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

[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.

This entry was posted on Wednesday, December 2nd, 2009 at 6:56 am and is filed under ex situ, Super Mario Bros.. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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

GamesOgre says:

January 11, 2010 at 4:20 pm

While we’re at it let’s figure out (1) where Mario stores all his coins and (2) what the probability is of a princess hooking up with a plumber. Ahh the wonders of reto gaming.

Reply
Kanna-Chan says:

February 11, 2010 at 12:12 am

‘blank look’ … wha?

Reply
Diana says:

February 17, 2010 at 2:31 am

And we thought things like Calculus and Statistical Analysis and Differential Equations would never come in handy… I deem this a mathematical masterpiece.

Reply
Chris says:

March 31, 2011 at 7:52 am

You have, it should be pointed out, assumed that each path will be taken with equal probability. I’m not certain that is true.

A Bayesian prior of some description, perhaps?

Reply
Uh says:

August 14, 2011 at 3:29 pm

My Math Logic:

There are 8 castles. Peach is in one of them. So, the castle you go to, there is a 1/8 chance she is in it.

That means 7/8 of the time, she is in a different castle.

Done.

Reply
mario e sonic says:

October 1, 2011 at 7:46 pm

mario e mario…

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free rider 2 tracks says:

October 12, 2011 at 12:21 am

free rider 2 tracks…

[...]Ex Situ Redux: Study Shows The Princess is in Another Castle 7/8ths of the Time « The Journal of Cartoon Overanalyzations[...]…

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