bonanova

*edit* Typo Izzy, you're close, of course. [note that 46 was written incorrectly: it's actually 45]. And I'll stipulate that the answer lies on some kind of smooth curve. So let me suggest the solution should either be exact to 5 decimal places [e.g. XX.YYYYY] where not all the X's or Y's are [necessarily] equal, and of course some of the trailing Y's could be zero, or a formula, say, that generates the numbers.

The sequence has been corrected - thanks Grey Cells for calling attention to it. 35, 45, 60, z, 120, 180, 280, 450, 744, 1260

I don't know how many there are. But if you find more than two, you can stop.

edit: typo Yup. That's the question.

Nope. The OP specifies just one probability, not two. It is an even money bet that both sisters got their wish ....

I'll make it short and sweet tonight, Alex said as he greeted the boys last night at Morty's, because yer gonna need every second of time you've got to figure this one out. That turned a few heads, but only Ian and Jamie bothered to sit down and listen. There's this puzzle contest I read about, and it has one of those sequences of numbers with one number missing. It's a shame to waste it on buzzards like yerselves, but I'll let you have a go at it, anyway. Here it is: 35, 46, 60, z, 120, 180, 280, 450, 744, 1260 Edit: '46' is incorrect. Here's the correct sequence. Thanks to Grey Cells for pointing it out. 35, 45, 60, z, 120, 180, 280, 450, 744, 1260 As Ian and Jamie stared at the numbers Alex had scrawled out in his dirty red ink, Alex continued, Kind of interesting that one of them doesn't end in 5 or 0 isn't it? They both nodded, but Jamie was about to venture a guess anyway. Well, there's an even bigger monkey wrench in the works, Alex concluded, and it's this: z is not an integer. At that, Ian and Jamie rolled their eyes and decided to go play darts. That leaves you to solve the puzzle. Enjoy!

The marbles in John's collection were of 13 different colors; each marble was of a single color. His sisters, Jane and Joanne, especially liked the red ones, and asked if they might have one each. Feeling generous, but not wanting to give them everything they asked, John let them each take one marble from the black velvet bag where he kept his collection, but he did not let them look inside. It is an even money bet that both sisters got their wish and pulled out a red marble. How many red marbles might have been in John's collection before his sisters chose? Edit: Of the 13 possible colors, obviously some were red. There may have been zero of some of the other 12 colors.

That would be the easy way. Nice going all.

I'm going to stick my head in a bucket of ice now

The blue H and T are the dots in the previous post - flips that didn't contribute to odd H plus T.

Impressive. Now lie down in a quiet place, put a cool washcloth on your forehead, and relax. No worry - that is not even close to the easy way. Or the hard way, for that matter. That is, there are [at least] two ways, one very quick, that are less cumbersome.

Interpreting an odd number of H followed by a T: H H T T H H T H T 9 flips H T 2 flips T T T H H T H H H H T T T H H H T 17 flips H H H H H H H H H H H H H H H T 16 flips T T T T T T T T H T 10 flips T T H H T T H H T T H H T T H H T T H H H T 22 flips At the point that you've "won" you have a T. Preceding that is an odd number of H Preceding that is either a T or nothing [you started with an odd number of H]

Here are some examples of what you want to have happen. It should clear up what the OP asks: Here the dots represent H or T but with the caveat that the dots don't contain the string of results you want. . . . . H H H T [8 tosses] . . . . . . . H H H H H T [13 tosses] H T [2 tosses]

I'm late to the game, but I had a free evening last night so I simulated the problem. All the results were near integers which suggested a direct calculation was in the works. Reading back through the thread I found EventHorizon did the calculation. Very nicely, btw, no infinite sums. FWIW, 3000000 trials each, with the desired number 0 1 2 3 4 or 5 steps away, gave average number of coin tosses as: 0 - 5.99899 / min = 2, max = 90 1 - 5.00305 / min = 1, max = 97 2 - 7.99791 / min = 2, max = 86 3 - 9.00089 / min = 3, max = 99 4 - 8.00015 / min = 2, max = 100 5 - 5.00985 / min = 1, max = 99 These sum to 41.0108 or average of 6.83514 coin tosses to reach a desired number after an initial randomizing spin. Differs by 0.0018 spins from 41/6.

Here's a simple question about probability and expectation. On average, how many times do you need to flip a fair coin before you have seen a run of an odd number of heads, followed by a tail? This puzzle was submitted, but not used, in the International Mathematical Olympiad twenty-five years ago. What makes it appealing for use as a test question [where time can be of the essence] is that it has an easy answer as well as a difficult answer. That is it tests for insight. As such it qualifies as a puzzle as well as a math question. Extra bragging rights go to the solver who solves it both ways.

Hi Striker, Sure it works, but so does this: "The prisoner in front of me has a white hat" "The prisoner in front of him has a white hat" "There aren't any more white hats after that until we've skipped 4 prisoners" etc. So you see that unless we restrict each prisoner simply to guess his own hat color, there's not much of a puzzle left. That means each prisoner just says "white" or "black" - nothing more or less. Hope to see some of your own puzzles posted here soon. Enjoy! - bn

I'm going to add my 2 cents at this point. I think that within the perimeters of the OP, all the salient analysis has been done, and the relevant conclusions have been reached. Congratulations to all. Peter Winkler found this gem of a paradox, surprisingly, from a question on an American High School Mathematics Exam back in the 1960s. The question simply asked, "If x^x^x^x^x^x .... = 2, what is x?" Questions of this type invite the student to show insight more than computational prowess. And in this case s/he was asked to see that removing the first x wouldn't change anything, so that x^2 = 2 could be written instead. All well and good. Until some genius kid noted the a similar substitution in x^x^x^x^x^x^x .... = 4 would produce x^4 = 4, and the same value of x namely 2^.5 produced different results. Translating this for BrainDen, I decided to ask it in two parts. the "=2" question in the OP, to get thinking locked into the notion of substituting by removing the first x, and in later post introduce the "=4" question to induce some head scratching. Which was sadistically satisfying to watch. I'm crediting armcie with raising the question of convergence. But I didn't declare it solved at that point. That would be cruel - to me of course - prolonging the scratching seemed like a good idea. So I glossed over the convergence issue [or tried to] by showing convergence [for x = 2^.5] to the value 2. But I didn't show [because it's not true] that it converged to the value 4. That was the part of the puzzle still to be solved - to find the range of x for which p(x) converged. And we finally got there, with contributions by many, you know who you are, and finally now very nicely by eventhorizon. The red herring in this delicious little gem of course is the equation x^y = y. Or x = y*(1/y). Thanks to Chuck Rampart for plotting it, and thereby fixing attention to it. And the search [hope?] for other magical integer solutions, given the nice result for 2. The imponderable was that plots of y(x) or x(y) can go far outside the region of p(x)'s convergence. Outside that region is meaningless for the purpose of this puzzle. Just as the student's notion that x^x^x^x^x^x^x .... =4 has a "solution." As we found here, the infinite exponential stack can never have a value greater than e. Yup, e's our man. The fact that x^4 = 4 does in fact have a solution [the red herring again] kept the discussion alive. The puzzle is solved, however, as soon as the range of x for which p(x) converges has been found, and anything outside that range is rejected. Everything outside that range is fair play in discussing the properties of x^y = y, but it's irrelevant for discussions of p(x). Anyway, kudos to all, and I hope it was fun. Here's Winkler's summary, and I quote: "The expression x^x^x^x^x ... is meaningful and equal to the lower root of x = y^(1/y) as long as 1 <= x <= e^(1/e). [sic] For x = e^(1/e) the expression is equal to e, but as soon as x exceeds e^(1/e) the sequence diverges to infinity. "This observation was first made by Leonhard Euler in 1778!"

The program returned an integer 2. That means it converged to a floating point value indistinguishable from 2. But it's not the computed p(x) value that's differing here, rather the value of the independent variable x for which the computed p(x) begins to oscillate between two different values. But your suggestion still demonstrates that the precision limitation involved in computing 2^.5 and 4^.25 did not deter finding the [exact] condition p(x) = 2 for those cases. I didn't think of that. The mystery [albeit a small one] persists.

The puzzle just asks for the shortest sequence of moves from the "solved" position to any "magic square" position