I would like to take a moment to consider an unanswered question in Magic. Consider the following gamestate:
I have an arbitrarily large amount of storm - say 1,000,000 (Tidespout Tyrant + 2 mox, or Salvager + lotus). Now I cast brainfreeze on my opponent, and discover they have Gaea's Blessing in thier deck.
- If we were to 'play it out' each time a copy of brainfreeze resolves that mills the Gaea's blessing, it will immediately trigger on top of the other Brainfreeze copies. So it will shuffle Mid-stack, and you will continue to resolve brainfreezes on the newly complete and reshuffled library (minus cards in play, hand, and RFG). As millions of copies of brainfreeze resolve, gaea's blessing will trigger hundreds of thousands of times.
- Assuming we continue to play out the million copies, at some point we will approach the "End State." The Graveyard will contain either 0, 3, 6, 9, ... , up to the decks greatest multiple of 3. Because there is a chance that the final brainfreeze will resolve and include Gaea's Blessing, and there is also a chance that the final brainfreeze will reveal 3 cards, with brainfreeze remaining in a tiny deck.
-- The person with brainfreeze is generally hoping his opponent ends on a small deck.
-- The person with Gaea's blessing is generally hoping to have a big remaining deck
--- In both cases the cards left in the deck are of interest to both players.
So the purpose here, is to take a moment and apply some basic Probability and Statistics theories to this example. The goal is to create a "Comparable Probability Model" to shortcut the End State.
Let’s start with Definitions - these are going to be important if you intend to follow my logic.
D = Number of cards in the deck + graveyard at the beginning of the loop. Often I refer to the "Deck" below, and really I'm talking about the remaining deck (Registered deck minus cards in play, hand, and RFG).
D0 = Number of cards in the LIBRARY when brainfreeze is actually cast.
D' = Number of cards left in the Deck when the final brainfreeze resolves. In essence: THE END STATE!
Event G = hitting a geae's blessing
Event G' = not hitting geae's blessing. In chains I'll denote this as "-" So for example the chain -G---G would be 1 miss, 1 blessing, then 3 misses and a blessing.
G and G' are mutually exclusive and exhaustive... because in 3 cards you either hit blessing or do not hit blessing.
I = the total number of brainfreezes to resolve
i = the number of brainfreezes that have resolved.
x = the number of Events G' since the last event G
Gx = hitting a geae's blessing on the x-th brainfreeze from the last time event G occurred.
A High level look at Dependence and Independence:
Prob (Gx) = f(D, D0, I)
* the probability of G is a function of I and D
Once the first Blessing is hit ...
Prob (G | X=x) is independent of i for all x
Using the variables:
Consider a deck of 6 cards, with 4 cards in the yard.
D = 10
D0 = 6
There are essentially 4 possible ends states:
D' could be: 10, 7, 4, 1
Consider I = 3
For i = 1, Prob(G) = Prob(G') = .5
Meaning of the 6 cards remaining in the deck Blessing is either in the top 1/2 or the bottom 1/2.
For i = 2, Prob(G) is dependant on the outcome of i=1.
Prob(G | i -> G) = Prob(G|x=0) = 0.300 = 3/10
Prob(G | i -> X) = 100%
For i = 3 Prob(G is dependant on X)
Prob(G | x=0) = 0.300
Prob(G | x=1) = 0.428 = 3/7
Possible Paths: End State
GGG: 10 -> 3/6 * 3/10 * 3/10
GG-: 7 -> 3/6 * 3/10 * 7/10
G-G: 10 -> 3/6 * 7/10 * 3/7
G--: 4 -> 3/6 * 7/10 * 4/7
-GG: 10 -> 3/6 * 1/1 * 3/10
-G-: 7 -> 3/6 * 1/1 * 7/10
Possible End state Probability Distribution with
Prob(D' | D0 = 4, D = 10, I = 3)
10 = 0.345 = 69/200
7 = 0.455 = 91/200
4 = 0.200 = 40/200
1 = 0
So what does this tell us? It says that in the given size 10 deck situation with 3 brainfreezes that
after you resolve all three brainfreezes you a 34.5% chance to end with all 10 cards in your library, a 45.5% chance to end with 3 cards in your yard and 7 cards (including blessing) in your library, and a 20% chance to have 6 cards in your graveyard, and 4 cards (including blessing) in your library. There is no possible way to end with brainfreeze being the last card of your library.
Here is an interesting observation. Lets examine the same scenario (D0=4, D=10) but with I = 4...
Prob(D' |I=4) = Prob(D'=d|I=3) * Prob(G |x=X)
Ok, the math syntax isn't 100% right on that, but the idea is that if you start with each D' of I = 3, you can multiply by the appropriate probability of G or G' to generate the I=4.
Lets take a look:
d -> Prob(D'=d|I=4) = ...
Endstate: Prob -> derivation
7: Prob(D'=7|I=4) -> Prob(G'|x=0) * Prob(D'=10|I=3)
4: Prob(D'=4|I=4) -> Prob(G'|x=1) * Prob(D'=7|I=3)
1: Prob(D'=1|I=4) -> Prob(G'|x=2) * Prob(D'=4|I=3)
10: Prob(D'=10|I=4) ->
= Prob(G|x=0) * Prob(D'=10|I=3)
+ Prob(G|x=0) * Prob(D'=7|I=3)
+ Prob(G|x=0) * Prob(D'=4|I=3)
An Alternate way to define D':
The exact position of blessing in the deck after each shuffle is uniform... there is equal chances that it will be the first card, the last card, or any specific card in between.
Lets look at new random veriable A. Defined as the number of cards flipped before you hit blessing. If D Mod 3 is 0, then the probability is uniform; but more generally The probability of A=a is a composite uniform distribution. For example in a 61 card deck, its uniform from 1 to 20 at prob(A=a) = 3/61. The final prob(A=21)=1/61.
Lets also consider the r.v. N ~ the number of times you hit gaea's blessing over the course of I.
Using these rv's we can express D' ~ f(N,A). I'm not sure that really helps us though... I think it could help us solve the expected value of D' - but that's not really useful to us since we are after a probability function not the moments.
So anyway, Lets take a look at some theory:
- If I > D'/3 then you -will- Blessing at least once.
- Once you blessing once, D0 really matter for future probabilities of G ... but it still could matter for the final result D'.
Deeper Dive into Probability:
Consider you are asked to bet a $1 on an arbitrary brainfreeze; say brainfreeze # 1000. No brainfreezes have yet resolved, and You are betting on the cards left in a 60 card deck after Brainfreeze #1000 is done. So your choices are number 60, 57, 54, 51, ... , all the way down to 3. What number do you bet on (assuming you win the same amount for each winning bet)?
The answer is 60. Essentially you are betting that brainfreeze #1000 -WILL- have a brainfreeze in it. Why is that? The reasoning is a bit hard for me to explain but here goes:
Think of the entire string of brainfreezes as a time series, assuming each brainfreeze is evenly spaced in time. you want to think about how much time the deck spends in each given state. So suppose you have the following 25 events in your serise:
---G- G---- ---G- ----- --GG-
*I broke it up by 5's just so it's easier to follow...
So consider the state of the deck:
60,57,54,60,57; 60,57,54,51,48; 45,41,38,36,33; 30,27,60,60,57
So the only way to get to 57 is to have a sequence that goes G- The only way to get to 54 is G-- and the only way to get to 48 is G----
and for every time you pass though to a number less than 60 you spend _exactly_ 1 interval at the number before it. 60 is a little different Every time you restart the deck you have to spend at least one interval of time at 60. For each time the sequence goes "GG" you are spending essentially double time at 60. Here is were you gain your advantage by betting on 60. For every unit of time you spend at 57 you spend one unit at 60 as well... however it _is_ possible to go from 60 to 60. And the small probability of "GG" is why you bet on 60.
Ok, same problem but now you cannot bet on 60. What do you bet on now? the answer is 57... by much the same logic. 57 will be reached more than 54 because of all the times you hit the sequence "G-G" You spend one unit of time at 57, and no time at 54. But every sequence that spends time at 54 will also spend 1 unit of time at 57.
This "time spent" concept is what drives your Bet. If you know its going to spend more time at a given place you should bet your $1 there.
So what does all this betting have to do with the problem? It is saying that the probability that D'=60 is greater than that of D'=57, which is in turn greater than that of D'=54... etc. *please note that this is NOT saying you will "always end with a full deck" its not even saying you will "Likely" end with a full deck (simulation shows that its around 10%). But what its saying is that the probability of ending on any selected D' will be smaller as D' decreases.
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So to restate purpose here: I want to examine possible comparable probability shortcuts to fast-forward (so to speak) through an arbitrarily large brainfreeze to the endstate. Because lets face it... taking into account complete deck composition there are practically countless possible endstates... some of which are favorable for the caster of brainfreeze (ei 21 cards in the yard which contain all the win conditions, will, and not blessing), some of which are favorable to the owner of blessing (ie 45 card in gy with all sorts of nasty mana and stuff but will and tutors still remain in deck), and many of which grant no particular advantage to either player.
Possible Models that I can think of ... none of which have probabilities associated:
Today) "The F-it Plan"
From what I understand the current method is just to shuffle up the geae's blessing into a finished deck and say that's the end state. This is supported by the "highest probability" of that to occur, but It basically provides no possibility of advantage to the person casting brainfreeze. In essence it makes Gaea's blessing a better card than it should be... because the brainfreeze player literally has zero chance to affecting you.
A) "Direct Endstate Probability Matrix"
Step 1 - Roll a D100 and compare result to a table based on size of Deck after the first blessing resolves. This will determine the final end state. So the table would look something like:
D100 -> Result
1-11 -> 0 cards in GY.
12-20 -> 3 cards in GY.
21-28 -> 6 cards in GY.
...
98-99 -> all but 6 cards in GY.
100 -> all but 3 cards in GY.
Then the strategy would be to remove the blessing from the deck. Shuffle it up, and mill the correct number of cards given the result. The shuffle the remaining cards in the library PLUS THE BLESSING together.
*limitation: tables of probability would have to be calculated for every possible # of cards left in the deck... because obviously the table for 10 cards remaining would be much different from 52 cards remaining. There may be some opportunity to define tables for given ranges. Also Judges would have to have access to these tables readily
*limitation#2: D100 only allows for a 1% variation... this would probably have to be a D1000 (3D10 of different color that represent defined digits) the D1000 would allow the tables to split on 0.1%.
*Advantages: You cut right to the end. And it would take little more than the time required for the judge to look up the correct table and roll some dice.
B) "Fast Forward to the Tail"
Lets say we prove that Storm = 10,000 is statistically the same as storm = 1,000. What we would have to do is back into the Storm require to reach a 'significantly equal' probability model. So lets say we can say ... with it 0.05% storm 28 is the same as all storm values larger than 28. So it would be fairly easy...
Step 1 - allow the first part of brainfreeze to resolve, and shuffle up the deck.
Step 2 - take the fully shuffled deck and flash forward to the last 28 copies of brainfreeze... and assume that your end states going forward will fit the distributions of the "tail end" of an arbitrarily large storm.
Step 3 - resolve the final 28 copies as you would if you got hit for a storm 28 brainfreeze... and empty the stack. If you hit blessing, continue to shuffle and resolve copies.
*Limitation - I can't even begin to imagine what this number will be, but I would guess its probably around 40 to 50...possible up around 100. So really, in the constraints of a 50 min round this may not be a 'short cut' at all. You still take lots of time, and also risk doing damage to cards because of multiple shuffling
*Advantages - It requires no tables ... just a magic number that applies to any deck with less than 60 cards left in deck.
C) "Hybridized Probability Matrix"
Basically generate a near tail-end probability table that would (using conditional probability) be similar to the end state probability. So something like:
d6 -> Tail End resolutions
1 -> 2
2 -> 4
3 -> 6
4 -> 8
5 -> 10
6 -> 12
Step 1 - resolve the first blessing and shuffle the deck.
Step 2 - roll a d6 an get the tailend resolution number.
Step 3 - with the shuffled deck including gaea's blessing, resolve that number of brainfreezes. So for example if you roll a 4, you simply resolve 8 additional brainfreezes, and then consider the stack empty. If you encounter blessing during these tail end resolutions resolve them normally.
Limitations: It may not even be possible to create this type of model. The complexity here could be enormous. On top of that proving the two models to be insignificantly different would be equally daunting. As a judge you would still need to refer to a table (unless its simply like # times 2 as illustrated above).
Advantages: This method shortcuts method-B because at most you have to resolve 12 brainfreezes, and typically you would end up resolving less. So its a time saver over method-B and not as "have faith in mathematics" as method-C does.
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Currently I have built excel based simulation tool which alows me to run thousands of trials while controlling each variable. So if my probability skills are not up to the task - maybe my statistics skill can help back into the probability distribution.
So far the simulations I've run support my claims - but no real numbers or percentages to report.
I'd love to hear more from some mathematicians, statisticians, and Tournement Judges on this!