koren

"Which of the two players has the advantage in this game?" My answer is Alice:

I suggest, we construct a kind of "triangles generator": We take a circle of any diameter, draw a chord AC from the endpoints of the chord we draw two line segments, the ends of which intersect on the circle. Symmetrically we draw two other segments on the other side of the chord and we got two triangles. When the chord is coincides with diameter the triangles are identical. The final touch – we connect the vertices of the triangles by line a-a. "Triangles generator" is ready let's start generate. For this purpose we are doing two things: 1) Moving the chord parallel to itself from the diameter stepwise with any resolution you choose. 2) At each step we move the line a-a together with the vertices of the triangles from point B to point A. By action 1 we iterate through all possible angles at the vertices of the triangles. (Look at nice animation here) By action 2 - iterate through all possible conjugate base angles. Eventually we generate all possible variants of triangles. At the resolution tending to zero we get close to infinite number of triangles, but an interesting observation for us, that in any case each specific obtuse triangle always has one corresponding acute-angled triangle. In other words: THE PROBABILITY OBTAIN OBTUSE TRIANGLE AT RANDOM CHOICE IS 50%. One must be careful when defining random triangles. There are many ways, yielding various distributions. The OP requires a particular method: selecting three points at random. = which means choose random triangle = This is easily done (for the unit circle) by selecting x and y values in the closed interval [-1, 1] and discarding any points for which x2+ y2>1. Inspecting a million or so triangles constructed from triplets of such points gives a clear answer to the probability in a finite region., Please pay attention - I prove clearly that any obtuse triangle corresponds to one of the acute-angled, which shows clearly that we have 50% chance to choose one or the type no matter in which way you do it.

I suggest, we construct a kind of "triangles generator": We take a circle of any diameter, draw a chord AC from the endpoints of the chord we draw two line segments, the ends of which intersect on the circle. Symmetrically we draw two other segments on the other side of the chord and we got two triangles. When the chord is coincides with diameter the triangles are identical. The final touch – we connect the vertices of the triangles by line a-a. "Triangles generator" is ready let's start generate. For this purpose we are doing two things: 1) Moving the chord parallel to itself from the diameter stepwise with any resolution you choose. 2) At each step we move the line a-a together with the vertices of the triangles from point B to point A. By action 1 we iterate through all possible angles at the vertices of the triangles. (Look at nice animation here) By action 2 - iterate through all possible conjugate base angles. Eventually we generate all possible variants of triangles. At the resolution tending to zero we get close to infinite number of triangles, but an interesting observation for us, that in any case each specific obtuse triangle always has one corresponding acute-angled triangle. In other words: THE PROBABILITY OBTAIN OBTUSE TRIANGLE AT RANDOM CHOICE IS 50%.

For me, it's not obvious. Imagine rain and two plots, one of which is twice larger than the other. Obviously, on a larger plot fall twice drops, i.e. probability for each drop to fall at a larger plot is approximately 0.66. And if the second plot is 1000 times larger, and if it is infinitely large? So, for me it's not obvious. And for you?

Depends on the technique of choice. If you choose first the point A and then B, the probability is zero, because A divides the axis on finite and infinite parts. (probability = N/infinity = 0) If you will simultaneously "throw" these points on the axis, then the odds that B is to the left or to the right from A are equal. The answer is 50-50. How you like it?

Look, I'm not sure in my solution too. Samething is wrong with the methodology. If we will take two arbitrary points A, B and begin to sort through all possible options by moving the third point C, we obtain an infinite number of obtuse triangles relatively a limited number of acute-angled (my solution). If we will take one arbitrary point and sort through all possibilities by moving two other points we will get completely different result. The funny thing is that according to the task conditions we have to "throw" on the plane the three points at once. (I always hated this probability stuff, now I know why )

I do not know who is right-handed, etc. To my question everyone raises the right hand when they say YES. For example, a question for the left-hander: "Will you raise your right hand on the question: "2>1?"?". In normal mode, for "2>1?" he will raise his left hand, in the construction of my question he, as well as the right-hander, raises his right. Let's see how it works: 1Q to #1: "Will you raise your right hand on the question: "are you A?"?" Right hand to 1Q - the person is A 2Q tu #2: "Will you raise your right hand on the question: "are you B?"?" Right hand - the person is B and #3 is C Left hand - the person is C and #3 is B Left hand to 1Q - the person is B or C 2Q tu #1: "Will you raise your right hand on the question: "are you B?"?" Right hand to 2Q - the person is B 3Q tu #2: "Will you raise your right hand on the question: "are you A?"?" Right hand - the person is A and #3 is C Left hand - the person is C and #3 is A Left hand to 2Q- the person is C 3Q tu #2: "Will you raise your right hand on the question: "are you A?"?" Right hand - the person is A and #3 is B Left hand - the person is B and #3 is A

Q really is the area between the two perpendiculars and it is really infinite but only in DIRECTION PARALLEL TO PERPENDICULARS. This direction is not relevant because the movement in this direction does not change the triangle to obtuse. The probability we are looking for depends actually on dynamics in the perpendicular direction and can be calculated by comparing two line segments outside the perpendiculars which are infinity.with the line segment Q which is tending to infinity NUMBER. So, the probability is INFINITY/NUMBER = 1. This solution is correct for one point let's say A, which means it will be correct also for B and for C, which in turn means it is correct for the entire case.

The amount of gas in vessel depends on three factors - the volume of the container, the gas temperature and pressure. In our case the volume and temperature are not changing. The pressure changes. As known the air pressure decreases with increasing distance from the ground. In the first glass position the pressure therein - P1 is the average betwin Pb - bottom pressure and Pt - glass top pressure. In the second - P2 = Pb , because this is the only level of interconnected spaces. Since P2>P1 the amount of air in the second position will be more. Oh! we have not given explanation regarding dice... This is to the author because it has nothing to do with this puzzle.

I guess you meant to say the probability is "1" My approach is very similar to above.. but there are couple of issues you might need to address.. "since Q is also INFINITY , you can't be doing INFINITY-Q" Q is not an INFINITY but number which varies from 0 to INFINITY having a fixed value in each test point. Of course I had in mind the probability = 1. Sorry.

The solution I propose is based on a certain question configuration: "Will you raise your right hand on the question: (yes/no question)?" A positive response in this case - the right hand raising , the negative - left and no matter to whom you addressed the question. All the rest is not difficult.

Ah, ok.. I misunderstood. As you have stated, it can certainly be solved with 3 questions. Further on in the same manner...

In other words, you can all the time ask only one person - in terms of this exercise it makes no difference.

The strategy is basically the same - first to find the R. The questions naturally differ. My third question, for example, to not R: " what would answer the other not R to the question "how would respond R to (any obvious question)? "?"

I misunderstood - after all English is not my first language (not even second). In any case, the second question to the robot №3 after white respond to the first question is unnecessary as the only possibility here is №2 - R. All the rest is OK.

I do not know if I violate some kind of procedure, but I would like to ask one more question. How do you handle the situation - the combination is robot 1-T, 2-R, 3-L. According to you: "Second question (to Robot #2 if the first response was Red, or to Robot #3 if the first response was White): "How would Robot #1 answer the question: 'Does 1+1=2'?" If the answer is White, then Robot #1 is Random. If the answer is Red, then Robot #1 is Truth or Liar and therefore the robot you haven't talked to yet is Random. " We talked with a third robot got red, but he was not R?

Сongrats! You got it.

If we mean "How would Robot #2 answer the question: 'Does 1+1=2'?" - it will depend on who is the second robot and in what state R is, but in any case you will have Red.

it still the same "question"

​My dear bonanova I created a puzzle and I want to see somebody to solve it

Edit: But I believe that last solution would work no matter how Random behaves. Please read the clarification I gave recently to k-man Edit: But I believe that last solution would work no matter how Random behaves. Edit: But I believe that last solution would work no matter how Random behaves. OK, let's see your last solution. The first question is OK and you find one robot that is not R. With the second question there is a problem. The robots are programmed to respond to yes-no question, which means to confirm the content of the question (expected event, statement, circumstances - whatsoever) by YES, or to deny it by NO. Your question has no content, which can be confirmed or denied. In fact it is one of those cases where the robots (including R!) are not responding. There is a problem with the third question too, but I think we can stop at this point.

Edit: But I believe that last solution would work no matter how Random behaves. Please read the clarification I gave recently to k-man