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30
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29
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Can you beat random?
Here's how you win in the long run
Expected number of losses
Thanks to Evargalo for fixing the last calculation.
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Sliding crosses in a 5x5 grid
I saw these questions late, so I decided to answer all 3 (3x3, 4x4, 5x5 grids) as well as the general case (nxn grids). The answer is
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Sliding crosses in a 5x5 grid
The answer is (maybe surprisingly)
Lets reverse engineer the problem:
So we can do so by doing the following:
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21
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Two prisoners and twenty marbles
Same number of nights as in PDT's solution but perhaps expressed in a simpler way.
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Can you beat random?
I agree with both answers, but the answer is a little simpler than their explanations. It can be simplified to:
Those two rules alone cover every possibility.
For completeness:
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19
votes
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18
votes
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Poison ivy chess: Can the king get to safety?
The king can get to safety via this simple strategy:
Look upon the squares on the main diagonal. Think of each square as a jumping target.
(Image: Light gray squares represent the position of the ...
- 2,468
16
votes
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Magic: the Gathering – Scry-sort
Assume the N cards are labeled with their desired order, 1 through N.
Compare the top two cards and place the smaller of the two on the bottom.
Repeat step 1 (N − 1) times.
Move the top card to the ...
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12
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12
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11
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Accepted
11
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10
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Can you beat random?
Here is a winning strategy:
Then at each move there are 3 possibilities:
Your expected win is
Your probabilities for round n (n>1) are:
Counting 1 for a win, 1/2 for a draw and 0 for a loss, ...
- 6,270
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votes
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8
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Think in a Odd way
The first person can't do better than 50%, as they have no prior information to act on, so the ceiling on survival is 9.5 prisoners.
In fact, we can reach that ceiling using the following strategy
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8
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Making an expression with the numbers 1 to 100 odd (or even)
On her first move, Anna can
Anna's subsequent moves depend on Boris' subsequent moves.
This works because
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Is it possible to fill an arbitrarily large hex grid completely given these rules?
I claim that the answer is
and here's why:
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Is it possible to fill an arbitrarily large hex grid completely given these rules? #2
I claim the answer is
because
- 146k
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votes
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How did she do that?
Sadin says he needs to open 2 boxes.
Sadin opens one box first (wlog, the one with an "apple" label).
Meanwhile,
Lena says she could do better...
She even has 2 solutions:
First :
...
- 6,270
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A game of 101 stones—win at (not quite) all costs
Time for an answer to the bonus question! We'll go straight to the generalized case. The theorem we shall prove is:
We have chosen to assume that $N$ is odd, since if $N$ is even, it introduces the ...
- 186
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votes
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Minimum K for detecting fake pearls in one weighing
As others have shown, it suffices to find a $10$-set whose $2^{10}$ subsets have distinct sums, and we want to minimize the maximum element of this $10$-set.
An upper bound from https://oeis.org/...
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Two Trains, One Track
Not sure if it's optimal, but here's a solution:
Note that the same number of steps would work with larger trains:
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You have magic, but sometimes the spell backfires: how to win regardless?
This "spell" is equivalent to any sports fan's totally ineffective superstition. It is the equivalent of a "lucky hat" - sometimes wearing it makes your team win, but sometimes it ...
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Magic: the Gathering – Scry-sort
This was getting too long for a comment, so I'm writing an answer.
Just a heads up for when you run this in your EDH deck, Prince is a blink, not a flicker. This means that the Felidar will return at ...
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The Alien Snails Experiment
The problem seems a bit under-specified, but as written, it seems the only possible valid answer would be
Furthermore:
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Two prisoners and twenty marbles
Minimising worst case waiting time, one can not do better than the number of nights presented by others.
One can not safely lower by one night, since the person going second then only has
So with the ...
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