Ex Situ Redux: Study Shows The Princess is in Another Castle 7/8ths of the Time

December 2, 2009

[Editor's Note: This is a republish of an old Ex Situ which was part of a set, but I thought with my addendum that it deserved its own article.]

Yonder at The Minus World, there was a cleverly-conceived, confusingly both underwritten and overwritten, and woefully trying-too-hard-to-be-funny report about the frequency of Princess actualization. The original link seems to be dead, but here is a reprint:

New Haven, Connecticut – Profoundly sidetracked Yale scholars have been working feverishly to determe the statistical accuracy of finding the hypothetically kidnapped Mushroom Kingdom Princess in a castle. It has been agreed that 87.5% of the time, prospective rescuers will not happen upon the Princess and will instead hold a brief conversation with a fungus man who will inform them vaguely that her whereabouts are elsewhere. While probability and numerical accuracy have been officially cemented, researchers remain consistently baffled as to why she keeps getting yanked to begin with, or if the bitch is even worth it at this point.

Addendum: The calculation to find the Princess Probability isn’t even accurate. Taking into account the Warp Zones, there are  many different paths Mario might take during the course of the game, each with a different total probability of the Princess being in a castle, hereafter referred to as P(Princess).

In World 1-2, there is a Warp Zone to Worlds 2-1, 3-1, and 4-1. In World 4-2, there are two Warp Zones, to 5-1, 6-1, 7-1, and 8-1 in toto. So let’s break the Mushroom Kingdom into two sets: A and B, where each possible path to get to World 4-2 is in A and each possible path to get from World 4-2 to the end of the game is in B.

Here is a chart enumerating each path in A, where an “X” indicates a completed castle, NC is the total number of completed castles for that path, and NP is the total number of Princesses being in a castle for that path:

Path World 1 World 2 World 3 NC NP Notes
A_1 X X X 3 0
A_2 X X 2 0 Warp from 1-2 to 2-1
A_3 X 1 0 Warp from 1-2 to 3-1
A_4 0 0 Warp from 1-2 to 4-1

And a similar chart for set B:

Path World 4 World 5 World 6 World 7 World 8 NC NP Notes
B_1 X X X X X 5 1
B_2 X X X X 4 1 Warp from 4-2 to 5-1
B_3 X X X 3 1 Warp from 4-2 to 6-1
B_4 X X 2 1 Warp from 4-2 to 7-1
B_5 X 1 1 Warp from 4-2 to 8-1

So the total number of paths possible is:

\displaystyle \sum_{i, j} A_i+B_j=20

(which really is just A\times B=20)

To calculate the probability of The Princess being in another castle, P(AnotherCastle), we can calculate P(Princess) and subtract it from 1 to give us:

P(AnotherCastle) = P(\overline{Princess}) = 1-P(Princess)

P(Princess) is equal to the number of Princesses being in a castle in all possible paths, divided by the number of castles in all possible paths:

\displaystyle P(Princess)=\frac{\sum_{i, j} NP(A_i+B_j)}{\sum_{i, j} NC(A_i+B_j)}=\dfrac{20}{90}=\dfrac{2}{9}

(Note that \sum_{i, j} NP(A_i+B_j)=20, since each possible total path A+B results in 1 Princess being in a castle.)

So instead of  P(AnotherCastle)=\frac{7}{8}, the correct calculation is:

P(AnotherCastle)=1-\dfrac{2}{9}=\dfrac{7}{9}

Furthermore, since each path has a \frac{1}{20} probability of being taken and the total number of instances of a Princess being in another castle summed over all paths is 90-20=70, we can calculate the expected values for the number of times the Princess is in another castle and the number of Princesses being in a castle:

E(AnotherCastle)=\dfrac{70}{20}=3.5

E(Princess)=\dfrac{20}{20}=1

To summarize: on an average full game of Super Mario Bros., one should expect:

  • The Princess to be in another castle about 78% of the time
  • The Princess to be in another castle 3.5 times
  • The Princess to be in a castle 1 time

We here at J. Cart. Overanal. feel this probabilistic description is far superior, assuming we didn’t screw up the math. Comments or complaints about wonky symbol useage are welcome below.

7 Comments | ex situ, Super Mario Bros. | Tagged: , , | Permalink
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Ex Sitibus: A Super Mario Triptych

March 24, 2009

mario_bar

Just to temper our postmodern ennui, we’re shaking things up a jot here at J. Cart. Overanal. Below we have three over-analyzationy links concerning Mario, who despite being a workaday fungal-plumber is really quite the Homo Universalis.

  • One: a brief, unwieldily image-based thesis on the symbology of Super Mario Bros., including a blunt, presumptive explanation for the meaning of life. Just click on the image to expand it legibly. (The true origins of this thesis have defied our gumshoeing; if anyone knows we’ll link to it properly.)
    mariodeep
  • Two: another masterpiece by the quasi-intellectuals at McSweeney’s, in the approximate vein of these other ones. Here we have an editorial by Dr. Mario himself, lambasting the Mushroom Kingdom’s corrupt and inefficient health care system.
    Dr. Mario Weighs in on Universal Health Care > Catena Ex Situ
  • Three: yonder at The Minus World, we have a cleverly-conceived, confusingly both underwritten and overwritten, and woefully trying-too-hard-to-be-funny report about the frequency of Princess actualization.
    Study Shows The Princess is in Another Castle 7/8ths of the Time     > Catena Ex Situ

Addendum: The calculation to find the Princess Probability isn’t even accurate. Taking into account the Warp Zones, there are  many different paths Mario might take during the course of the game, each with a different total probability of the Princess being in a castle, hereafter referred to as P(Princess).

In World 1-2, there is a Warp Zone to Worlds 2-1, 3-1, and 4-1. In World 4-2, there are two Warp Zones, to 5-1, 6-1, 7-1, and 8-1 in toto. So let’s break the Mushroom Kingdom into two sets: A and B, where each possible path to get to World 4-2 is in A and each possible path to get from World 4-2 to the end of the game is in B.

Here is a chart enumerating each path in A, where an “X” indicates a completed castle, NC is the total number of completed castles for that path, and NP is the total number of Princesses being in a castle for that path:

Path World 1 World 2 World 3 NC NP Notes
A_1 X X X 3 0
A_2 X X 2 0 Warp from 1-2 to 2-1
A_3 X 1 0 Warp from 1-2 to 3-1
A_4 0 0 Warp from 1-2 to 4-1

And a similar chart for set B:

Path World 4 World 5 World 6 World 7 World 8 NC NP Notes
B_1 X X X X X 5 1
B_2 X X X X 4 1 Warp from 4-2 to 5-1
B_3 X X X 3 1 Warp from 4-2 to 6-1
B_4 X X 2 1 Warp from 4-2 to 7-1
B_5 X 1 1 Warp from 4-2 to 8-1

So the total number of paths possible is:

\displaystyle \sum_{i, j} A_i+B_j=20

(which really is just A\times B=20)

To calculate the probability of The Princess being in another castle, P(AnotherCastle), we can calculate P(Princess) and subtract it from 1 to give us:

P(AnotherCastle) = P(\overline{Princess}) = 1-P(Princess)

P(Princess) is equal to the number of Princesses being in a castle in all possible paths, divided by the number of castles in all possible paths:

\displaystyle P(Princess)=\frac{\sum_{i, j} NP(A_i+B_j)}{\sum_{i, j} NC(A_i+B_j)}=\dfrac{20}{90}=\dfrac{2}{9}

(Note that \sum_{i, j} NP(A_i+B_j)=20, since each possible total path A+B results in 1 Princess being in a castle.)

So instead of  P(AnotherCastle)=\frac{7}{8}, the correct calculation is:

P(AnotherCastle)=1-\dfrac{2}{9}=\dfrac{7}{9}

Furthermore, since each path has a \frac{1}{20} probability of being taken, we can calculate the expected values for the number of times the Princess is in another castle and the number of Princesses being in a castle:

E(AnotherCastle)=\dfrac{70}{20}=3.5

E(Princess)=\dfrac{20}{20}=1

To summarize: on an average full game of Super Mario Bros., one should expect:

  • The Princess to be in another castle about 78% of the time
  • The Princess to be in another castle 3.5 times
  • The Princess to be in a castle 1 time

We here at J. Cart. Overanal. feel this probabilistic description is far superior, assuming we didn’t screw up the math. Comments or complaints about wonky symbol useage are welcome below.

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1 Comment | Dr. Mario, ex situ, philosophy, Super Mario Bros. | Tagged: , , , , , , | Permalink
Posted by The Editor

Ex Situ: The “Art” of Seth McFarlane

October 2, 2008

Seth McFarlane’s intermittently amusing Family Guy is a bit of a controversy around the offices here. On one hand, it constantly uses repetitious non sequitur shock gags to fool semi-stoned brains into thinking they are watching something funny. On the other hand, it is terribly, lazily animated. A Manichean dilemma!

Via Cartoon Brew, we found this nicely thorough piece by Kyle Evans analyzing the style and content of Family Guy, with extra emphasis towards Mr. McFarlane’s Cavalcade of Comedy‘s Super Mario Bros. parody. It’s like Mr. Evans was able to extract a beautiful geode of pure reason from our collective consciousness. (i.e. we agree with everything he says) Here’s a vivisection:

Also exactly the same is the character designs, with Mario looking like a cross between Brian and Peter, while Princess Peach looks like Lois in a pink dress. Apparently Seth never learned to draw from any other perspective than a three quarter front on. This insistence on keeping all characters at this angle creates an offputting effect when you have characters conversing – which is the majority of Seth’s work; endless, mind numbing conversations. Having two characters side by side on a slight angle talking to one another creates this bizarre effect as though they’re staring just past one another – there’s absolutely no sense that these characters are truly engaged in conversation. It doesn’t help that you hardly ever see a non-talking character animated or that these conversations are carried entirely by the moving of the lips.

The full article has diagrams and audio-visual aids. Interesting, even if you disagree.

The “Art” of Seth McFarlane > Catena Ex Situ

2 Comments | American Dad, ex situ, Family Guy, Super Mario Bros. | Tagged: , , , | Permalink
Posted by The Editor

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