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Stay put for about 45 days, after which the pirates would have circumnavigated the globe and returned to your current position.


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 one point you will definitely find the locket of the coffin you first opened To prove this lets look at the other possibility and try to enter a loop without ...


That's easy


Here is the strategy:   It works because:


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 not necessarily flipped the same way). Replace $H$ with $0$ and $T$ with $1$. In each cycle, you flip all $1$'s to $0$'s, until Satan flips a $0$ to a $1$. Once ...


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 only 2 players left, they must annihilate each other. So if there are 3 left, your only hope of survival is to annihilate both your opponents, which can only ...


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 opponent chooses, the expected value will favour the strategy given above. By the law of large numbers, the average amount gained from a single throw The reason this ...


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 origin of a coordinate system. We will describe our position in polar coordinates, as a function of time: $(r(t),\theta(t))$ (where we have arbitrarily chosen a ...


In the original game, sometimes the new tile that pops up is a 4. Let's ignore that possibility for the sake of simplicity. We've 16 squares in total and considering the new tile popped up at the position we chose, to make it to 4096, we need minimum these tiles: 2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 2 These can be combined from right to left ...


Tic-tac-toe has been solved. The optimal first move is to go in the corner. As always, there is a relevant xkcd.


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 player puts their first coin 1 inch to the left of the center coin, the first player mirrors this by putting their coin 1 inch to the right of the center 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 flip of player $i$, with tails $1$ and heads $0$ (it's arbitrary which is which). Let $s_i$ be $1$ if player $i$ survives the round and $0$ otherwise. Then, ...


Strategy: How this works:


The first move can be made anywhere without sacrificing the game. If the opponent plays perfectly, any first move leads to a draw. However, if the opponent does not play perfectly, then the optimal place to go is the corner, since that leaves only one spot (the center) for the opponent to go to get a draw, increasing their chance of making a mistake.


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 produced. The winning player is using the following strategy:


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 first player can make a move that eventually leads to a win in ten moves or less. I took the tree and trimmed it down to only the optimal moves for the first ...


The solution: The reason:


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 $n=m=1$, then the first player automatically loses because there is no valid move. So lets assume $m>1$. When $n=1$, then the first player will win simply by moving ...


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 $2^i$, but none of the numbers $2^0,2^1,\dots, 2^{i-1}$. If this were not true, then the amount you decreased $2^i$ would be at most the amount you decreased the ...


Answer Because


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 third term. If Satan ever decreases the third term, then that makes it easier for you. So to prolong as long as possible, he would always increase the third term....


Another strategy for that works for any odd number of squares:


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.


There is a winning strategy for



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, $f[1] = 2$ since he can ask a single question. Now we can find a recurrence for $N$. Bob can choose to ask for a subset of size $i$. Since Alice is not ...


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 than 92 is a losing position, and all other numbers less than 92 are winning positions. From any number greater than or equal to 92, it is possible to remove a ...


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 win with perfect play.


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 right this second. After 20 hours, your distance from the starting position will be $r=20t+20$ while your angle will be... more difficult to find. If we ...

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