Hot answers tagged

95 votes

Chasing pirates

Stay put for about 45 days, after which the pirates would have circumnavigated the globe and returned to your current position.
user avatar
  • 1,044
78 votes
Accepted

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 ...
user avatar
62 votes

Picking up stones

That's easy
user avatar
  • 2,586
60 votes
Accepted

The lion and the zebras

Here is the strategy:   It works because:
user avatar
  • 7,678
59 votes
Accepted

Coin Flipping Game with the Devil

Satan should stick to fiddling. You will win, and here is a simple proof. Consider the game $n$ turns at a time. After each cycle of $n$ turns, all the coins are in their original position (though ...
user avatar
  • 7,678
58 votes
Accepted

Left coin, right coin, last coin?

It doesn't matter which option you choose, because Your probability of survival if you're one of n players left is as follows: Informal proof It was established in the question that if there are ...
user avatar
58 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 ...
user avatar
  • 69.3k
56 votes
Accepted

Chasing pirates

If we assume the ocean is flat and extends indefinitely in all directions, there is a strategy that guarantees we can catch the pirates in at most 800,000 years. Put our current location as the ...
user avatar
  • 13.9k
53 votes
Accepted

Coins on a table

The winning strategy for the first player is to put their coin in the dead center of the table. Then whatever move their opponent makes, they exactly mirror it, around the center. e.g. If the second ...
user avatar
  • 2,329
52 votes
Accepted

Say 100 and win

because:
user avatar
  • 2,819
41 votes

Left coin, right coin, last coin?

The answers of rand al'thor and Callidus are great; I just want to give a different argument for the result. Claim: After each round, the number of surviving players is even. Proof: Let $f_i$ be the ...
user avatar
  • 23.7k
38 votes
Accepted

A lonely pawn on the chessboard

Strategy: How this works:
user avatar
35 votes
Accepted

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 ...
user avatar
30 votes
Accepted

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 ...
user avatar
  • 6,329
28 votes

Finding digits that sum to 15

The solution: The reason:
user avatar
  • 139k
27 votes
Accepted

Rook game on chessboard

I will assume that the square in the lower left is painted red as well. Note that an $m\times n$ board is the same as an $n \times m$ board. So WLOG we will assume that $n \le m$. When $n=m=1$, ...
user avatar
  • 8,841
26 votes
Accepted

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 $2^i$...
user avatar
  • 31.5k
26 votes
Accepted

A game with 330 pebbles

Answer Because
user avatar
26 votes
Accepted

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 ...
user avatar
  • 7,333
26 votes

A Tic-Tac-Toe type game

Another strategy for that works for any odd number of squares:
user avatar
  • 21.4k
25 votes
Accepted

Knight on a 5 by 5 board

As shown below Alice gives the red moves and Bob gives blue moves... Moves 2 and 10 have symmetrical options which end up with same result. All other moves are forced.
user avatar
25 votes
Accepted

Face Up Poker with Alice and Bob

There is a winning strategy for
user avatar
  • 13.9k
24 votes
Accepted

Who would be the first to defeat an abundant number?

user avatar
23 votes
Accepted

Guess my number! Pay for the answer!

The answer is that only x=11 yields a fair game. I solved this using a recurrence. Let $f[N]$ be optimal expected amount of money that Bob has to pay Alice to find the best answer. Clearly, $f[...
user avatar
  • 2,438
23 votes
Accepted

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 ...
user avatar
  • 33.3k
22 votes

Chasing pirates

Drive 20 hours in a direction we will denote with as having $\theta=0$. Drive in a spiral pattern such that you are $20t+20$ nautical miles away from you starting position. $t$ is time in hours from ...
user avatar
  • 9,012
22 votes
Accepted

20 coins on the table

Always leave a multiple of 4 coins. So if Brian takes 1, Gabe takes 3. 2-2, 3-1. Starting with a multiple of 4, player 2 can always win with perfect play. Starting with any other number, player 1 can ...
user avatar
  • 7,300
22 votes

Knight on a 5 by 5 board

The winner is: The winning strategy is to
user avatar
  • 23.7k
22 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 ...
user avatar
  • 22.3k
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 ...
user avatar
  • 44.8k

Only top scored, non community-wiki answers of a minimum length are eligible