Prime

You are not allowed to use "()" -- only "[]". See the statement of the problem.

Fair enough. My appologies for not figuring out where to post it. You know, I am a programmer.

How about 1+5+6+7+1+1=21? Or did you mean use each number exactly once and each operator as you pleased?

How many computer programmers does it take to screw in a light bulb?

In your scenario (SWITCH/?) the probability of Monty's last remaining door to be A (the door initially selected by player) is 7/16, and the probability of the last remaining Monty's door to be door C is 9/16. But that does not help your final selection at all. For the probabilities of a particular door to be the last unopened Monty's door and probability of HS hiding behind it can not be mixed or combined. You set a scenario where Monty has possession of HS and therefore the probabilities for the doors A and C are 2/5 and 3/5 respectively. But overall probability of that scenario is 5/8, so overall probabilities of the doors are (2/5)*5/8=1/4 and (3/5)*5/8=3/8 -- something that we know already. That is before Monty opened his last door. When he does, his choice is FORCED NOT RANDOM in your scenario. Therefore remaining door simply inherits the probability of 5/8 of hiding HS.

Oops! Configurations 5 and 6 in my illustration above are one and the same.

You are right. I didn't notice 0001 in the second and 4th row would start painting. I like your representation of how that combination would paint -- each next line obtained from previous by one or more boolean AND operations. That's a step toward formalization of painting process. Here are 6 patterns that could be constracted with intial 4 squares and paint entire 4x4 area. They have different number of distinct rotational and mirror symmetries. Some will produce different number of duplicate solutions. E.g., pattern 1 and 2 could be derived from the same painted diagonal. But you can cover all possible combinations with 2 rectangular shapes. In other words, there are no 3 rectangles that would paint the entire board, which you wouldn't be able to represent as just 2 rectangles.

Color code is difficult to follow. However, is it possible that Bayes' theorem is not applicable here (whatever it is)? Especially, if it yields an incorrect result. The two doors that the host has after the initial switch in your scenario have collective probability of 1/4 + 3/8 = 5/8. And Monty's consequent manipulations do not change that probability for his remainin doors. If you break that situation into 3 uneven scenarious, then in one of them (when B hides the prize) the probability for B is 100% -- no further calculations are needed. Whereas, in the other 2 cases, the probability for B is 0 and the probability for Monty's remaining door to hide the prize is 100% (and he did not choose randomly in those cases) -- no theorems need to be applied here either.

Actually, the probability of A is still 1/4 at that point, and probability of B and C is 3/8 each.

I would have to accept your solution as the only correct one, if I were any less specific in my statement of the problem. (And Taliesin helped me to make it even more specific with his question.) I clearly stated that A has only 30% chance of hitting his target. So what happens if he aimes at non-lethal area of his body and misses?

I said "correct answer" -- not "correct solution". You seem to imply that a man may get two shots in one on one duel against B.

Do you mean which one of many bill collectors he sends home? If he chooses at random -- then his choice gives you no additional iformation. But if he wants to tease you, or have some other criteria that you can guess -- then yes! Or do you mean the fact that he chooses bill collector -- not the prize? That is the very thing that allows you to improve your chance over the original choice.

Monty's possibility for random decision destroys the uneven distribution among his two doors. Unless, you know of his teasing nature, and expect that if he has two doors with bill collectors, he will leave your original choice unrevealed -- just to tease you.

Here is a general solution for the problem of this type. And it hardly uses any math at all. And no calculations, I promise.

You must have misread the statement of the problem.

You have guessed the correct answer, but did not solve the problem.

[ Some of the variations are explained well here. However, you have missied the right answer.

No. Good point, though.

A, B, and C have a three-way duel. The rules are as following: A shoots first, then B, then C, and so on. Each man can shoot at whoever he wants on his turn. However, when one of them gets hit, he does not shoot anymore, and no one shoots at him. The duel continues until only one man is left standing. It is well-known fact that A hits his intended target 30% of the time, C – 50%, whereas B never misses. What is the best strategy for A? Small hint:

C must keep shooting in the air. This way A and B will keep shooting at each other and then C will get first shot at the winner, and, therefore, better than 70% chance of survival. On the other hand, if he shoots at A, for example, then if he misses -- it's the same as above. But if he hits him -- then B gets the first shot at C and C has got less than 20% chance of surviving that duel. I'll post another version of this puzzle, which actually requires some calculations.

Bona, Your 54 combinations to fill 4x4, are missing some variations, some are incorrect, and, possibly, some are counted more than once. (I did not check close enough to be sure.) The last combination you showed (1000,0001,0100,0001), will not paint. Also, consider 0100 1001 0000 0010 or (4,9,0,2) in hexadecimal. This one has 4 90-degree roatational variations, then a mirror image and four rotational variations of that -- 8 more variations in all. Another one I don't see in your set: (2,4,8,1) with 4 rotations. A method for traversing all possible variations while keeping track of all duplicate solutions at the same time -- would be an algorithm to find the answer. The challenge in that approach is to design an algorithm that would only go through those variations that satisfy the condition of painting the board. And, of course, keeping track of duplicate solutions. Then a computer program could come up with solution. Although, I still think there might be some analytical way. Perhaps, binary representation of the position on the board could be used somehow? Or rotations of the board?

If the three conditions, Clue set forth, are NOT met, the board will not autopaint entirely. However, meeting those conditions does not guarantee filling the board: Note, each and every column and row on that board has a square, no two painted squares touch with their sides. But no auto-paint. The probability question, is really the question of how many different collections of 8 initially painted squares will autopaint the entire board. The total number of combinations of 8 from 64 is 4,426,165,368.

Oh, I got the notation. It’s an illustration without using “Paintbrush”. To simplify we could represent each row of a rectangle as a binary number with one in the position of each painted square and zero otherwise. Then list set of resulting numbers starting from top row. For example: one painted diagonal variation of the 3x3 square could be represented as (100, 010, 001); or if you convert it into hexadecimal: (4,2,1). I don’t think it is feasible to solve this problem by trying to traverse all possibilities by hand. (I already tried solving “Dr. P, S, and N problem on this forum without use of a computer, and just got a headache.) Combinations of 8 from 64 are over 4.4 billion – a respectable number. The algorithm could be: 1. Find all possible ways to break down a rectangle into two other rectangles that would paint the entire area with the minimum number of initial squares. 2. Repeat (1) for each rectangle found. But that would be a programming heuristic solution. More elegant way would be finding how some common property of each and every configuration of initial squares limits the number of such configurations.

I don't agree with your formulation. It seems to imply that it's all the same to you if you serve 50 years while your partner is serving 100 as if you serve 2 years while your partner is serving 4. But even if I'm to interpret it that you feel as bad for each year that you serve as you feel for every two years served by your partner, still something's amiss there. It turns out it's all the same to you if you serve 5 years and 3 months while your partner is serving 14 years and 9 months, as if you serve 11 years and 11 months while your partner is serving only 1 year and 5 months. I say, you should try to be less concerned about your crime partner. It was his idea to rob that liquor store, anyway.