MIT Mystery Hunt 2020/Food Court: Difference between revisions

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{{Infobox puzzle
⚫
'''Food Court''' is a [INSERT PUZZLE TYPE] puzzle from the [[Yesterdayland]] round of the [[MIT Mystery Hunt (2020)|2020 MIT Mystery Hunt]]. It [INSERT BRIEF DESCRIPTION]
|title = <!-- Puzzle title, defaults to subpage name, but be sure to include in case of slashes or weird formatting -->
|hunt = <!-- Hunt name, defaults to page parent. Do not link with [[]]; it'll automatically do so -->
|round = Yesterdayland
|image = <!-- Should start with File: -->
|image_width = <!-- include "px", defaults to 200px -->
|image_caption =
|author = Chen Chen
|answer = SEARS
|solves = 44
|guesses = 133
|link = https://puzzles.mit.edu/2020/puzzle/food_court/
|solution_link = https://puzzles.mit.edu/2020/puzzle/food_court/solution/mathful.pdf
|checker_link =
|stats_link =
}}
⚫
'''Food Court''' is a [[List_of_STEM_puzzle_topics#Mathematics|mathematics]] puzzle from the {{l|Yesterdayland}} round of the [[MIT Mystery Hunt 2020|2020 MIT Mystery Hunt]]. It [INSERT BRIEF DESCRIPTION]


==Solve Path==
==Solve Path==
[INSERT SOLVE PATH]


The puzzle body consists of twelve probability-based math problems and two directed graphs, along with some flavortext.
[INSERT PICTURE]

{{spoiler|label=Solution}}

Naturally, the first thing to do is to solve all the math problems, obtaining a probability from each.

{{spoiler|label=Mathematical aside}}

p1 requires knowledge of [https://en.wikipedia.org/wiki/Bertrand%27s_ballot_theorem Bertrand's ballot theorem], resulting in an answer of (15-4)/(15+4) = 11/19.

p2 can be derived into a geometric series with a = 1/8 and r = 1/8, for p2 = 1/7.

p3 involves taking the complement of an event occuring with probability 6*5*4*3/6^4, for an answer of 1-5/18 = 13/18.

p4 can be visualized along a line segment of length L. p4's event occurs if the cut occurs past (8/9)L or before (1/9)L, for a total probability of 2/9.

p5 is just the [https://en.wikipedia.org/wiki/Monty_Hall_problem Monty Hall problem], with a well-known answer of 2/3.

p6 is a conditional probability problem, with answer (1/10)/(1/10 + 2/5 + 3/5) = 1/11.

p7 requires knowledge of the behavior of an [https://en.wikipedia.org/wiki/M/M/1_queue M/M/1 queue]. The equation in question results in ρ = 1/2 and π2 = 1/8.

p8 can be simulated as a Markov chain, but the result can also be found [https://en.wikipedia.org/wiki/Gambler%27s_ruin#Fair_coin_flipping here].

p9 is a Nash equilibrium problem.

p10 is an [https://en.wikipedia.org/wiki/Branching_process#Example_of_extinction_problem extinction problem], a kind of branching process.

p11 is another conditional probability problem with somewhat different framing.

p12 is a simple example of [https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle the inclusion-exclusion principle].

{{spoiler-end}}

Having obtained all the probabilities, the solver can now turn their attention to the graphs. These can be identified as [https://en.wikipedia.org/wiki/Markov_chain Markov chains]; after plugging in the probabilities, the solver can now solve for the steady state in said Markov chains.

Each of the probabilities in said steady state can be exactly expressed with two decimal digits of precision (in other words, they are integer multiples of 1/100); moreover, they all land between 0.01 and 0.26. Converting to letters yields a cluephrase that yields the answer.

{{spoiler-end}}

[[Category: Pages that need images]]
==Puzzle Elements==
==Puzzle Elements==


{{spoiler|label=Spoiler-y Elements}}
* [INSERT ELEMENTS]
{{Element|Mathematics|The first part of the puzzle body is twelve probability problems, each one requiring a different set of theorems from probability. The resulting numbers then have to be plugged into a stochastic process, and the steady state of said stochastic process derived.|Knowledge Required||[[List_of_STEM_puzzle_topics#Mathematics|Mathematics]], Probability}}

{{Element|Marked Elements|Every node on the Markov chain is marked with a distinct number between 1 and 20, which serves as an ordering.}}

{{Element|Alphanumeric Substitution Cipher|The resulting probabilities on each node can be run through the A1Z26 cipher for a letter.}}

{{Element|Final Cluephrase|This yields CHARS SURFACE OF A STEAK, which is a crossword clue pointing to the answer.}}
{{spoiler-end}}

Latest revision as of 15:16, 10 March 2023

Food Court
MIT Mystery Hunt 2020
Yesterdayland
Author(s)Chen Chen
AnswerClick to revealSEARS
Statistics
No. solves44
No. total guesses133
Links
PuzzleLink
SolutionLink

Food Court is a mathematics puzzle from the Yesterdayland round of the 2020 MIT Mystery Hunt. It [INSERT BRIEF DESCRIPTION]

Solve Path[edit | edit source]

The puzzle body consists of twelve probability-based math problems and two directed graphs, along with some flavortext.

 

Naturally, the first thing to do is to solve all the math problems, obtaining a probability from each.

 

p1 requires knowledge of Bertrand's ballot theorem, resulting in an answer of (15-4)/(15+4) = 11/19.

p2 can be derived into a geometric series with a = 1/8 and r = 1/8, for p2 = 1/7.

p3 involves taking the complement of an event occuring with probability 6*5*4*3/6^4, for an answer of 1-5/18 = 13/18.

p4 can be visualized along a line segment of length L. p4's event occurs if the cut occurs past (8/9)L or before (1/9)L, for a total probability of 2/9.

p5 is just the Monty Hall problem, with a well-known answer of 2/3.

p6 is a conditional probability problem, with answer (1/10)/(1/10 + 2/5 + 3/5) = 1/11.

p7 requires knowledge of the behavior of an M/M/1 queue. The equation in question results in ρ = 1/2 and π2 = 1/8.

p8 can be simulated as a Markov chain, but the result can also be found here.

p9 is a Nash equilibrium problem.

p10 is an extinction problem, a kind of branching process.

p11 is another conditional probability problem with somewhat different framing.

p12 is a simple example of the inclusion-exclusion principle.

Having obtained all the probabilities, the solver can now turn their attention to the graphs. These can be identified as Markov chains; after plugging in the probabilities, the solver can now solve for the steady state in said Markov chains.

Each of the probabilities in said steady state can be exactly expressed with two decimal digits of precision (in other words, they are integer multiples of 1/100); moreover, they all land between 0.01 and 0.26. Converting to letters yields a cluephrase that yields the answer.

Puzzle Elements[edit | edit source]

 

Knowledge Required (Mathematics, Probability) - The first part of the puzzle body is twelve probability problems, each one requiring a different set of theorems from probability. The resulting numbers then have to be plugged into a stochastic process, and the steady state of said stochastic process derived.

Marked Elements - Every node on the Markov chain is marked with a distinct number between 1 and 20, which serves as an ordering.

Alphanumeric Substitution Cipher - The resulting probabilities on each node can be run through the A1Z26 cipher for a letter.

Final Cluephrase - This yields CHARS SURFACE OF A STEAK, which is a crossword clue pointing to the answer.

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