bonanova

Awesome. This the solution to beat, currently. Nice job.

Thank you -- for reminding me where I saw this problem. I looked for it several other places [and of course did not find it.] I read the book, after PT made reference to it and found most of it interesting, especially the terse and clever writing style. I took the "10-minute" answer as more or less obvious, it had a certain conservation to it - 5 is average, so 0 and 10 must be the extremes - but I didn't look [yet] at a proof. I might. At least I can end my search for the original source.

There is no such assumption. Could be all black for example.

Kern9787: 16 rows of 3, and 1 row of 4, I count. A slight adjustment could create a 17th [vertical] row using 3 of the 4 leftmost points. Still one short....

A square is a figure with four equal sides. How do you draw equal length lines using the available tools? A straight edge is only that - don't assume it's a ruler. But OK, I'll give you that - use the entire length of a finite-length straight edge. If you can use a straight edge in any meaningful way, it must have finite length. A square has another property: four right angles. How do your construct right angles using the available tools? A line drawn from the center to any side is perpendicular to the side iff it's parallel to two sides. How do you construct a parallel line with the available tools? But more to the point, your post would only construct an isosceles triangle. We need an equilateral triangle.

All well and good. Those who reject 2/11 for the first 6 subjects have no question to answer: For them, p[7] = 1/6 for all seven subjects. Good. Let's not hear from them again. But for those who believe p[7] = 2/11 for Subjects 1-6: where does the argument Subject 7 gives that his p[7] is also 2/11 break down? Or does it?

And I give my regards to all cristians for Noel, from world of muslims. Nice job. Not a simple problem. I thought about pointing out that it could have been spades or diamonds or clubs. I any case the probability increases; then asked why? But the increase in odds in this case is difficult to see, and one gets lost in the calculations. So I did it in terms of dice, where one's intuition leads to problems much more quickly. Thank you for the seasonal wishes ... most gracious.

Assume your premise that for example 1-6 and 6-1 are not separate cases. What do you then calculate for the probability of Subject 7's dice totaling 7? Assume there are 36 possible outcomes of rolling 2 dice. Write them out on a piece of paper. Cross out the cases that are eliminated by the statement "at least one of the dice is a 6." How many cases reamain? How do you explain that?

Not quite - that would free those who stood behind a prisoner with the same color hat. If the colors alternated, no one would survive. If the even number prisoners state the color in front of them and the odd prisoners state the color they've just heard, 10 are guaranteed to survive. Probably you meant to say that.

For the cube, I checked up to 5000 x 5000 [pyramid] base without solution.

Draw three large circles that intersect in such a way that 7 regions are defined. Three regions are inside just one circle; three others are inside two; and one, the central region, is in all three circles. Place seven coins, one in each region, so that all of them show HEADS. Now play this game: Select one of the circles [recall each one contains 4 coins] and either . Turn all its coins over: H->T and T->H; orReset all its coins to H: H->H and T->H.. Repeat this process until finished. The object is to have the central coin show TAILS and all the others show HEADS. What is the minimum number of moves needed to win? Or can it be won?

But, doesn't counting 6-6 twice [to get 12 cases instead of 11] say that [1] Die A=6 and die B=6 [2] Die B=6 and die A=6 are somehow different? Subject #1 [at least one of whose dice, call them A and B, was a 6] reasoned these were the possible cases: A-B = 6-1, 6-2, 6-3, 6-4, 6-5, 6-6, 5-6, 4-6, 3-6, 2-6 and 1-6; of which the first and last total 7. Similarly for Subjects 2-6.

Amazing creativity has produced the following results: I may have counted wrong here or there; if so let me know. xucam with 19 trees: 5 rows of 2 / 11 rows of 3 / 5 rows of 4 / 2 rows of 6 Alice JH with 15 trees: 15 rows of 3 / 2 rows of 4 / 2 rows of 5 Charles1317 with 13 trees: 12 rows of 3 halfbrain with 15 trees: 19 rows of 3 / 1 row of 5 Prime with 19 trees: 9 rows of 3 / 6 rows of 4 / 3 rows of 5 Prime with 17 trees [and creative landscaping] 20 rows of 3 Halfbrain [with 15 trees] and Prime [with 17 trees] have produced the requisite 18 rows [and more!] of three trees. Both layouts lack the bonus row of [exactly] 4 trees. All layouts so far exceed the minimum number of trees. Here's a hint: Charles missed it by a single tree [and 6 rows.]

A favorite, posted here.

The gardener's official statistician has determined that's more than needed. Also, to clarify: a row of 4 trees is only that - it's not 1 or more rows of 3.

The eccentric old man decided to plant a new garden. This time, he told his gardener, I want 18 rows of at least 3 trees. But don't waste my money. If you buy more trees than are needed, you'll be fired. But I am a kind man: if the garden includes an additional row - one having 4 trees - I'll give you a raise. [1] How many trees did the careful gardener need to buy? [2] In what configuration did he plant them? [3] Did he get his raise?

A magician blindfolded seven subjects, then for each of them he rolled a pair of fair dice and asked them the probability of their dice totaling 7. He said to the first, here's a hint: truthfully, at least one of your dice shows a 6. The subject counted 11 cases of at least one 6, two of which, 1-6 and 6-1, total 7. So he answered, my chances of having 7 are 2/11. Very good, said the magician. He then said to the second, I've looked at your dice, and at least one of them is a 5. This subject counted 11 cases of at least one 5, of which 2-5 and 5-2 made 7. So he answered 2/11. Very good, said the magician. He told the third subject, I see at least one 4 on your dice. That subject also found 11 cases of 4, of which 3-4 and 4-3 made 7. So he answered 2/11. Very good, said the magician. The next subject was told at least one of his dice was a 3. Like the others, he found 11 cases, and of them only 4-3 and 3-4 were favorable. So he answered, my chances of having 7 are 2/11. Very good said the magician. The next two were told their dice showed at least one 2 and one 1, respectively. They found 5-2, 2-5 for one, and 6-1, 1-6 for the other, among 11 cases gave 7. They both answered their chances of 7 were 2/11. Very good said the magician. The seventh subject had been listening to all of this. And before the magician could speak, he said, I don't need a hint. I know that you're going to tell me some number appears on at least one of my dice. And you've already confirmed what the right answer is in each case. So whatever you were going to say, I know most certainly what the odds probability of my dice totaling 7 are is. My answer is 2/11. But we know his odds are probability is 1/6, right?

Not exactly... A2-B2 was changed to (A+B)(A-B)

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Your tools don't include a means of constructing a perpendicular. Your slanted line must be at 60o. How will you ensure this?

Confused. Lower? or higher?