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20

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 shows the following (with a lot of spurious accuracy!): [0 0 9] 0.08969140375863138 [0 2 7] 0.033696054860265535 [0 4 5] 0.029747584762232965 [0 3 6] 0....


15

I think Oray got the right answer. Here are some drawings to illustrate the solution.


15

I would suggest an alternate (simpler) strategy:


15

Up to symmetries of the board, there aren't very many possible moves for the first player: Does this strategy work?


13

The sequence continues because they are


11

Suppose $t(n)$ is the average number of spins you get if you start with $\$10n$. We want $t(2)$. If you start with $n$ ten-dollar bills, put $n-1$ in your pocket and play until you are broke, then take the next \$10 out and play until you are broke, and so on: this is exactly equivalent to just starting with $\$10n$ and playing until you're broke, so $t(n)=...


10

This is not a proof but basic intuition; The value of $d$ should be (considering they are wise and choose not random points but the best possible point where they have the advantage) where it makes with the strategy of Lastly,


10

The player who can win is by this strategy:


10

Nash equilibrium This task, including the uncertainty about what the opponent would pick, would be a classic fit for game theory, and it must have a solution that's locally optimal i.e. where anyone deviating would make their chances worse (the Nash equlibrium). It's not a question whether such a solution exists, it's well known and proven that it does. Of ...


9

First player wins Example:


8

The result is that


8

Answer to the non-bonus part: Proof: Bonus part: From 12 Hence


8

They decide that and the leader When


7

First of all, lets present a strategy for $N=4$. $N=4$ $O$ can win in 7 moves (13 total). To start, the first two moves are arbitrary due to rotation and stretching, so let us assume we have the following; 1. $O_1=[0,0]$ 2. $X_1=[-1,-1]$ Now, $O$ will pick a line not on the $O_1,X_1$ line and place another point. $X$, meanwhile, should place a ...


7



7

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


7

You should To prove, let us first do the following


6

I think that the solution is Strategy Reasoning Why I chose these numbers


5

The player with the winning strategy is Strategy


5

I think you can Here's my reasoning: Well, any two points can be connected by a line, right? Then you can see that The challenge is that your opponent can block the connections with a point of their own. But as you remember about Because of the {important} knowledge above, you see that Making the equation Let's do 4. By putting {Note: In this step ...


5

The key fact is as follows: Therefore this particular game ends when the numbers on the board are So the conclusion is


5

To take another approach:


5

Then, Additionally,


5

This is an extension of Dr Xorile's observations, which fully answers the question that "there is no optimal strategy as long as knowdledge(and/or guessing) of your opponent distrubution is missing". But if we knew the opponents strategy, then a) Maximisation of wins b) Maximisation of enemy causalties It is interesting to see that even if we knew the ...


4

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) All games of exactly N piles will end in a finite number of steps. Induction step: If H is true for some N, it is also true for N+1. This is because out of the N+1 piles, one is the rightmost one. When a chip is removed ...


4

This is because


4

A simpler answer.


4

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 11 Therefore the only starting numbers that win for Alice are Bonus Part: 12


4

Let's name the lines. +---+---+ | K | +---+ L + J + | H I | + G +---+---+ | | +---+---+ F + | D E | + C + A +---+ | B | +---+---+ The winner can be found by analyzing the game regardless of the position. Now that we know the winner, who is which color? And Bob's last move would ...


4

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)


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