Prime

Also, in the original puzzle the man dividing up gold was a voter as well. Here I asked for clarification whether the off-the-ship candidate votes and the answer was -- no. If that's not the case, the amount of bribes changes, but not the essense of the solution.

I read the original problem where 5 pirates divi up 100 gold coins. Captain gets to keep 98 quite obviously. The logic in that puzzle is simple, straightforward, unambiguous and it does not apply here at all. In that puzzle, correct analysis leads to a single round. There is no such thing as "the pirate in the last position knows he will get a bribe in the second to last round". You must have missed that bit in my analysis. I devoted a whole paragraph to it. When there are 2 men left beside the captain, the captain choses one to throw off the ship and his own vote is enough to do so according to your conditions. The last man shares the same fate. None of them gets any money. So in the round before that when there are 3 men left beside the captain, any pirate must accept bribe, as it is the last chance to get paid. If the bribe is offered to the first mate who stands to become captain, should captain fail -- he should accept the bribe too, if he knows what's good for him. For if he refuses, the captain will simply walk over to the next pirate and offer bribe to him. And for 2nd and 3rd men that's the last chance to make any money as well as for the first. So there is no guaranty, which pirate gets the bribe. No pirate can count on anything here. And pirate's succession position bears no significance in this setup. Unlike the original pirate problem.

I don't know the original game. However, for this one I've shown that it does not matter, which of the pirates captain chooses to vote off the ship, nor does it matter which pirates he choses to bribe. Captain flips a coin to chose who to vote off and who to bribe, spends total of 9 coins, and keeps his position. If he tries to spend less -- he walks the plank. But that does not happen, as he is a game expert. Is there anything else to solve here? For example, if you are a first crewman in succession, and the captain offers you a bribe -- you should take it and consequently vote with the captain. Refusing would only lose you money and not gain anything at all. Same for all other pirates. That works as long as all pirates are game experts and understand how to maximize their win. Being closer or further away in succession order carries no advantage in this set-up. "Benevolence" has no place in this arrangement. It means less chance of bribe to pirates, so it should never be a consideration.

I think, in general, good strategies should deviate toward making the rodent to tread the same squares as much as possible. And when you no longer can send mouse down the same path and make it return, when closing path, build the new perimeter as close as possible. Preferably on top of the path, which mouse has traveled already. Per each square blocked make mouse traverse as many squares as possible. The total ratio is hard to estimate with a good strategy. Perhaps, recursively. First finding best patterns for a smaller board, then using the entire board as an entity for a larger board calculation. Looks like a computer simulation problem to me. I can't imagine how a simple formula could be derived here.

There is no need for latter proof. It was already demonstrated by example. This is how mathematical induction proof works. 1). Demonstate, the rule works for some specific case using specific numbers. (That was done here with n=2) 2). Assume there exists such n for which the rule holds. (Perfectly legal assumption in view of the preceding demonstration, that already proved that assumption.) 3). Show mathematically that assuming the rule is true for some n, it will also hold for n+1. (And that's the end of it.) Here is a popular example of the inductive proof: Prove that all men are bald 1). Suppose a guy had just one hair. Is he bald? Of course, he is! 2). Regard a guy who has n hair, while still being bald. 3). Now say, that guy grows just one additional hair and has n+1 hairs. Is he no longer bald? No way! One more hair won't do it. QED

1). We are civil. That's just our brand of civility. 2). Mojail's proof is called a method of mathematical induction proof. Mojail first demonstrated that the theorem holds for one specific case. Then made a supposition that the theorem holds for some case n-1. That supposition is valid, because of the preceding demonstration. Then proved that the theorem also true for n. Since n-1 is an arbitrary number, the proof is valid.

So....how will your strategy continue? And what does your score end up as? I'm not sure how that would work...

Here is a fresh idea. Make mouse "pay" while you're building the wall, while preventing it from even starting to traverse the structure.

One way, or another you need a third state.

What stops a pirate from taking a bribe and voting against the captain anyway? This way he gets money and fun. His position for succession is the only thing to take into consideration then. Or do pirates have some kind of honor code requiring them to follow through on an agreement?

Couple more things need clarification. Does off-the-ship candidate participate in voting? Is captain the only one who can bribe other pirates?

That's too simple, although hard to argue with. This proof involves guessing the equation. Consider your 2-d analogy where the points are described by y = 1 - x function. If I gave you a pair of points (x1, y1) and (x2, y2), then you'd have to go on solving a system of equations to tell me what is the equation for the straight line passing through those two points. For example, given (.2, .8) and (.5, .5), after solving it for a straight line, you'd also need to notice that line's funny property of x+y=1. "A plaine is uniquely determined by 3 points" is one essential part of it. Another, that an equation of type x+y+z=a describes a straight plane. There is also a somewhat more involved geometrical proof. But that aside, I don't see the problem (2) the average number of random picks as being solved here. There was a hypothesis that it can be expressed as Taylor series approximating e, but no reasoning to support that.

I don't quite understand the conditions. How is the maze set up initially? How can solution be independent of the initial setup? What if there was but one path to the cheese? Then the mad scientist could not set any new walls. Do you have a picture of the maze?

Do pirates know their succession order? I mean…

I take it, you cannot solve those problems. 1. Whatever math book you are looking at to get your answers, may not show the proof/derivation of "x+y+z=1 is the equation of the plane passing through (1,0,0), (0,1,0), and (0,0,1)", because it is an 8-th grade school problem. I'm just curious, can you manage to solve it on your own? 2. I didn't ask you why Taylor polynomial approximates e. I asked, how do you see that Taylor polynomial describes the average we are to find in this problem? I am beginnig to suspect, that you are borrowing "your solutions" from some reference, while not understanding clearly the meaning thereof. You also show very poor deductive skills by refering to Prime as "her". If you feel the 2 questions I posted here are below your station, I invite you to solve my restatement of another Unreality's probability topics. Since I constructed that problem on my own, there is a possibility that it can't be googled up, or found in a book. I think the first of the questions there is relatively easy.

The induction proof looks simple and sound. Here is the same, slightly re-shuffled, which may or may not make it more clear.

I'm still curious about: 1. Proof that every point lying on the surface that passes through (1,0,0), (0,1,0), and (0,0,1) of the unit-cube has its xyz coordinates adding up exactly to 1. (That shouldn't be all that difficult.) 2. Even more interesting, how do you derive that series consisting of number of trials times their respective probabilities is represented by Taylors's polynomial?

What about...

Great work! You have found correct reference weights for both 4 1 -- 4-lb rocks and 7 1 -- 5-lb rocks. I should add, it's the only reference weights that can work. However, not quite the criteria for finding the reference weight. Note, I specified how many rocks were there. And I figured out the minimum necessary to make the problem solvable. So when you use that number in a formula for finding reference weight, you are relying on my giving you the optimal numbers. True criteria tells you both things: 1) How many rocks you need for a given weight range and 2) what should be the reference weight. So, moving on, here is my solution with the proof for 7 rocks ranging from 1 to 5 pounds and just one reference weight. I'm giving the solution, but holding back on my criteria/formula for now. If you take time to go through that solution to verify it, you could also get a notion on what was my criteria for finding reference weight and minimum number of rocks for a given weight range. I belive the weight range 1 through 6 pounds is the maximum you can solve with just one reference weight. Past that you need more than one reference weight. For an encore and to finish off this puzzle: 1).How many rocks (at the minimum) and what reference weight do you need to solve the range of 1 through 6 pounds with just one reference weight? 2).What is the largest weight range that could possibly be solved with 2 reference weights? Is the original "Weghty thoughts" problem with 26 rocks in the 26-lb range solvable with 2 reference weights? (No need to go through all permutations to prove solution. Just show genral reasoning and test for the cases when purported solution may fail.)

Still, I don't understand the cube analogy. Or the formula for the probabilities for the consecutive steps.

Here is my simulation program: Sub randpick() Randomize picksum = 0# totpicks = 0# For i = 1 To 1000 picksum = Rnd totpicks = totpicks + 1 Do While picksum < 1# picksum = picksum + Rnd totpicks = totpicks + 1 Loop Next i ActiveCell.Value = totpicks / 1000 End Sub[/codebox]

How and why would that correspond to the probability in question? Consider two probabilities: 1). You have just picked 1 random number and it is less than 1. What is the probability that your very next pick will get the total to 1 or over? 2). You have just picked 10 random numbers and their total is still less than 1. What is the probability that your next (11th) pick will take total to 1 or over? Shouldn't probability (2) be greater than probability (1)? Wouldn't it be very likely that when you haven't made the total after 10 picks, you'd be somewhere very close? According to your estimates the probability to make total with the second pick (after the first pick didn't make it) is 1/2. Whereas, the probability to make total on the third pick (after previous 2 picks didn't make it) is 1/3. That is the more picks you made, less likely the next pick is to make the total. That just cannot be.

Unfortunately, the "brute force" enumeration missed some variations. In particular, the cases when you have all 4 rocks 4lb and all 4 rocks 3lb each. I don't see how your chosen reference weight can solve that. Your reasoning for reference weight criteria is lingering dangerously close to the solution. You just need to give a bit more attention to the details.

I wrote a quick simulation in Excel and got an average result of 2.7283 over just several thousand trials.

What kind of market is that where farmers buy wolves? I am not asking what use farmer has for a wolf who craves his livestock. But what kind of wolf guards cabbage patiently while his she-goat is sailing away? Wouldn't wolf go scouting for another she-goat? Can wolf row? This puzzle is more puzzling than it initially seems.