Check out part 1 where I describe hypergeometric probabilities here.

Sylvan Messenger

I could once again provide hypergeometric probabilities for the scenario in which we remove one Sylvan Messenger and consider the remaining deck. However, I’ll go a little deeper because the number of Elves in the deck changes if you’ve drawn more than one Sylvan Messenger by the time you can cast the first one, and I’d like to capture that effect. So instead, I coded a simulation of millions of games for a deck with 4 Sylvan Messengers, 22 lands, and a certain number of other Elves, and asked my computer to analyze the outcomes for the first Sylvan Messenger that is cast each game.

Here are the outcomes:

SylvanMessenger_1 SylvanMessenger_2

To be happy with Sylvan Messenger in my deck, I’d want to hit at least 2 Elves on average. This way, it’s like a Harmonize where one of the cards drawn is a zero-mana 2/2, and Harmonize was a Constructed-playable card. Looking at the numbers, this means that I recommend at least 26 other Elves (besides 4 Sylvan Messenger) in a deck with Sylvan Messenger.

Sphinx’s Tutelage

Good old Grindstone is back! Precise calculations on how many cards you can expect to mill are complicated as it depends on your opponent’s deck size and you have to go through all the permutations. But we can get a good approximation by considering an infinitely large opposing deck with a fixed fraction of colored cards.

If you play against a mono-color deck with 2/3rd spells and 1/3rd lands, then you’ll repeat the milling process if you hit two spells, which happens with probability (2/3)^2=4/9, and you’ll stop with the complementary probability of 5/9. The expected number of milling processes is geometrically distributed with an expected value of 1 /(5/9)=1.8, and since you mill two cards per process, you can expect to mill 3.6 cards per trigger.

If you play against a two-color deck with 1/3rd spells of one color, 1/3rd spells of the other color, and 1/3rd lands, then you’ll repeat if there is a match, which happens with probability 2*(1/3)^2=2/9, and you’ll stop with the complementary probability of 7/9. Here, the expected number of milling processes is geometrically distributed with an expected value of 1 /(7/9)=9/7, which comes down to milling 2.57 cards per trigger in expectation.

In reality, the numbers will be a little lower because if you keep on hitting matching colors, then the density of lands in the remaining deck goes up. But overall Sphinx’s Tutelage’s numbers remind me of a Nephalia Drownyard activation, and I could see it as a slow-bleed win condition in certain decks.

Managorger Hydra

It looks unassuming, but it triggers off of both players’ spells, and it has trample to boot. A 4/4 trampler for 2G would be Constructed playable, and Managorger Hydra is usually going to be better than that if it comes down early. This card may be a sleeper.

But now the fun part: Suppose you have a Mass Hysteria (or Concordant Crossroads for the old-school players among you) on the battlefield. Moreover, you’ve pulled off some kind of bizarre combo that left you with infinite mana and an infinite number of Managorger Hydras in your hand. In this situation, for how much can you attack after playing the Nth Hydra?

To solve this, we can start by noting that one Hydra attacks for 1 damage and two Hydras attack for 1 + 2 = 3 damage, since the first Hydra triggers off the second. Three Hydras makes for 1 + 2 + 3 = 6 damage. And so on. In general, N Hydras will attack for 1 + 2 + … + N damage—a sum that is more easily expressed as N*(N+1)/2. So, for N=100, your Hydra team will crash in for 5050 damage. These Hydras do grow heads quickly!

Timberpack Wolf et al.

Gotta catch ‘em all! But how many copies of each of these cards can you expect to see in a draft?

Well, if each booster contains 10 different commons drawn from the pool of 101 commons, then the probability of finding one specific common, say Timberpack Wolf, in a booster is 10/101= 0.099. Further, assuming that boosters are independently distributed, the expected number of Timberpack Wolves per draft is 24*10/101=2.38. But expectations don’t tell the whole story. The probability distribution is as follows:

Draft Table Distribution

So although it will be rare, it’s not impossible to have someone draft a deck with 5 or 6 Timberpack Wolves. (And yes, you can put all of them in your deck! The 4-copy maximum rule only applies to Constructed events.)

But how high should you should pick these cards?

Before answering that question, let me ask you a different question. Given that the expected number of Timberpack Wolves per draft is 2.38, how many Timberpack Wolves can you expect in the 23 remaining boosters if you open a Timberpack Wolves in your first pack? Is it 1.38, 2.38, or something else?

The answer to that question is 23*10/101 = 2.28. In other words, if you see a Timberpack Wolf in your first pack, then that observation alone updates the expected number of Wolves in the draft from 2.38 to 1 + 2.28=3.28.

Since 2 Timberpack Wolves in a deck is lackluster but the bonus ramps up when you get your hands on 3 or 4 copies, my draft strategy would be to keep an eye on these “collectable” cards for picks 1-4 or so. I wouldn’t pick them right away, but I’d keep a running count. Then, if I passed at least one Timberpack Wolf in picks 1-4 and see a fresh one in picks 5-8, I’ll probably pick it up and hope that the other one wheels. If it does, then I’ll start picking Timberpack Wolf much more highly in subsequent packs. If it doesn’t, then I’ll probably forget about it. I expect that this strategy will maximize payoff while limiting exposure.

Finally, I’ll move to Constructed, where Faerie Miscreant is an interesting option for Blue Devotion. Could it be better than Gudul Lurker or Hypnotic Siren? To answer that, the number I’m most interested in is the expected number of cards drawn per Faerie Miscreant played after six regular draw steps when playing first. This comes down to ( (1/2) * Prob[2 successes in 13 draws from a 60-card deck containing 4 successes] + (2/3) * Prob[3 successes in 14 draws from a 60-card deck containing 4 successes] + (3/4) * Prob[4 successes in 15 draws from a 60-card deck containing 4 successes] ) / ( 1 – Prob[0 successes in 13 draws from a 60-card deck containing 4 successes] )=0.19 cards.

This is nice, but not terribly exciting. I think the card draw from Faerie Miscreant will come up just about as often as the late-game bonus from Gudul Lurker or ability of Hypnotic Siren to protect Thassa from Dromoka’s Command. Nevertheless, Faerie Miscreant is a valid option, especially in a Collected Company version.