bonanova

OK, what would you want them to be?

Using standard math operations and four zeros, get 120.

How many squares are there?

What are the next three numbers in this series? 4, 6, 12, 18, 30, 42, 60, 72, 102, 108, ...

Using only a straight edge, construct a perpendicular to the drawn diameter of a circle, from a point P outside the circle.

There was a young lady quite bright, Whose speed was far faster than light. She went out one day, In a relative way, And returned the previous night!

Or do the words have to form a sentence?

Sorry. Meant to say the deck is shuffled. But since it's a random draw I guess you could consider they're drawn from a box instead of dealt from a deck. In any case, assume you encounter the cards, sequentially, in random order. By symmetry, the piles will be expected to have N/2 cards each, but that's not the puzzle question. The question asks for the expected value of the highest-numbered card in the left pile. In the example given, the highest-numbered card in the left pile is 8.

You draw cards at random and without replacement from a deck of N cards, each of which is marked with a unique integer from 1 to N, inclusive. You create two piles, as follows. The first card is placed on the table in front of you. If the second card drawn has a smaller number, you place it to the left of the first; if it's larger, you place it to the right of the first card. You now have a left pile and a right pile. You continue until all N cards have been drawn. Each card is placed on the left pile if it has a smaller number than the previous card, and on the right pile if its number is larger than the previously drawn card. Note, there is never a tie: each card has a unique number. What is the expected number of cards in the left pile after all N cards have been drawn? OK, that answer is not difficult to determine, since the definitions of the piles are symmetrical. Here's the real question: what is the expected value of the largest number in the left pile? Example for N=10 and the order of the shuffled deck is 2 7 5 3 6 1 4 9 10 8. 2 is placed on the table7 is placed to the right of the 2 (7>2) So 2 has become the left pile, and 7 is the right pile. 5 is placed on the left pile (5<7)3 is placed on the left pile (3<5)6 is placed on the right pile (6>3)1 is placed on the left pile (1<6)4 is placed on the right pile (4>1)9 is placed on the right pile (9>4)10 is placed on the right pile (10>9)8 is placed on the left pile (8<10)The left pile contains 2 5 3 1 8, and its highest-numbered card is 8.What is the expected value for N=1000?

Just to finish this one up, using TSLF's clue, here is a planar three-match solution.

Hi sw121290, and welcome to the Den. I'm wondering what question you are answering. Roolstar has asked, "Can you find a way to guarantee the freedom of some prisoners ...?" Technically I guess the question calls for a yes/no answer. If it's No, then there is no algorithm to discuss -- you cannot find one. But if, and only if, your answer is Yes, he then asks how good your algorithm is: how many prisoners will it save, with certainty? By answering "zero" are you saying you found an algorithm, and it guarantees the survival of zero prisoners? It would seem that would require (a) determining with certainty the colors of all the hats (b) communicating to each prisoner the color of his hat and (c ) convincing each prisoner to give the wrong answer. All three of these tasks are formidable. Just to focus on (a), it seems the first prisoner to speak cannot know the color of his hat under any strategy, and so he can't be effectively persuaded to give the wrong answer. That is, it would not be possible to guarantee his answer is wrong. Or by answering "zero" are you saying there are "zero" (no) algorithms that guarantee the lives of even a single prisoner? But several algorithms have been given that guarantee 19 survivors. Or by answering "zero" are you saying that prisoners are dumb, selfish or untrustworthy to the extent that there is "zero probability" that they can or will correctly follow a procedure, even if it gives on average a 97.5% (19/20 + 1/2 x 1/20) instead of 50% survival probability? Selfish prisoners would take those odds, even if it requires them to think. Anyway, and Rainman addressed this, the question isn't about the IQ or character of prisoners; it's about about the effectiveness of algorithms. Honorable Mention, anyway, for clearly thinking outside the box! I hope you have some puzzles for us to solve. Give us a try!

Three logicians enter a bar. Bartender says Do all of you want a drink? First logician says, I don't know. Second logician says, I don't know. Third logician says, Yes.

Ah, yes... I had a notion of changing the right side rather than the left, but didn't follow through. Nice one.

Just nit-picking. Heh.

Confirmation

Einstein, Newton and Pascal are playing hide and go seek. It's Einstein's turn to count so he covers his eyes and starts counting to ten. Pascal runs off and hides. Newton draws a one meter by one meter square on the ground in front of Einstein then stands in the middle of it. Einstein reaches ten and uncovers his eye and sees Newton immediately and exclaims "Pascal! You're it!" _________________

It's not clear whether the back story to this post is the escape from constraint of anyone else's opinion, or simply the claim of absolute superiority of one's own thinking. This uncertainty is a meta-puzzle in itself, and it is one that may merit its own thread. However, this forum is for questions about logic and math, so if such a thread is started it should be placed in the Others forum. Further posts of this type are subject to deletion. Regarding the current thread, the author of the OP put "red ends" in his drawing, and he confirmed, when the question was raised, they are part of the required symmetry of the solving configuration.

Just for fun, I programmed up a little script. Using a suitable underlying GMP library for Euler's totient function, it checked up to 10 million in 13 seconds, with a result of 60.793% distinct pairs relatively prime. That checks with 6/pi2 = 0.607927101854.....

Most of you got it. BMAD made a close estimate of the probability, which is actually just slightly greater than 60% to be co-prime. I'm marking her answer, therefore. This time the skeptics missed a chance to win some coin. Edit: Since the coprime probability is so close to 60% (slightly larger) I pondered making the odds 3:2, forcing a calculation in order to be sure. It's not a simple calculation. But I decided to be nice.