79
votes
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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 ...
- 5,876
59
votes
Accepted
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 ...
- 74.1k
52
votes
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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
- 145k
38
votes
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38
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 ...
- 50.3k
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 ...
- 6,449
28
votes
26
votes
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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 $2^i$...
- 32k
26
votes
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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 ...
- 7,428
26
votes
26
votes
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23
votes
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 ...
- 22.7k
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 ...
- 33.6k
22
votes
Accepted
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 ...
- 45.3k
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 ...
- 21.6k
21
votes
Accepted
It'd be on an infinite board, but he can't fit one in his hideout
The winning strategy is
and in fact,
- 145k
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 ...
- 22.7k
20
votes
Accepted
99 numbers on the blackboard
I believe this is the answer.
The strategy is below.
Edit: Slightly clearer response with strategy.
- 1,026
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 ...
- 5,260
19
votes
Accepted
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 ...
- 129k
18
votes
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 ...
18
votes
Accepted
18
votes
Accepted
Blindfold Tic-Tac-Toe
You can't do this going second:
The strategy going first is:
Interestingly,
- 145k
17
votes
Accepted
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?
- 18k
17
votes
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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),
- 33.6k
17
votes
Accepted
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, ...
- 1,187
16
votes
Only top scored, non community-wiki answers of a minimum length are eligible