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Applications of mathematics for the design and analysis of games and puzzles
9
votes
Accepted
Finite subsets of $S\subseteq \mathbb{N}$ such that $S\setminus\{s\}$ can be partitioned wit...
As conjectured by bof, the answer is all odd $n \geq 7$.
Proof. Let $S$ be a set of positive integers such that $S \setminus \{s\}$ can be partitioned into two sets of equal sum for all $s \in S$. By …
7
votes
Accepted
Questions about "The best card trick"
The answer to Question 1 is $52!/48! - 4 \binom{13}{2} \binom{50}{2}= 6115200$.
That is, I claim that the number of $4$-tuples that cannot occur is $4 \binom{13}{2} \binom{50}{2}$. To see this, no …
8
votes
Math puzzles for dinner
You have 1000 bottles of wine. Exactly one of the bottles contains a deadly poison, but you don't know which one. The killing time of the poison varies from person to person, but death is imminent i …
4
votes
Math puzzles for dinner
An evil sorceress is holding 100 princes captive. Right now they are all in the same prison cell and can discuss strategy. However, in a moment, they will each be taken to individual prison cells, w …
3
votes
Math puzzles for dinner
Here's one that I like that I just heard a few days ago. Alice and Bob play the following game. Alice is randomly dealt 5 cards from an ordinary deck of cards. She is allowed to show Bob 4 of the 5 …
7
votes
Accepted
Separating Heavier from the Lighter Balls
In the comments, Emil Jeřábek and Terry Tao have given an asymptotic lower bound of $4n / \log_2 n$.
For the upper bound, $2n$ is trivial, obtained by weighing each ball individually. Here is a so …
18
votes
2
answers
4k
views
Turning pants inside-out (or backwards) while tied together
An entertaining topological party trick that I have seen performed is to turn your pants inside-out while having your feet tied together by a piece of string. For a demonstration, check out this vide …
11
votes
Fairest way to choose gifts
Here's an idea. For any partition $(A,B)$ of $[2n]$, where $|A|=|B|=n$, we can ask each child if they prefer $A$ or $B$. If one prefers $A$ and the other prefers $B$, then we are done. Otherwise, t …
3
votes
Escaping from infinitely many pursuers
In the case that the pursuers have to actually catch the fugitive, this was answered in the article Escaping an infinitude of lions by Mikkel Abrahamsen, Jacob Holm, Eva Rotenberg, and Christian Wulff …