bonanova

Cheers! Kewl. But do you take the bet?

Nope. No such assumption. They can be all black, all white, anything in between.

So I flew to Las Vegas and taught the pros out there a thing or two, boasted Alex last night at Morty's. His friends came over to hear the details. So what did you play? asked Davey, in only a slightly skeptical tone. Well, it could have been anything, was the reply. Since I've got the whole gambling thing down to a science, it didn't really matter. I passed the roulette table first, so I played there the whole time. Ian knew something about that game. For example, the ball can drop in slots numbered 1-36, but there are two other slots, 0 and 00, which usually belong to the house. If you bet $1 on a number - 5 for example - the payoff is $36, but your odds of winning are 1/38; and your average loss is $1 - (1/38)x$36 = $1/19, or about a nickel. You can't win, he said. If you play all night, the house will always beat you! Ah, and that would be the case for lesser mortals, for sure, Alex beamed. Myself, well, I won't say how I made out, that would spoil things. Instead, I'll make you a wager. Here, take a look at this. And he unwrapped the large box that lay at his feet. A genuine Las Vegas Roulette Wheel now lay on the table before them! Here's the deal, boys. I've got a stack of 105 $1 bills - just like I had last week in the casino. I'll choose a number - say 18 - and bet each of these dollars on that number, - 105 spins of the wheel - until they've all been placed into the hands of chance. But here's where the fun begins. I'll wager you $200 - even odds - that at the end of the 105 spins I'll be money ahead. If I'm not, you win the $200. If I'm ahead, I win the $200. What do you say? Any takers? A moment passed, and neither of the two answered. OK then, and just because I have a generous nature, I'll give you a bonus. You'll be the house for the 105 bets. You get the house odds on all my bets, and if you come out ahead, you get an extra $200. Davey began to stroke his beard while Ian couldn't believe what he'd heard. Alex had finally lost it. Before Davey could answer, he called out, I'll take that bet! Did Ian go home a happy man? or did he leave poorer but wiser? Would you have taken the bet?

Thanks. Good to know that. It was also on an exam at MIT. So it's been around, as it should - it's a good one. It can be calculated. But lateral thinking came to my rescue when it was posed to me sometime around 1982 by an MIT grad and colleage who'd had to answer it as an undergrad.

The answer is an odd number, spelled out. Like thirteen. Only that's not it, because, for example, eighty-five comes before thirteen in the dictionary. Of the first 10 billion integers, which one, when spelled out, is odd and comes before all other odd integers less than 10 billion?

Not scrambled. Numbers are the count of pieces of each length - 3 pieces of length 1 for example. They add up to 36 = 12 words x 3 pieces / word. Use each of them once and only once. If something is repeated [ there are two E's ] you use them both once.

Each of the dozen words was divided into three segments. They are listed for you, sorted alphabetically and by length. The letters in each piece are contiguous and in order. For example, if one clue was "high places" and three of the pieces were PS RO OFTO you could make RO.OFTO.PS. And you'd use the 2nd letter O along with the 2nd letters of the other 11 words to find tabby's song. OK? Take three of the pieces and make a word. Take three more and make another word. When you've used up the last three, you'll have a dozen words. The words you have to find are described in the first part of the OP. One is an adjective, two are maladies, one is a musical instrement, and so on. If they don't fit that list, you've not found the words that I divided. But when you've done all that, take the 2nd letters of the dozen words and find a song tabby would sing.

Hi biosteve, What's unrealistic? A plane flies from point A to point B and returns. And, if the plane flies backwards instead of forwards, then it returns backwards instead of forwards. [1] There's no wind. It takes t1 minutes. [2] There's a wind. It takes t2 minutes. Compare t2 to t1. What's your answer?

It's Saturday night and being a puzzle writer I have no social life. So I make a list of a dozen things: a conference a multiplier a musical instrument a person who celebrates a holiday twice a superlative means of communicating some clothes some fibers some names someone who hides people who hide two maladies I lazily let my pencil wander down through the list, twice, thus dividing each word into three pieces of lengths 1, 2, 3, 4 and 6: E E T [3] AD ES ES ES FI MA MO OW PO RS SA WW [11] AYS CRI CZE GRA LAN NDD OLD ROU STO VEF WAR WAW XOP [14] DROB GHAG GUAG HONE NIKE STOM SPES [7] ACHACH [1] Finally, I notice that if I take the 2nd letter of each word, I get some music that tabby might sing. Name the tune.

A sufficient condition is for x1 consecutive numbers to exist. The first time that happens [the lowest x1 consecutive numbers] determines the least Y. In this case the lowest 6 numbers are 43 44 45 46 47 48 - any higher number is one of these plus a multiple of 6. How you construct Y [not just find it] is an interesting question - depends at least on whether the x's have common factors.

Everything hits the ground eventually. Point is, the triple play was complete first.

Think of a napkin ring - of some diameter - that's 6" high.

Nice job. This holds for collinear winds [as the OP asked]. Can you generalize for cross winds [at angle x] as well?

MaestroOD and qwert have it ... nice going. Grey Cells - welcome back!

Why is it indeterminate? What does L'Hôpital say about this? Is 10% of infinity not 10% just the same? What percent of all integers are even? Is that indeterminate? Alex gave an understandable and meaningful set of choices ... which one is right?

You're a magician, and you've just learned the neatest trick. The trick begins with a number of bags, each containing a different number of marbles. You have no bags or marbles, so you visit the local magic store. There you buy 15 black velvet bags. How many marbles do you need to buy if you use all 15 bags in your trick? Remember, this is a magic trick

Catbert, apologies for the time interval preceding the clarifying edit. My bad; altho I don't think it was more than a few minutes. Out of curiosity, if blanks are included, and if blanks precede a's in the sort, is Catbert's answer correct?

edit to show answer as a spoiler ops Hi Karl Mac, and welcome to BrainDen - you have it. But leap year is irrelevant.

Thanks! I have no chance of solving this, but - have a great trip, and don't drink the water!

Octopuppy has it. His method works for even or odd prisoners and can be extended for hats of k different colors, as well. So much for the sadistic Warden Smith.

Duh Puck rings the bell ... nice one! Either one works. Anyone wanna say why?

And remember that the 7th line of the OP says ... [Edit: spaces and hyphens are ignored.] I'm glad you people have had so much fun while I had a long night's sleep. Usually I'm up by the crack of noon, but today I slept in a bit. Reading the posts so far, it looks like ... normdeplume - good to see you back.

Right. Your earlier answer was better, and ten billion is the largest number you wrote out.