# Tag Info

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### How to beat Count Dracula

The lockets and coffins are always found in loops. open a coffin look at the number of the locket in the coffin go to the coffin with the same number as the locket open the coffin repeat from 2 At ...
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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 ...
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because:
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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
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### A lonely pawn on the chessboard

Strategy: How this works:
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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 ...
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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 ...

### Finding digits that sum to 15

The solution: The reason:
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### Blackboard cleaning

Dr Xorile's answer of is optimal. Suppose that the numbers 1, 2, 4, 8, ... 512, and 1024 are all on the board. In order to eliminate the number $2^i$, there must be some step where you decrease $2^i$...
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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 ...

### A Tic-Tac-Toe type game

Another strategy for that works for any odd number of squares:
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### Blackboard cleaning

It can be done in: "But how?" you ask. Well, I'm glad you asked: Convert the numbers to binary, and subtract the bits. Or, to put it another way, subtract 1024 from all the numbers that can ...
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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 ...
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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 ...

### 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 ...
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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,

### 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 ...
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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.

### 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 ...
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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 ...

### How to beat Count Dracula

Source: TheDarkTruth's answer There are 3 parts to this answer: The algorithm to find the locket marked with number N(1) Ignoring the fact that he has to stop after 500 coffins, this is the ...
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This is how:
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### Blindfold Tic-Tac-Toe

You can't do this going second: The strategy going first is: Interestingly,
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### Ninety-nine non-negative numbers

TL;DR The rest of the post describes how I came to the conclusion above. Closer look Changing the rules Winning position How to win?
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### Pop the Last Balloon

A strategy is: Because
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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),
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, ...