bonanova

Customer: "How much is a large order of Fibonaccos?" Cashier: "It's the price of a small order plus the price of a medium order."

A wizard selects three excellent logicians and places hats on their heads. He explains to them he has written a positive integer on each hat, and that one of the numbers is the sum of the other two. Each logician can see only the numbers on the other two hats. A prize is offered to the first person able to be certain of the number on his own hat. The wizard starts questioning the logicians in order, starting over again if none of them can be certain of his number. There is no guessing. Each logician must answer: "My number is ___" or "I don't know." [1] Can any of the logicians win the prize? [2] If so, which one? [3] How many rounds of questions will it take?

A standard cubic fair die has 6 faces with equal probability of 1/6 for each face to show. Let die A be numbered 1 2 3 4 5 6. Let die B be numbered 1 1 2 4 6 7. Let die C be numbered 1 1 3 4 6 7. If A is rolled against B, in 16 cases A is larger; in 15 cases B is larger; in 5 cases it's a tie. If C is rolled against A, in 16 cases C is larger; in 15 cases A is larger; in 5 cases it's a tie. If C is rolled against B, in 15 cases C is larger; in 14 cases B is larger; in 7 cases it's a tie. Thus, A is said to beat B; C is said to beat A; and C is said to beat B. Now, Is it possible for the faces of three dice D, E,and F to be numbered in a way that: D beats E; E beats F; and F beats D? [Edited to make it clear we're talking about three new dice.]

Good try Nikyma, but I think you got tangled up in Question 2.

A number is said to be rotated when its leftmost digit is moved to the right end, and the other digits shift to the left. 3117 becomes 1173, 1731, 7311 and finally 3117 again, on successive rotations. It turns out that the Centigrade representations of certain temperatures are the rotation of the Farenheit representations. For example, if 413 degrees F were 134 degrees C - but it's not. What is the lowest temperature for which this is true?

Pinning star on itachi-san's lapel....

Nikyma has it.

1. The first question with B as the correct answer is: A. 1 B. 4 C. 3 D. 2 2. The answer to Question 4 is: A. D B. A C. B D. C 3. The answer to Question 1 is: A. D B. C C. B D. A 4. The number of questions which have D as the correct answer is: A. 3 B. 2 C. 1 D. 0 5. The number of questions which have B as the correct answer is: A. 0 B. 2 C. 3 D. 1

If you know a little about the game of Clue, you can solve this. In four successive games of Clue the following statements became true: [1] In one game Miss Scarlet used the wrench, but not in the Library. [2] In one game the rope was used in the study, but not by Colonel Mustard. [3] In one game the gun was used in the conservatory [4] In one game Professor Plum was the culprit, but not in the Library [5] In one game Colonel Mustard was the evildoer, but not in the conservatory [6] In one game Mrs. White was the perp, but she does not like ropes. [7] In one game the lead pipe was the weapon, used perhaps but not necessarily in the kitchen. Name the culprit / weapon / location combinations for the four games.

202 122 232 425 262 728 ==?== ==?==

Sure you can. The expected number of couples is an expression involving R and B. Here's a thought question as you think about the solution: Does the expected number of couples double if you double both B and R? If you want numbers, suppose R=3 and B=4. Or another: R=6 and B=10. If you get these answers, you probably know the general formula. Edit: R+B is even.

Suppose you have a bag with R red marbles and B blue marbles. R + B is even. You reach in and pull out two marbles and place them side by side. You continue pulling pairs until all the marbles have been taken from the bag. What is the expected number of Red-Blue "couples"? Edit - removing spoilers for the moment ... Edit - to make it clear the couples have one Red and one Blue marble.

Hi Tink, Check your math in red above. That may help. Or, look at it this way. Of 12 red balls, none are transparent; 5 are glass. Of 18 balls that are not red, there are the other 8 glass and all 14 transparent. Does that help?

I'm thinking itachi-san has it,

Great job, all ... ! Hands calculated so far: By Dackombe: Total hands - 2 598 960 Straight Flush - 40 Four of a Kind - 624 Full House - 3 744 By andreay: Three of a Kind - 54 912 Two Pair - 123 552 Pair - 1 098 240 High Card - 1 317 888 <- very close, but slightly off. Seventh Sage is correct about the most difficult one to calculate. High Card. Three numbers left to find. Flush - fairly easy Straight - easier High Card - hardest What we know is that these three must total 2598960 - [40+624+3744+54912+123552+1098240] = 1 417 848.

Dackombe has the denominator. Welcome to the Den! The numerators are the challenge, for the reasons he mentioned. Hint: you can discuss the probabilities one at a time, as if it were eight puzzles. You might be surprised which one is the most complicated to describe and calculate. Have fun with this. To me, the interesting part is that it's fairly easy to look at five cards and not confuse 4 of a kind with 2 pairs for example. But if I'm a little bit sloppy with my math, I might count it that way. Or even worse, count 4 of a kind simply as a pair. What's needed is to describe what all 5 cards must or must not be.

Commemorating the post that matches my member number, let's try a bit of useless [this side of Vegas at least] calculations. In a standard deck of cards there are 52 cards: 1-10, J Q K of four suits S H D C. In draw poker, 5 cards are dealt to each player. Hands are valued in increasing degree as follows. Edit: Examples of each type of hand shown in red. 0 - High card: Highest ranking card. This hand is none of the following types. - 3D 6H 8C JH KD 1 - Pair: 2 cards of same rank; 3 cards of different ranks - 4C 4D x x x 2 - Two Pair: 2 cards of one rank, 2 cards of another rank, 1 card of different rank. - 6H 6S 9D 9H x 3 - Three of a kind: 3 cards of same rank; 2 cards of different ranks. - KC KD KS x x 4 - Straight: 5 cards in rank sequence, but not all of same suit. - 3D 4D 5C 6H 7S 5 - Flush: 5 cards all of the same suit, but not all in sequence. - 3H 6H 7H 9H JH 6 - Full House: 3 cards of one rank; 2 cards of another rank. - AH AC AS 8S 8H 7 - Four of a kind: 4 cards of same rank. 7S 7H 7D 7C x 8 - Straight flush: 5 cards in sequence and all of same suit. - 4H 5H 6H 7H 8H. Prove that the valuation of 0-8 is reasonable by computing the probabilities of being dealt each type of hand. Express the probabilities as fractions where the denominator is the number of possible unique 5-card hands dealt from a single deck. Easy check - they add up to unity. The prize for being first with all correct answers is 10 "attaboy"s. The current value of an "attaboy" is that you may combine 100 of them with $0.85 for a cup of coffee at Dunkin Donuts.

Stop groveling ... I didn't mean anything negative by my comment, only that we both have to think a little more about this. Writersblock is pretty shrewd at times, he may have us going for awhile. And we apparently did note the same pattern. I'm stumped for the moment ... hitting the sack.

grey cells, it wasn't correct when I posted it, so I'm guessing it's not correct now either. But it's nice to know two great minds think alike ... !

Hey my friend, LTNS. This one's a challenge. My first answer does not fit the clue so I'll keep at it:

I'll assume that the decimal point has been ruled out - being a symbol. However, if it were allowed, there'd be 27 additional answers.

The OP hasn't logged on since 22nd August 2007 - 05:50 PM Looks like you're free to choose the best answer from those already given.

A word of guidance: When criticizing a puzzle, or insisting that one answer must be right, or another answer can't be right, try to keep in mind exactly what a puzzle is intended to be ... It's in the nature of a puzzle to place an answer out of plain sight. The puzzle solver's task is to determine where it could be, so long as a provision of the problem is not explicitly violated. In other words, don't treat a puzzle as if it were a blueprint or an engineering specification; treat it as a challenge: find an answer. If the OP were to have made it absolutely clear that the character string i-n-c-o-r-r-e-c-t-l-y was meant instead of the word itself, there would be no puzzle. Now try this one: Is one of the words in this sentence misspelled? The dictionary would say No. The puzzle maker would say Yes.