bonanova

The easy way is to suppose the answer is the same for any sphere [with diameter not less than 6 inches], and calculate the answer for a 6-inch diameter sphere. The answer is 36pi cubic inches - exactly the volume of the sphere. The hard way, and why the easy way works, can be found earlier in this thread.

Yah, but now CR is gonna get all over you for counting ace pairs instead of counting decks. See post 17. Courtesy of a brilliant colleague, the proper calculation goes something like this: Imagine a well shuffled deck in this way: ......A...[k]...A...[l]...A...[m]...A...[j]... where the dots are non Ace cards. i, j, k, l and m are the numbers of non-Ace cards around the four Aces. Consider the case of zero Ace pairs: at least one card separating the Aces -- k, l and m are nonzero. How many ways can that happen? N[zero ace pairs] = sum(i=0,45) [ sum(j=0,45-i) [ sum(k=1,46-i-j) 47-i-j-k ]] since there are 47-i-j-k ways we can assign l and m. Permitting ace pairs simply means allowing k, l and m to be zero: N[permit ace pairs] = sum(i=0,48) [ sum(j=0,48-i) [ sum(k=0,48-i-j) 49-i-j-k ]] - now there are 49 ways to assign l and m. p[at least one ace pair] = 1 - p[no ace pairs] = 1 - N[no ace pairs]/N[permit ace pairs] p[at least one ace pair] = 1-(C(49,4)/C(52,4)) = 1201/5525 = .217... which is a little less than 3/13. And .217... agrees with my 10-million simulated shuffle results. See post #17.

Hi Duh Puck, I mentioned your objection to Alex this morning. Always the mind messer [well, mess isn't precisely the appropriate word] he said, A stool that doesn't exist isn't built like an airplane just cuz it doesn't exist. I think he was trying to suggest that a nonexistent stool could have four nonexistent legs. And then he walked away, kind of muttering to himself. Hope that helps.

Yeah, explain that for us...

So, what's he smiling for, all smug like? Jamie wondered. He said he's goin' to mess with our minds tonight, Davey replied, but I'll tell you, he's not gonna fool me. I'm pretty much onto his tricks by now. The subject tonight is logic, Alex began, and as usual I'll keep it nice and simple so you all can understand. That caught Ian's attention, and now the three of them were ready to take another challenge. Morty's was never so lively as when the boys squared off in a battle of wits. Tell me something that's true, Alex began, true today, true tomorrow, true here, true down the street, true without regard for time and place in any way. Davey scratched his beard, as he often did, to give the appearance of deep thought. But Jamie chimed in first. Well, take that stool yer sittin' on, he said, it's got four legs. Four today, four tomorrow, four where ever you take it. The number of legs on that stool is four. There's a true statement for you. Excellent! cried Alex. Now let's talk a little bit about contradictions. How would you say just the opposite of that? What statement could you make that absolutely can't be true if that statement is true? Can ya answer me that? Ian was all over this one. C'mon, give us a challenge will ya? It's simple! A statement that can't be true is just the opposite of what Jamie said. The number of legs on that stool is NOT four. You lads bring tears to me eyes, ye really do, continued Alex, you're exactly right. There's no way those two statements can ever both be true - no place, and no time. One has got to be true, and the other has got to be false. By this time Davey was wondering why he had bothered to scratch his beard. Ya know, so far I'd have to say I'm not really overwhelmed with the depth of this discussion. Have ya got something for us, or not? And he looked wistfully across the room at the dartboard. I was just getting to it, Alex said. It's this: I'll put up a pound against a month of drinks and tell you that I have a statement that's true, just like Jamey's, any time, any place. But it's different. Cuz when you make the contradictory statement like Ian-boy here did, well, that statement is just as true as the first! Now here's the challenge. You can take my bet, and end up buying my cold ones for a month, or ... you can come up with a true statement of your own, whose contradiction is every bit as true, and I'll do the buyin'. The beard scratching resumed - now in earnest. And Davey suggested quietly that a sporting man just might drop a tiny clue. Alex smiled, compassionately, and said, OK then, the statement goes like this. The number .......... is nine. And the contradiction goes like this: The number .......... is NOT nine. And the part where the dots are is exactly the same in both statements. For you Yankee fans, it's not about spring training.... Davey slumped down into his chair, motionless. Ian quietly moved back to the bar. Jamie closed his eyes for a moment and thought. Yes, he said, I think I have it! What was Jamie's idea? Would you have taken the bet? ---------- "Use the spoiler, Luke."

sunburntpixY, Thanks for the inquiry to my puzzle. I don't have a diagram that i can post. But this may be helpful... It will be helpful to visualize the following ideas if you draw a circle on a piece of paper and think of it as a cross section of a sphere. The center of the circle is the center of the sphere. Now draw parallel lines -- equal distances from, and either side of, the center of the circle. Make sure the lines go all the way through the circle. The lines represent a cross section of a hole drilled through the sphere. The distance between the lines is the diameter of the drill. The area between the lines is a cross section of the hole. The area outside the lines but inside the circle is a cross section of the portion of the sphere that remains after the hole is drilled. And, most importantly to understand, the length of the lines between the points that they intersect the circle, from end to end, is the length of the hole. Note that the length of the hole is less than the diameter of the circle [sphere]. Repeat. The length of the hole is not the same thing as the diameter of the sphere. Finally, note that if you had drawn the lines farther apart -- go ahead and draw another pair of lines that are farther apart -- the length of the hole decreases. OK - now think through these statements. [1] For there to be anything left of the sphere, the diameter of the drill better be less than the diameter of the sphere. [2] For the hole to be 6 inches long [length is measured in the direction the drill traveled while it was making the hole] the sphere had to have been at least 6 inches in diameter. You can't drill a 6-inch hole through a 4-inch-diameter sphere. But you can drill a 6-inch hole through a 7-inch-diameter sphere. [3] In all cases, we are talking about drilling a hole completely through the sphere. You can look through from beginning of the hole to its end, and see all the way through the sphere. You could string the sphere onto a rope, so long as the rope is thinner than the drill, of course. And, in all cases, what we mean by the length of the hole is the distance between the beginning and the end of the hole. You have to see this length as being different from the diameter of the original sphere. [3] If the diameter of the drill is negligibly small, a negligible amount of the sphere is removed, and the remaining volume is essentially the original volume: [4pi/3]r^3 and the length of the hole is essentially the diameter of the sphere. Only in this case is the length of the hole the same as the diameter of the sphere. Think of a pearl, with the tiniest of holes, string on the slenderest of threads. In the case of a vanishingly small [diameter] drill, the length of the hole approaches the diameter of the sphere, and the remaining volume approaches the initial volume. [4] If the diameter of the drill is negligibly less than the diameter of the sphere, the remaining portion of the sphere is a tiny band of material at the "equator" of the sphere - think of the drill entering the sphere at its "North Pole" and exiting the sphere at its "South Pole", and the length of the hole - the height of the remaining volume - is now much less than the diameter of the original sphere. In the case of a vanishingly small difference between the sphere's diameter and the drill's diameter, the remaining volume approaches zero - the entire sphere has been drilled away - and the length of the hole [obviously] approaches zero as well. This would be the case if you drew your two lines tangent to your circle instead of intersecting the circle. If you really understand this point, you can see how a 6-inch hole could be drilled [somewhat disastrously to be sure] through the Earth! Remember, the length of the hole is defined as the height of the remaining portion of the sphere, not as the length of the drill. Or think of it this way: if you crawled inside the hole and painted the inside surface of the hole, the painted surface would be a circular cylinder. The length of that cylinder is the length of the hole. Now if you can visualize all of that, you understand what "length of the hole" and "remaining volume" mean. The puzzle is cryptically worded. Intentionally. It's meant to make the solver think through all of these things so that the question is even understood. That's part of the challenge of the puzzle. If you can visualize the "length of the hole" then the question at least is understandable: If you drill a 6-inch hole through a sphere, what is the volume of the remaining portion of the sphere? And, if you've read this thread, you've heard it asserted that in every case - for every sphere whose diameter is at least 6 inches - the remaining volume is the same. A person with calculus skills can compute that volume and see that the answer can be expressed in terms of the length of the hole - only. An amazing result! But another person - one who does not like tedious calculations - simply takes note of the fact that an answer has been requested even though the original diameter of the sphere is not given. The only way that can be a reasonable request is if the answer does not depend on the original diameter. The person who reaches that conclusion smiles, rubs his/her hands together, and computes the answer for the simplest case: a 6-inch-diameter sphere. Remember, we said that's the case where the hole has zero diameter and thus removes zero volume, so that the remaining volume is the original volume: For a 6-inch diameter sphere, V = [4pi/3]r^3 = 36pi cubic inches. And that's the correct answer. Hope that makes this thread easier to follow. You're not alone in asking about what it means. - bn

By definition of prime number the square of any prime number is divisible by itself, the prime number and 1.

It's not footballs - they use leather.

A person dies unless he speaks the color of his own hat. Saying the color of the hat of someone else could get him killed...

Can't believe you'd post a shaggy dog puzzle ...

Let me try. I can easily show that 0/0 can be seven: 7x/x = 7. Let x->0. 7x->0 and x->0. 0/0 = 7. If 0^0 and 0/0 are identical, then it can be shown that 0^0 can equal 7. Care to try?

Hint: Take a long look at the number 9.86 - you'll recognize it as pi2. Put pi [and N] into the denominator of my fraction. OK now? http://brainden.com/forum/public/style_emoticons/#EMO_DIR#/wink.gif

You da man ... !

I can't respond to that ... INS may monitor these boards. APL. You prob never heard of it. Very handy, not free.

Can we assume that the road from A to B has a constant grade?

Hi Chuck, I'll meet you halfway on this one. Multiplying by 51 is entirely correct in that it gives the probability of Spade ace followed by Heart ace anywhere in the deck. And it works to multiply this result by 2 to have those two suits in either order. When we limit things to those two suits, there is no possibility of double counting - there are no other simultaneously occurring events of relevance. But when it's opened up to the other two suits, multiplying by 12 does allow multiple counting. These are the multiple-counting cases: [1] .....AA.....AA..... [two occurrences] [2] ....AAA.....A...... [two occurrences] [3] ......AAAA......... [three occurrences] I've tinkered with enumerating these cases to get a closed form solution. While I'm doing that I computed 10 million shuffles, which provides a fair degree of confidence in the results. Out of 10 million shuffles, 7825733 had 0; 2041909 had 1; 130578 had 2, and 1780 had 3 consecutive-ace occurrences [CAOs]. Summing the CAOs gives 2308409 for a CAO probability of .23084..., which agrees with 12/52 = .23077... Summing the decks that had at least one CAO gives 2174267 for a probability of .2174... So the number of decks that have at least one CAO is about 6% fewer than the number of CAOs. That is, ignoring the multiple occurrences over-counts by about 6%. Thanks for a lively discussion, and apologies for not seeing the flaw in my thinking until now.

x^y is well defined everywhere except the origin. That is, [limit as x,y -> 0] x^y is not well defined - it depends on how x and y approach 0, or which one gets there first. But that does not mean 0^0 can have any value whatsoever, the way 0/0 can: For example, [any_value]x=0 -> [any_value]=0/x. let x->0. [any_value]=0/0. The 0^0 case is more well behaved. [1] [limit x->0] x^0 = 1. [2] [limit y->0] 0^y = 0. Look more closely at case [1] and invoke the equation x^0 = x/x: [limit x->0] x^0 = [limit x->0] x/x = [limit x->0] 1 = 1. That is, x/x is well behaved [continuous] at 0; even tho it's 0/0 there, its value is unambiguously 1.

It involves the letters F and K. And, no, it's not a dirty word.

What a morbid path to a cheap answer. I love it!

Confirm - you get to the finals with a .500 probability. Your opponents are Earl [.667 of beating him x.055 probability that he's in the finals] or Queen [.5 x .189 prob] or King [.25 x .471 prob] or Duke [.333 x .284 prob] Your chances of winning the final match is .3436 and being champ with half that - .171 We have differences in the 3rd decimal place, but I think we're getting the same result basically. --------- It's still slightly better with Earl in your half of the bracket and make the finals with a .556 probability. You than have a .341 chance of winning the final match and a .191 chance at the championship. Which seems to make it advantageous to play weak opponents as long as possible and then fight some statistical combination of the strong ones at the very end.

Bingo!!!! Confirmed.

Yup .. I spoke before I thought. I'll run that simulation now.... thanks. --------------------- Edit: I ran that simulation. It gives you a 0.130 winning chance. The problem with that approach is that you have to fight through some strong opponents to reach the finals. The good part is that you then have a 50% chance of winning the final match. IF you get there. With You against Earl, you're in the semi's with a .667 probability. But in the semi's you have a .667 probability of playing the Duke and a .333 probability of playing Queen. That takes your odds of reaching the finals down to .259 so your winning chances are .130. With my lineup: ---- Me - 100 Prince - 0 Queen - 100 Earl - 50 ---- King - 300 Duke - 200 Peas - 100 Jest - 0 [loses to peas] ---- I get through Prince Queen and Earl - i.e. I make the finals with a .556 probability. My opponent in the finals is: King [against whom I have a .25 chance against his .450 probability of being there], or Duke [against whom I have a .333 chance against his .267 probability of being there], or Peasant [against whom I have a .5 chance against his .280 probability of being there], or Jest [0 - he's dead] My championship chances are thus: .556 x [.25 x .450 + .333 x .267 + .5 x .280] = .556 x .341 = 0.191. So I only have only a .341 chance of winning the final match, but my .556 chance of making the finals more than offsets that.