bonanova

Ian waved to Jamie and approached the bar. What's wrong with Alex? he said. He's sitting in the corner fumbling in his pockets and talking to himself. I think he's had a few too many, Jamie replied. Plus, a friend gave him a birthday present - a puzzle - and it has him stumped. You remember the friend, the one with two children, one being a boy, who kept asking the probability the other's a girl? Anyway, he's been there a while. Let's see if we can help. So I've got three cards, Alex explained... * a black card - black on both sides, * a white card - white on both sides, and * a mixed card - black on one side and white on the other. I pull one at random from my pocket, and place it on a table, like this. There. As you can see, the side facing up is black. Now what are the odds that the other side is black, also? After thinking a moment - and downing a couple O'Doul's, Ian smiled and whispered his answer to Jamie. What would you have said?

Explanation [i think this post was lost over the weekend]

Happy Birthday to me ... HBTM ... etc sang Alex as he strolled into Morty's last night and then announced, Boys, I've got a real treat for ya tonight! First, take a look at this. And he hung from the ceiling a complicated, interconnected set of five scales, whose 10 balance pans he'd labeled A, B, C, ... H, I, J. Now the barkeep gave me one request, bein' it's my birthday on Saturday, and all. He's gonna draw out fifty-five cold pints. Then, if we can place a different number of pints into each of these balance pans, we'll have a real party, cuz the drinks will be on the house! Every pan gets a pint to start with, and all five scales have to balance. Alright boys, let's get at it ... we have until midnight to get this done. Otherwise we pay for the drinks. Feel free to help Alex, Jamie, Davey and the boys celebrate! Edit for clarity The marks on the bars are equally spaced.

Barkeep, A pint for the smiley one here!

You are the director for the annual Open city-wide Men's Singles Tournament. So you're in charge of drawing up the brackets and seeding the entries. Your job will be easiest if the entrants number a power of two. As luck would have it, 97 locals sign up. But at least they all have distinct rankings, so the seeding is straightforward; and, since it's a single-elimination format, there is no losers bracket. But since there's an odd number of entrants, you'll need some play-in matches to start, and maybe some byes for the higher seeds. Eventually you can reduce the number of surviving players to a power of two, and the rest of the pairings are straightforward. You spend a sleepless night working on it and finally come up with a set of brackets that work. How many matches are in the brackets that you drew up?

Alex hadn't been to Morty's for a few days, because, it was later learned, he was home trying to get brainden to come up on his aging computer. When Jamie called to see if he was OK, he told Alex it was down for maintenance, and the boys missed him. And so, even tho a little red-faced [ ], Alex trudged in last night. I've been thinking, he said, about using this old watch chain in my pocket for this Friday night's poker game. And he held up a tiny gold chain with 21 links in it. OOOOOOOOOOOOOOOOOOOOO The problem is that I want to be able to make smaller groups of links. Look here, I'll show you what I mean. Suppose I were to cut the 8th link. And he sketched this picture: OOOOOOO C OOOOOOOOOOOOO [making chains of length 7 and 13 links, and a single, cut link] Then I could make bets of 1, 7, 8, 13, 14, 20 or 21 links. But I want to be able to bet any number of links, from 1 to 21. Ever since last Thursday I've been thinking about how to do that cutting as few links as possible. Now, I'll buy a pint for the first one of ya that can match my best idea. Tell me how many links to cut, and which ones.

Precisely in accord with the OP. A tolerant person encounters a person who possesses the implicit quality of intolerance. A person who already possesses a quality would hardly be said to encounter it [within himself]. You may say "feel", but the OP said "encounter". ...mumbles to himself: One thing I can't tolerate is people who can't tolerate my perfect English usage ....

Good one ...

Apologies for the friggin bad clues. Here it is.

you are contradicting yourself in the question. How can you encounter intolerance if you are tolerant at the first place? By encountering people. Not all people in the world are tolerant people. A tolerant person might meet a person that is not tolerant. And it may be impossible to simply walk away. For example [a] business colleague family member

How would he face intolerance?

I am a tolerant person. What should be my reaction when I encounter intolerance? [1] Tolerate it, since I am a tolerant person; thus condoning something I do not believe in. [2] Not tolerate it, since it goes against my beliefs; thus ceasing to be a tolerant person.

Clue summary

Not bad. Right track, but doesn't answer to all the clues, including now this one. Use only one arithmetic function - [the simplest one] Then there's one final step. This is fun ...

And the answer is ...

Maybe ...

OP meant to imply the answer was a single-digit number. Here are the 16 numbers you need to find the answer. Remember to think in groups of four.

Wow. Looks interesting. Now if I just had another life to spend on this.

A spoiler is buried in the OP. It may identify the number, but to win you have to find it from the grid results.

hints

Guys guys, I think this math stuff is going crazy. You guys are saying it is 100% to contain a number at least one digit 3. Well, lets look it this way. What about a number to contain 3 OR 4. According to the argument it is going to be 200%. No, because a number can contain a 3 AND 4. That is very funny ... because there are infinite numbers that contains neither 3 nor 4. (Hope no one will ask for an example). An example is easy, mentioned in a previous post. The sequence of numbers 1, 11, 111, 1111, 11111, 111111, 1111111, ... is infinite and contains no 3's or 4's. Still, the fraction of all numbers that contain a 3, or that contain a 4, or that contain any specified digit is 1 - [.9]**N where N is the length of the number. As N -> infinity, [.9]**N -> 0 and the fraction -> 1. By the same reasoning, the fraction of all numbers that contain all the digits, 0-9, is also 1. Mind bending, isn't it?

I know what No Limit and Pot Limit mean, what does High Stakes mean? i.e., how much of their stake is a player allowed to bet on each hand? Are we told each player's starting stake? If it's $1000, say, then all players save one could be wiped out on the first deal. If the stakes are unlimited, so that no one is ever wiped out, then the single player among the ten with the most chips is the winner and all others, still having chips, only fewer, are the losers. Is that the question? Which has the most chips after 1000 hands? Nobody ever goes All In?