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60 votes
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Rock, Paper, Scissors and Trump

Looks like One strategy would be to The opponent will naturally soon realise what's happening. But it won't help. Here are the possible results from Alicia's point of view: So whatever the ...
Bass's user avatar
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52 votes
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Say 100 and win

because:
astralfenix's user avatar
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43 votes
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Any fans of The Big Bang Theory?

The key to solving this question is noticing This is a hint to what the dartboard actually represents: So the scoring is given by
Deusovi's user avatar
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37 votes
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Making a 9 digit number divisible by 11

Note: This answer assumes that the non-zero restriction only holds for the first move, not for any subsequent digits, i.e. that the restriction was imposed only to ensure a valid 9-digit number was ...
Jaap Scherphuis's user avatar
31 votes
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Titanic Tic-Tac-Toe

The first player can always force a win. I wrote this Python script to analyze the tree of possible game states reachable from an empty starting board. No matter what the second player does, the ...
Kevin's user avatar
  • 6,459
29 votes

Finding digits that sum to 15

The solution: The reason:
Deusovi's user avatar
  • 146k
27 votes
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Who would be the first to defeat an abundant number?

Daniel Mathias's user avatar
26 votes
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A game with 330 pebbles

Answer Because
Alessandro's user avatar
26 votes
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Polynomial game with Devil

It is always possible for you to force the polynomial to have the root $-2$: $$ x^2 + (a+2) x + 2a = (x+2)(x+a)$$ Your strategy is to increase your term until it is slightly higher than half the ...
user3294068's user avatar
  • 7,498
26 votes

A Tic-Tac-Toe type game

Another strategy for that works for any odd number of squares:
ffao's user avatar
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24 votes
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Two-Move Chess Game

fblundun's user avatar
  • 1,536
23 votes
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Removing chips from the table

From any multiple of 4 less than 92, any move leaves a number of chips that is not divisible by 4. Then, removing 1, 2, or 3 chips results in another multiple of 4. Therefore, any multiple of 4 less ...
f'''s user avatar
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22 votes
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The maximum number of SETs with six cards

For any two distinct cards, there exists aunique third card that completes the pair into a SET. If I pick three distinct cards $A,B,C$, then I can add three cards $X,Y,Z$ such that ABX and ACY and ...
Gamow's user avatar
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22 votes

Strategy to beat the Casino

By using Joel Rondeau's strategy as a base, plus some kludgy patchwork, we can get a strategy that always gets at least 6 wins. For reference, this is his strategy: His strategy clearly nets at ...
ffao's user avatar
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21 votes
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It'd be on an infinite board, but he can't fit one in his hideout

The winning strategy is and in fact,
Deusovi's user avatar
  • 146k
20 votes

The 100 soldier problem

I'll kick off with some observations. Determining a Nash Equilibrium for such a large solution space is not trivial. So here are some numerical attempts for much simpler problems: For problem a, it ...
Dr Xorile's user avatar
  • 23k
19 votes

A game with 52 cards

An upper bound Bob picks a random permutation of the deck and then rotates the subset of his deck that does not match Alice's. The worst case is when the whole deck is a rotation of his initial ...
Carl Löndahl's user avatar
19 votes
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99 numbers on the blackboard

I believe this is the answer. The strategy is below. Edit: Slightly clearer response with strategy.
cmxu's user avatar
  • 1,016
19 votes
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Noughts and Crosses puzzle

The position is as follows: No two of Eques's counters occupy the same line of 3 (satisfying the never-threatening requirement), and no matter where Knott (O) places their next O, Eques (X) has a ...
Stiv's user avatar
  • 138k
19 votes
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Blindfold Tic-Tac-Toe

You can't do this going second: The strategy going first is: Interestingly,
Deusovi's user avatar
  • 146k
18 votes
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Heaps of marbles

This is how:
Ivo's user avatar
  • 11.2k
17 votes
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Pop the Last Balloon

A strategy is: Because
Ian MacDonald's user avatar
17 votes
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The game of 1036

A player has a winning strategy if and only if They can write $0$, or They can write a number where their opponent does not have a winning strategy. If $N$ is $1036$ (12 divisors),
f'''s user avatar
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17 votes
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Addition hangman

The Guesser can guarantee a win with 4 tokens. Guess $1$. If it's a match, but the solution is still ambiguous, the problem must contain a $3$, $5$, $7$, or $9$ so those can be guessed in order, ...
Miles's user avatar
  • 1,197
16 votes

Rubik's cube two-person game

Answer: Reasoning:
xnor's user avatar
  • 26.6k
16 votes
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A Tic-Tac-Toe type game

I think the answer is that the Strategy
hexomino's user avatar
  • 135k
16 votes
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Fight Battle 21

I'd Then Then So
SteveV's user avatar
  • 15.8k
16 votes
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Masyu-making game

Up to symmetries of the board, there aren't very many possible moves for the first player: Does this strategy work?
Deusovi's user avatar
  • 146k
16 votes

The Game of Barranca

I'll address whether values of N exist such that if the target of the game is N, there is a winning strategy for the second player[.] The answer is First, a lemma: This seems straightforward ...
msh210's user avatar
  • 13.1k
15 votes

Clear board in Othello (Generalisation)

Here is one way you could begin to prove that the board can be cleared for all values of $m$ and $n$. Proof by induction. [incomplete] Case 1: $k = 1$ Here we take k = 1 to mean the smallest ...
Tim's user avatar
  • 1,018

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