bonanova

True, but can anyone answer that question?

OP says "I am thinking about one of my cards." That is an act of selection. It followed an algorithm of some sort. Flipping a coin, shuffling them face down then turning one over, taking the one closest to her as I tossed them across the table, saying eenie meenie money mo ... not thinking at all, praying, in some way, deterministic or not, she made a selection. And, inescapably, how that selection was made affects the answer. So the answer as the OP stands is, "depends." Without knowing her selection algorithm, how does one arrive at 3/7 as 'right'? If she thought of the lower-ranking card, the probability of both cards being aces is unity: spade ace plus one of the red aces are the only possibilities. OP doesn't say the thought-of card was the lower one in rank, true; it says nothing about the thought-of card. 3/7 is 'right' if the thought-of card was the result of a coin toss. But OP does not say she tossed a coin. I understand your basic question; I have wondered it myself. Absent any description of the reporter's algorithm, is there a least-restrictive default assumption that becomes preferable? I could, along with you, make a case for random choice (coin toss) to be that preferred, assumed algorithm. But unless that premise is understood and generally accepted among problem writers and solvers, and not just a few of us, it does leave answers uncertain. In my boy-girl puzzle I had some fun making this point, by imposing a random selection of reporter algorithms. Added in edit: 3., 4., and 5. I believe, all apply to picking either card with equal probability, along with the fact that the answers are the same for the two cards on that basis of choice: of the 56 permutations, of 14 hands the statement could have been made; and of those 14, six have two aces of any color.

Yes. See my spoiler above.

Subtle variation on the You are transferred to Killville, by your employer, for a one-year temporary assignment. Its population comprises 10 Killers and 10 Pacifists. A Killer immediately guns down any other person he meets. A pacifist never harms a soul. When you arrive at Killville, you are required to register, either as a Killer or as a Pacifist. If you register as a Killer, you are issued a gun and required to act as a Killer. If you register as a Pacifist, you are given the town's best wishes for good luck and advised to careful. You desire to live long and prosper. Which way do you register? Assume all encounters are 1-on-1 and random. Use spoilers please.

Wikipedia has a helpful elementary discussion of resistance [unit = Ohm] of an object as it relates to the resistivity [unit = Ohm-cm] of a material. The easy relationship is that the resistance in Ohms of a 1-cm cube of material is equal numerically to its resistivity in Ohm-cm. Resistance = resistivity x length / area in compatible units. It thus makes sense to speak of a material having a resistivity that is [or, as in the case of the OP, is not] constant. Thus, in the formula R(p) = 2*sqrt(p), R refers to the resistivity of the material. After you've specified the material's shape, i.e. created an object, it's meaningful to speak of resistance. Again, the distinction is that resistivity is a property of a material, independent of its shape; resistance is a property of an object. Resistance depends on the object material's resistivity and on the object's shape.

That's the number I had in mind.

What is the "oddest" prime number?

Let's restate to avoid misinterpretation (given the conditions set forth in OP about 1's +, x and parentheses): 1. What is the smallest number that cannot be expressed using fewer than 16 1's? 2. What is the largest number that can be expressed using 16 1's or fewer? Since we can add any leftover 1's if fewer than 16 are used, precisely 16 1's applies in the second case.

Nice approach. I had thought of constructing a set of stepwise overlapping regions to visually find back paths, but it got complicated.

A new BD record! The process can continue. For example, fourteen points are possible. But there is a limit, after which it's not possible to continue. Is there an intuitive proof/reason why there would be an upper limit? If you have a program to place N points, can you verify the limit is 17?

Great! Curr3nt has solved for 8 points. The bar is now raised to 10 points. The interval [0, 2520] gives integer boundaries for 1-10 divisions.

Has anyone noticed that

Draw a line segment, and place a point on it anywhere you like. Now place another point, but be sure the two points are in opposite halves of the line segment. Continue by adding a third point, but make sure they all are in different thirds of the line segment. And so on. After placing eight points you'd have something that looks like this. Here are eight points, each in a different eighth of the segment. But in what order were they placed, and in what part of the 8 intervals must they lie in order for each successive group of points to have comprised a legal arrangement? It may seem a trivial task: it's certainly trivial to describe. Have a go at it, with some graph paper or a calculator, and see how many points you can place. Let's standardize the problem as follows: Define the line segment as the number interval [0, 1000], and give your answers as integers. For example, for the pretty easy case of four points you might get {200, 800, 450, 700}. And so on. Have fun.

That was a reply to your question. Is the answer the obvious "almost all" of them?

TheChad: