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I agree with the answer some others have given but I don't think their analysis is correct. First of all, So, let the expected number of coins you get, when the urn starts off with N coins in it, be f(N). We want f(100). On each turn, the billionaire chooses a number m from a geometric distribution with parameter 1/2 (note: there are two different things &...


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I think this arrangement of mines will work (red squares are mines)


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The answer is and here's why:


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Apart from the solution that hexomino found, there is another solution: According to my computer program, there are no other solutions up to symmetry (so 4 solutions if we count the rotated/reflected pattern as distinct).


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This is equivalent to asking how to place kings on a 6x6 board so that each empty square is attacked by 3 of those kings. This is a special case of the problem posed here. That web site gives hexomino's 16-king solution for the 6x6, among -king solutions for the 2x2, ..., 8x8 boards.


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After seeing hexomino's solution and your comment, I started playing with (manually generated, but automatically checked) ideas in Excel, with the goal of and quickly found I also found/verified the other known solution... But other avenues explored have which would be fairly obvious to @jaap-scherphuis, who I now see posted while I was typing up mine... ...


4

The value of $n$ is important as follows: The value of coins in the pot at the start of each turn will always be equal to The game ends Currently this is a partial answer, as I don't currently have time to work out, and my google-fu failed to find the formula for Update: I later realised I abandoned a further edit and forgot to undelete this answer after ...


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The maximum expected number of coins you can get is How can you get that much money?


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It can be done using Outline: Detailed example showing a few rerfinements:


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Let $n\gt 2$. I can do it by using no more than: The strategy that establishes this bound is: Notice that step 2. is only needed when $n$ is odd. If Steve's argument that $g(n)\gt n$ is correct, then $n=3$ is optimally solved with this. I'm not sure if this is optimal or if we can find a better strategy for $n\gt 3$ ? Below is a deeper analysis of my ...


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Attempt at a bit more rigor in an answer. Consider first the strategy of taking 1 coin each turn: For other strategies: Comments:


3

What is the maximum and minimum expected number of coins you can extract from the game?


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First, observe that: as such During the procedures a surface of a glove can be in one of 3 states: If it's possible to generalise to $n$, it must necessarily be possible to do 3, so I'll start with that simpler case, where doctors X, Y, Z must all operate on patients A, B, C. Without loss of generality, X performs the first operation on A. There are two ...


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Here is one example


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I expect there are many. e.g.


2

With N teams there are Edited to improve solution:


2

I can do ten lines. It starts nicely but ends funny. Or eleven lines. Or even twelve lines. Xelia owes Alice three magnets now...


1

The picture seems to demonstrate an ambiguous problem with 2 possible answers. We're given that Ram is 18th from the left end of the row, and Sumit is 23rd from the right end, and that there are 5 other boys in between Ram and Sumit. It is possible that Ram is to the left of Sumit, with 5 boys in between, and there are 18+5+23 = 46 boys in the row. This is ...


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Well, mean is infinite (easy enough to see): Median is finite though. In code I slightly simplified the problem - function [n,m,k] = so_coins(n) % n=number of coins left in the urn. m=0; % number of coins taken so far k=1; % steps performed - end condition. while (n > 0 && k < 1000000) % stop after 1m steps of taking coins out. r=rand(n,1)...


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This answer draws freely from other answers, in particular, @Gareth's and @Steve's. As pointed out by others our strategy Let's assume that the billionaire gets cold feet once there are a total of M (M>100) coins in the pot. We can give a closed form solution for the expected time $E_{100,M}$ until either this happens or the game ends regularly. This ...


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Between 500K and 500M gold coins, depending on the exact net worth of the billionaire As others have explained, on average one gold coin is replaced for every coin you take. So the expectation is that the game goes on as long as the billionaire can continue replacing the coins. A one-ounce gold coin sells for around 2000 US dollars according to this website ...


1

For the bonus bonus,


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