bonanova

I agree. Which side of the would you take?

[spoiler=Looks like]The obvious challenge is to create an opposite face for the center square. But that means it could not touch the center square, as they all do. So at the moment, my correct answer is no. I assume the 1x1 is a typo for 1x1x1.

When they find the time, they can't find the energy. And v.v. Does this explain the scarcity of quantum physicists?

my OPs give all the clues necessary to solve. Let me point out two for this case. Note that: The OP contrasts clever math - like which I think is very clever - with the answer needed to solve this puzzle. So the answer is not of that type. It asserts the equation can stand, with the addition of two things. Could it be as simple as saying 1 ten-dollar bill = 2 five-dollar bills? That would be adding two things, but those two things are so similar it really would not make a puzzle. The "aha" moment should be forthcoming shortly.

Another difference in people's answers might stem from the "no backtracking" stricture in the OP. Bushindo interpreted that to mean only up and only right [no down and no left] moves may be made. A less restrictive interpretation would give a higher answer. One such might permit revisiting a previously occupied square but not to retracing any piece of your path. That is, not reversing any previous move between squares. No backtracking, therefore, should be clarified.

The OP specifies that you get three rolls in that next minute. Every dollar in your stake gets to be must be wagered each minute. Each dollar thus has an experience that is independent of but statistically identical to all the others. Does that suggest an approach to analyze the problem? But the answer has to be the same as if you considered a simple infinite sequence of wagers. Since all the events are independent, and we're looking for a stop condition. So there are at least two analytical approaches to take.

Of course. But your decision must be justified by a correct estimate of failure. BTW, your winnings are all in Brain Den currency. Send Rookie a PM with the address to send the check.

Ah, but the game can go beyond that point, as you note in your spoiler title.

Right. The problem properly sated is to be the first to draw three numbers that total 15.

You glue two small square pyramids together at their bases to form an eight-sided object that becomes a fair die. You mark the opposite pairs of faces with 0 1 2 and 3. You begin the game with $1. You roll the die; the number that shows is your new stake. That is, with equal probability you lose you dollar, you keep your dollar, you double your dollar or you triple your dollar. A minute later, you bet each of your dollars, if any remain after the first roll, with another roll of the die; one roll for each dollar that you have, and collect your winnings, if any. After another minute passes, each of your dollars, if any, suffers the fate of another roll. To be clear, in all cases each dollar is wagered individually. As the minutes turn into years, you eventually become rich, or you go bust. What is the probability that you go bust? I like the conditions that Prime gave in his last game. You can simulate the game if you like. But a submitted solution must comprise an answer and a method, both of which are correct. Enjoy!

, my opponent chooses N (not equal to 5), and I choose 10-N Excellent. And yes the chips are taken out of circulation once selected. But ... I neglected to mention one other condition. The person wins whose total is 15 after picking up three chips. I changed the OP to include that condition.

Now you have it. Good solve.

A king tours an nxn chessboard without visiting a square twice. Your opponent begins by placing the king in any corner. Thereafter, you and she alternate making legal moves of the king. 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 legal king moves only?

Wouldn't a polynomial be continuous?

Good point. But that was not what the puzzle meant to ask. OP Has been appropriately edited.

Close, but ..

Nine white poker chips have the numbers 1, 2, 3, ..., 8, 9 written on them, one number on each chip. You and a friend alternately select a chip, with the aim of being the first to draw three numbers that sum to 15. You may play first or second. Do you have a winning strategy?

An odd number of guests at a party play a game. When the music stops, the guests stand still while their host measures and confirms the distance between each pair of guests is unique. They are then instructed to keep an eye on the nearest other guest. Every guest is thus watching some one. But is every guest being watched?

Hi rmgx, and welcome to the Den. Thanks for submitting this riddle. If you want to submit others or solve even more of them, you can visit the Word Riddles Forum, where I will now move this one. Enjoy!

I brought up the idea of median, since the distribution is skewed and I needed a way for both 13 and 14.7 to have significance. Bushindo stated exactly, and clearly, why it suggested itself. Prime explained well why and how wait times and <p> = .5 are different conditions. I've thought about it for a day now and I haven't decided for sure how a puzzle like this one might be worded in order to require finding a skewed median. but i think it could make an interesting puzzle. For sure, the wording would need to be precise. For this puzzle, I had in mind the cute 14.7 calculation. But I didn't word the OP to ask for it. I was surprised to confirm by simulation that 13 is the answer, as I worded it. Prime is way too kind to say a mistake on my part is anything close to a noteworthy event. But it is clear that (using his example) since betting on a "1" in four rolls wins, wait time (six rolls) isn't the answer for an even bet. Well guys, I enjoyed this puzzle too. Thanks for teaching me something. Footnote: sorry plainglazed, the solution post goes now to SP!

Thanks plasmid for taking us out of the box. With the pitcher and goggles, and infinity tho, I lost track of where the equation fit in. You might be on the right track, but we're not quite there yet.