bonanova

A is true. [probability is not mentioned or estimated or inferred in the statement.] B is true. [probability is not mentioned or estimated or inferred in the statement.] Therefore May's chance of winning is C.

Excellent question. Where do you get all these questions ... ??? All I can say is, I recall a day when a regular malt was $0.25; with extra scoop $0.35. Maybe my puzzles are getting outdated like me.

Good start. But consider that the probability of getting a 2nd roll is 35/36 - a pretty good chance.

Yes. You can assume, as the OP says, p[12] = 1/36 and p[7] = 1/6.

Here. Can you sometimes do better? Yes. (4/3)x takes you from inscribed to circumscribed circles in a hex tiling. Can you always do better? No. Not in case 2.

The moderator checks his NY resolution list. Yes - there it is ... number seven: Do a search before posting ... even in the Jokes forum. Locking his own thread ...

The puzzle is cleaner if it simply asks the probability that May wins. My bad, that's the way I intended to word it. The entire problem has been calculated by MrAscii. fleminator has offered way of looking at the puzzle that has a natural extension to N and N+1 coins. Can the proof be reduced to two statements of certainty, i.e. that don't reference probability at all?

Until yesterday I had only one blonde joke worth telling. Now I have two: Sally, the blonde, calls her boyfriend one afternoon to ask for help. Please come over, Joey, I have a killer jigsaw puzzle here, and I'm just not making any progress at all. What's it supposed to be when it's finished, he asked. Well according to the picture on the box, it's a rooster, she answered. Joey agrees to help, and a few minutes later he's at the door. Sally lets him in and shows him where she has the puzzle spread all over the kitchen table. He studies the pieces for a moment, then looks at the box, then turns to her and says, This is going to be harder than I thought. I have an idea. Let's take a break, pour a couple drinks, sit down, and relax a bit. OK, Sally agreed, but what will we do after that? Then, he said, we'll put all the corn flakes back into the box.

May said to Moe, take all the coins out of your pocket, and I'll get all the coins I have in my purse. We'll throw them all, at the same time, and whoever gets more heads buys the other a chocolate malt, OK? Moe agreed, but was disappointed to find that May had 6 coins to his 5. What's the probability that May won the bet?

Curious about this place called Morty's, mentioned more than once here at the Den, you wandered in last night to meet the boys and have a cool one. You must not have blended too well, because Alex saw you coming and was striking up a conversation before you could say Sam Adams. Ya look like a smart lad who likes a challenge, he began. I've got a pair of dice here, and I'm sure you know the chances of rolling a 7 are 1/6, but to roll a 12 they are only 1/36. Well, then; even up, I'm willing to bet you'll roll a 12 before you roll two 7's in a row. Keep in mind two dice are thrown each roll. It sounded like an even bet; after all, 1/6 x 1/6 = 1/36. Do you take the bet?

Hi Delfian, welcome to the Den. Check out old puzzles, not ones you've written, on search. Posted before.

Cute, but don't build yer house using them... <_<

Yes.

While the Breakfast Club was detained in the school library, one of them started a small fire. When Mr. Vernon questioned them, he heard the following replies: Allison Reynolds: It was Andrew or Brian Andrew Clark: Neither John nor I did it. Brian Johnson: Both of you are lying. Claire Standish: No, Brian; Allison or Andrew is telling the truth. John Bender: No, Claire; that is not true. Mr. Vernon knows that two of them cannot be trusted. Believing the others, who does he conclude is the fire-starter?

Take a cube with 5" edges, and paint it blue. Then slice it mathematically [zero saw blade thickness] into 125 unit cubes. Throw away any cube that has no blue faces, and construct a rectangular brick with an all-blue exterior. How large [volume] can that brick be?

Five points lie on a line. When the 10 distances between each pair of them are computed and listed from smallest to largest, we obtain 2, 4, 5, 7, 8, k, 13, 15, 17, 19. What is the value of k?

Let's get my definition of "formal validity" out into the sunlight, for understanding and discussion. Then I'll invite you to do the same. Let x = "This statement is false" be an element of the set S of all self-referential statements of the form "This statement is B". Let y = "This statement is true" be another element of S. Let T be the property of Truth. Let F be the property of Falseness. Then x becomes x = F(x). and y becomes y = T(y). Then, because F and T are formally defined over the set S, and x and y are elements of S, both statement are formally valid. y is formally valid and decidable: T(y). x is formally valid, but not decidable: F(x) leads to ^x or ^F(x). So we cannot conclude F(x), and we also cannot conclude T(x). x is formally valid and undecidable. --------------------- In the realm of arithmetic, Let p = "1 + 1 = 1 + i" be an element of the set E of equations of the type A = B where A and B are expressions. Let q = "1 + 1 = 1 + 1" be another instance And r = "1 + = x 4 - x 3 - / i 13 / = 45 / -" be considered as be a third. Let F and T be the properties defined above over the set E. p is formally valid and decidable: F(p). q is formally valid, and decidable: T(q) r is not of the form A=B [formally invalid]; it's not an element of E. Therefore F and T do not formally apply to r. r cannot be discussed with respect to F and T. Now you have my definition of formal validity. In this case it means + and x and / and - are binary operators, taking prefix and postfix arguments. It means elementary terms, 1, i, 4, 3, 13, 45 are proper arguments of these binary operators. It means = is a binary relationship taking two numbers or expressions and asserting their equivalent value. p and q follow these formal rules. One happens to be false, the other true. r does not; it is therefore formally invalid. It's not the case that r is undecidable with respect to truth: T(r) and F(r) don't apply. From this, see that r and x [above] are not similar. In a previous post, you said in effect that because r is invalid, so is x. Since my definition of valid does not permit this, let me ask for a definition that does: What's your definition of validity? You seem to define valid as: if F(a), then a is invalid. Is that how you define valid: if a is false then it's invalid? Why use two terms [false and invalid] confusingly, in different ways and for different purposes [false means not true and invalid means we can dismiss it] that in the end have the same definition?

Beat me to it.

Nice puzzle JiYoonIt, and welcome to the Den. Since it's been posted before here, and again with a varation, further discussion should be on those threads.

Yeah, I think the pieces are all there. Great work. Choose one ... Puzzles decrease industrial productivity and distract students from study. Puzzles are like a workout: nothing useful is done, but useful work becomes easier. Here are two nice solutions, for which I do not claim credit. [Guy Kindler and Peter Winkler] A table the size of the original is covered with 4x plates.