bonanova

In anagrams, you just rearrange the letters. Here are six historical anagrams, with clues. Have fun! VIEWING A STIR WESTERN VIDEO MAYOR'S FILM TYPE ECHO IN MUDDY LANE IF TUNES DO SO MUCH I AM A WEAKISH SPELLER One is a writer of some note, born in the 1560's One is a U. S. State - admitted in the 1860's One is a an old '50s tune - one of the most recorded songs of the 20th century. One is a '60s movie - set outside Salzburg Austria One is a unique singer - he made the scene in the '70s One is a Clint Eastwood movie - also from the '70s

Where's Alex? asked Davey. Dunno, said Ian, but I found this crumpled paper written with his red pen. Lemme see, said Jamie.

Blindfolds on? Ready .... Go!

For the fun of it I marked two quarters with 3 black sides and looked at 50 cases. 10 blacks -> 7 black on other side 10 blacks -> 10 black on other side 10 blacks -> 6 black on other side 10 blacks -> 6 black on other side 10 blacks -> 5 black on other side ======================== 50 blacks -> 34 black on other side => 68%

What if there were two contestants?

Amends ... have one on me. http://tbn0.google.com/images?q=tbn:AeD0ew...egalbeermug.jpg

I'll see your Newton's Third Law and raise you a Solar Powered Truck! Gas ........? What gas? Good one.

I'd love to hear your solution. But consider this, regarding mine.

Further clarification. This differs a little from the OP for clarity - it does not change the solution. Jamie is blindfolded. The four mugs are placed on the corners, randomly up or down. Each move, Jamie touches only two mugs [not all of them]. Here is what constitutes one move: Davey turns the lazy susan an arbitrary multiple of 90 degrees. Touching only the lazy susan, Jamie chooses any two adjacent corners or any two diagonally opposite corners. He picks up those two mugs. He puts them back on the same corners, individually flipped or unflipped, i.e. individually up or down, as he chooses. Jamie gets 5 moves. At the end of any move: if the mugs on the lazy susan are all up or all down, Alex rings the bell and Jamie wins.

Clarifying. Jamie is blindfolded. The mugs are placed on the four corners, randomly up or down. Move #1 - Jamie knows where the four corners of the lazy susan are and can pick up [1] two adjacent mugs - or - [2] two diagonal mugs Then he can feel whether they are up or down. He can flip either, neither, or both and replace them. Davey randomly turns the lazy susan to a new position. Jamie doesn't know what Davey did. Move #2 - Jamie [still blindfolded] does the same thing again. Davey turns the lazy susan again. Move #3 ... same. Move #4 ... same. Move #5 ... same. At this point [or before] all the mugs must be up or all must be down. At any time, if all the mugs are up or all the mugs are down, Alex rings the bell, and Jamie wins. Can Jamie ensure a win - without relying on luck?

It seems, said Alex the other night, that no one likes questions about odds. They're either too simple to care about or too subtle to believe. Ian, Davey and Jamie all agreed, and the four of them downed their ale. Which brings me to tonight's challenge, said Alex. It has nothing to do with odds -- just this little lazy susan thing. And he laid it on the center of the table, giving it a spin to show how it worked. Now here's the deal, boys. Each of us puts his mug on one corner of this lazy susan thing, upside down or not, doesn't matter. Then, one of you has to turn them all upside down or all right side up. There's only three simple rules: [1] You pick up any two mugs, feel them, and then put them back, whichever way you like - up or down - they don't have to be the same. [2] Then someone else turns the lazy susan to a new position each time, after the mugs are touched. [3] And ... you have to do it blindfolded. Whenever you get them all up or all down, I'll ring this bell here, and you win. But ... you must do it in 5 moves or fewer. Davey rolled his eyes and declined; so did Ian. Jamie thought for a long time and then agreed to try. Is there any way Jamie can ensure he will win the bet, without just being lucky?

Two caveats:

If you reason that because the only cards that can show a B face [the BB and BW cards] are drawn with equal likelihood the answer must be 50%, then try the experiment. Do precisely what Alex did, ignoring the WW card for simplicity. Make two cards: BB and BW. Place one from your pocket onto a table. [1] If it shows B, add 1 to your possible outcomes total. [2] If the reverse side is B, add 1 to your favorable outcomes total. Repeat until you have 30 possible outcomes. Now, are the favorables closer to 15? or to 20? If it's 17 or 18, do another 30. And maybe another 30. 1/2 or 2/3 will eventually come into sharp focus. And when it does, keep in mind that you drew the BB card 1/2 of the time! If I were a betting man, and I'm not, I'd bet an entire donut on 2/3. Why? Because of the equal likelihood requirement. You see a B face 100% of the time when the BB card is drawn but only 50% of the time the BW card is drawn. 2 cards x 2 faces = 4 equally likely events. 3 events show a B face. 2 events are favorable. Basically, you count the BB card twice - because it can show a B face two ways; and count the BW card once - it can show a B face only one way. Does that make sense?

No. And I hope he brought a parachute.

You make a valid point. I hope that doesn't sound condescending ... The answer depends on the premises of the set-up. The way you describe the set-up I agree the answer is 1/2. But look back at the story. I created a narrative that presupposed nothing except that a card was placed on the table. And then it was observed that the top face was black. It wasn't presupposed to be black. And if it doesn't matter which color it would have been, a card needn't even be drawn; only to ask the question: if I pull a card at random from my pocket and place it on the table ... what are the odds that the other side will match? Where the answer is clearly 2/3. Presupposing the color that's visible [as in your analysis] makes the answer 1/2. Part of the fun I get in constructing these stories is providing enough clues for an unambiguous answer. Sometimes I succeed , Sometimes I get it wrong.

Get Alex, Jamie and Davie to remove 2 or 3 glasses at a time, keeping things balanced at each step.

Intuitive probability fails when it compares favorable to possible outcomes that are not equally likely. I buy a lottery ticket. There are two outcomes - it's a winning ticket or it's not. One outcome is favorable. My odds of winning the lottery are 1/2. Ooops ... reality check! The times that this is really fun is when the faulty result is feasible. With the black and white card problem, you can easily make the cards and do the experiment say 30 times. A valid intuitive solution notes that 1/3 of the cards have opposite color on the other side. The probability of opposite color is 1/3 and same color [black or white doesn't matter] is 2/3.

I think you're considering each side as a separate entity ... That's it, exactly.

kewl.

If you're referring to the God I know, the answer is this. God cannot fail.

What if it was side 2 that was black?

It can't be.