16
votes
11
votes
Making a 2n-digit number divisible by 9
I'll add on to SQLnoob's answer and say Bob can't win if n isn't
So she chooses
$$
m = \begin{cases}
7-k & k < 7 \\
1 & k = 7, 8
\end{cases}
$$
For example,
- 10.1k
9
votes
Moving a knight to a new square each turn: who wins?
Here is a solution without the 4x2 grids (although it is a generalization of that solution).
Pair up the squares such that the two squares in each pair are separated by a knight's move.
On each of Bob'...
- 3,674
8
votes
Accepted
Paper, pencil and a bunch of bars
This game is
So in this case
If we take "rational" to mean that when either player had a winning move they took it then
Now
So
- 121k
8
votes
Accepted
8
votes
Accepted
Best strategy for stick-taking game
Pretty much all Nim games can be solved by starting at the end and working backwards.
Let's first solve the basic Nim, with only one pile, and simple actions (take 1-4). Let's enumerate the endgame ...
- 78.6k
7
votes
Accepted
7
votes
Accepted
A pile of chips involving primes
Here's a little Python program to test it yourself:
https://repl.it/repls/ScientificIdenticalPixels
And here's C++ code written by user @im_so_meta_even_this_acronym
https://ideone.com/SfMHqC
- 2,377
6
votes
Single-pile Nim with Three Players
Even though the question says I'm Bob, I'd like to start by stating my intention to be the Quetzalcoatlus, thanks very much.
I believe that the key to this game is that
Example game:
However, ...
- 32.3k
6
votes
Single-pile Nim with Three Players
Partial strategy up to 11 candies. A is the player whose turn is next, B is the next player, and C is last.
Don't think I can do it for 40 without turning this into a novel. But I have to say that, ...
- 79.7k
6
votes
The 15 Pebbles Game
As with pretty much all the nim variants, this one can be solved by starting from the end and working backwards. With the original total number of stones being an odd number (15, as given in the title)...
- 78.6k
5
votes
The 15 Pebbles Game
The strategy is to do a move that
In particular, for $15$ pebbles, your first move would be
The reason this works is more interesting than with other single-pile nim variants.
- 55k
5
votes
Accepted
Another variation of the game of Nim
1. Proof that all games will end after a finite number of steps
Proof by induction:
Induction hypothesis H: (not known to be true yet)
...
- 78.6k
5
votes
Accepted
Single-pile Nim with Three Players
I will assume that
Given several equally desirable options, the players will randomly choose one.
Doing so, it is easy to work out what happens when there are $n$ candies;
If $n=1$, then player 1 ...
- 32.6k
4
votes
Can you help me understanding the Stones game?
Here's a clearer statement of the rules. Each player has two options:
Remove a stone on this turn.
Remove a stone on the next turn.
This is because not removing a stone causes one to be removed next ...
- 32.6k
4
votes
4
votes
Single-pile Nim with Three Players
With this information:
To be perfectly rational and impartial, only wanting to maximise their own chance of winning.
and it is assumed that
If there were 5 candles
If there were 6 candles,
If ...
- 30.4k
4
votes
Double or Take game
Let’s see: I assume that all numbers must be greater than or equal to zero, otherwise the possibilities become too painful. If that was the intention, I’ll look harder.
1
2
3
4
5
6
7
8
9
10
...
- 32.3k
4
votes
Accepted
4
votes
Accepted
A pile of chips involving powers of 2
Code to find the pattern: https://ideone.com/O5S4Qu
(The third number printed out on each line is the winning move if the player to move is in a winning position)
- 938
4
votes
4
votes
Accepted
The 50 game between two players, selecting numbers between 1 and 10 inclusive + variations
Question 1: The status of a game can be encoded as a pair A-B, where B is the sum including the last number said, and A is the sum excluding it.
Question 2:
- 12k
3
votes
Marbles on a Mancala Board
User hexomino already figured out the puzzle, and managed to actually find the very complicated path that was exactly how I came up with the game. To recap:
The game itself is a lot easier to play ...
- 78.6k
3
votes
Three-player Nim
If everyone is trying to maximize their points:
If both of your opponents work together to minimize your chances to win the game itself:
- 6,879
3
votes
Accepted
The Box of Tic-Tacs
Partial (assuming I understood the requirements correctly):
My assumptions: You (captive) start.
You and the kidnapper always return on the even turn the same number of tic-tacs and you decide the ...
- 18.4k
3
votes
Accepted
A short nim game
If you have to take exactly 5:
If you can take a number from 1 up to 5 (which is what I'm assuming)
If you can take a number from 0 up to 5
For the alternate game (which actually isn't too much ...
- 9,924
3
votes
Accepted
Game Night at the Binomial Elks Club
Any permutation of the rows and columns is treated as an "equivalent" board.
Part 1: Matt chooses to go second.
(Case A) If Ben takes 2 or 3 tokens, then Matt can turn it into a 2x2 board and win.
...
- 146
2
votes
Removing green marbles from the table
As described above, the first player loses if $N$, the initial number of marbles, is a Fibonacci number, and wins for any other integer $N > 1$. Proof follows.
First, a definition.
The remainder ...
- 7,518
1
vote
Accepted
Zigzag nim game
As you suspected, the correct approach is to start from the end.
First, let's invent a notation for the game positions. Let's view each position from the POV of the player whose turn it is, so that &...
- 78.6k
1
vote
Three-player Nim
First of all:
Taking this:
With more marbles:
Following that logic:
List of winners:
Patterns:
Applying to the problem:
Minimum rounds:
- 2,436
Only top scored, non community-wiki answers of a minimum length are eligible