bonanova

To your question: I prefer a single universe, with an identity that is more or less persistent and choices [free will]. That's the game and rules I accepted early on, and with which I am more or less comfortable. However. As I read current thought about the multiverse, there seems to be a universe for each outcome. That's beyond me at present, both in comprehension and in comfort.Also, it probably reduces "identity" in a physics sense, to be an attribute only of elementary particles, not conglomerations of them, like myself. We could also take an intuitive sense of identity [without invoking multiple universes.] If someone asks me "who" I am, I could take that as an identity question.I would probably start with name, address, phone number [two of them], Social Security # [being a successful colonist, but what if the British had prevailed?], hair color, height, weight, ethnicity, distinguishing marks, habits and so forth. In short, the kinds of information that might be in a police dossier. But I could move. Would a different address change "me"?I could change my name, phone number, hair color, have moles removed, lose a limb or two in combat, have plastic surgery, and so forth. Would any if this change my "identity"? My intuitive answer here is No. But in the sense of the OP, I could turn right or turn left at the next intersection.Would my identity be the same in either case as it would be if I stop my car and walk home? Here, in the physics sense, I would say the choice did change me. Three universes are necessary to house the three different outcomes. In the physics sense, moving and changing my name, etc. also would change me.In that discussion I was taking identity as an intuitive thing: my identity = the "real" me. Identity is complex, needing a frame of reference. Determinism, just to address the point, had its funeral with double-slit interferometry. Enough rambling.

If Rich uses "No God" as an adjective, that names his god. Faith, fighting, defenses, dogmas, intransigence and playing cards don't affect this.

Outside the box; off the table. Honorable Mention for sure.

Sometimes it's Martin, a semi-idol of mine, but not this time. It's gardener, lower case, with two "e"s. Correct. A row of four is not also two (or four) rows of three. A row is a row, and its size is the number of trees aligned with it. Clue: If the rows of length greater than three are excluded, there are (only) eighteen rows.

No, Sherlock, but great investigative reporting! Oh that makes things different. In that case I think there are several solutions. Your statement is correct. Your question is valid. But since you don't yet have a unique answer, it's your question to answer. Btw, no offense intended by the Sherlock comment.

No, Sherlock, but great investigative reporting!

Nice, and extremely well ordered. But the gardener was more frugal. He used fewer trees.

I remarked to a friend how surprised I was to learn that as few as 23 randomly chosen people give better than even odds of a shared birthday. She agreed, saying three of her friends shared her birthday. The product of their ages is 2450 cubic years, she said, and their sum, remarkably, is twice your age [meaning mine.] It was a statement, to be sure, but I knew from the twinkle in her eye it was also a challenge. I scribbled on the back of an envelope for a moment. Well? she smiled, got you stumped? In fact I could not answer. I'm afraid I need another clue. OK, she said, I am older than any of the three, and my age is equal to the product of the ages of the two youngest. Triumphantly I announced the ages of her three friends. p.s. That's a challenge.

Your solution is better than my statement of the puzzle. I meant, of course, to enquire as to the oddness of the grains in his shovel. Before he buried me. OP modified accordingly

I put a handful of pennies on the table and I wondered whether I could group them so that each penny touched three and only three of the others. It took a while but I found it was possible. I then wondered whether I could do it with fewer pennies. It was clear that I could not. How many pennies were in the group?

Thanks Prime, I overlooked that post somehow.

At the beach last summer my grandson filled his yellow pail with sand. He then scooped out a toy shovel full of sand with the intent to bury me. At the point in time, how "odd" was his pail shovel? His pail shovel was more likely than not to have an odd number of grains of sand. His pail shovel was less likely than not to have an odd number of grains of sand. His pail shovel was equally likely to have an even or odd number of grains of sand.

The gardener planted his orange trees in a superbly ordered manner. Not counting pairs of trees as a "row", he created nineteen rows of trees. How many trees was it necessary for him to plant?

You have it. Here is a succinct approach.

English, not proper, not acronyms It would count twice. We are looking for how many of the sixteen possible three-letter strings are words. There are solutions greater than 13.

A velvet bag contains 9 black marbles and 6 white ones. You draw them from the bag, one at a time. When you are finished you have, of course, more black marbles than white. What is the probability that, in addition, after each draw you have drawn cumulatively more black marbles than white?

may have involved more work and exhibited less beauty than a good puzzle should. So let's allow N E W S moves only, like a Rook, but limit moves to one square at a time, like a hobbled Rook. A hobbled Rook tours an nxn chessboard without visiting a square twice. Your opponent begins by placing the hobbled Rook in any corner. Thereafter, you and she alternate making single-square N E S W moves. A player loses when there is no previously unoccupied square available for a move. For what n do you have a winning strategy? Is there an n for which you have a winning strategy in the modified case where your opponent can make any move on her turn, while you are constrained to make single-square N E S W moves only?

It may be that only one distribution of colors is possible, leading to a unique answer. We'll see. Edit after reading revised OP: Nope, we have to eliminate [preferably] or analyze each case. More sleepless nights.

A list of color combinations might be helpful. They can be solved individually.

It may be fruitful to assume a distribution and then see how it can be determined.