a land, or
22/53*31/52+31/53*22/52=2*(22*31)/(53*52) ≈ 0.495
And if this is starting to look familiar to you, then good, you might just be a mathematician after all. You see, once we take out the three lands and four spells out of the library, all we’re basically doing now is drawing balls out of an urn again, just this time it’s an urn that has 22 white and 31 black balls in it. Once again, we can calculate an expectancy of
E(L')= Σ n * (C(22,1)*C(31,n-1)/C(53,n) ≈ 8.413.
Again, remember, this is just a mathematical construct, it doesn’t mean you have to wait 8 turns to draw another land. But what IS interesting is when you compare that to the expectancy if you start with 2 lands and 5 spells.
E(L'')= Σ n * (C(23,2)*C(30,n-1)/C(53,n) ≈ 84.15
Which is, in case you didn’t notice, a considerably larger number. In other words, if you have an opening hand with two lands in it, on average you’ll have to wait quite a bit longer to draw up to four lands than you will if you start with three lands.
Shocking, I know! But we’ll get into how this can actually be of decision-making help to us next week.
- What is the expectancy, without looking at your opening hand, of drawing 3 lands? Of five?
- What are the expectancies of drawing four lands for the other numbers of lands in your opening hand?