danydandan Posted March 23, 2019 Author #251 Share Posted March 23, 2019 4 hours ago, Golden Duck said: It took me a few months to figure out; ie, the entire season. It's the question on limits I was alluding to earlier. You are going to have to enlighten me GD. Link to comment Share on other sites More sharing options...
Ozymandias Posted March 23, 2019 #252 Share Posted March 23, 2019 1 hour ago, danydandan said: You are going to have to enlighten me GD. Me too! This is not an area of maths that has ever appealed to me nor is it something that I have considered before. Link to comment Share on other sites More sharing options...
Golden Duck Posted March 24, 2019 #253 Share Posted March 24, 2019 9 hours ago, danydandan said: You are going to have to enlighten me GD. 7 hours ago, Ozymandias said: Me too! This is not an area of maths that has ever appealed to me nor is it something that I have considered before. The Binomial Probability Distribution gives us the probability of observing y success from n trials. The Negative Binomial Probability Distribution gives the probability of observing y trials before serving r successes. The Geometric Probability Distribution is a special case, of the Negative Binomial Distribution Function, where r = 1. The combinations rule equals 1 in this case. P(n) = p(1-p)^(n-1); where p is the probability The sum of P(n) is S(n) = 1 - (1 - p)^n The expected value, mean, or average is μ = 1/p. The median is when S(n) equals 0.5; ie, when (1 - p)^n = 0.5. Therefore the median is v = log_1-p(0.5) = -1/[log2(1-p)] Dividing the median by the mean starts to converge on a number as p gets smaller. As p approaches 0, what is the limit of -1/[log2(1-p)] ÷ 1/p 1 Link to comment Share on other sites More sharing options...
danydandan Posted March 24, 2019 Author #254 Share Posted March 24, 2019 3 hours ago, Golden Duck said: The Binomial Probability Distribution gives us the probability of observing y success from n trials. The Negative Binomial Probability Distribution gives the probability of observing y trials before serving r successes. The Geometric Probability Distribution is a special case, of the Negative Binomial Distribution Function, where r = 1. The combinations rule equals 1 in this case. P(n) = p(1-p)^(n-1); where p is the probability The sum of P(n) is S(n) = 1 - (1 - p)^n The expected value, mean, or average is μ = 1/p. The median is when S(n) equals 0.5; ie, when (1 - p)^n = 0.5. Therefore the median is v = log_1-p(0.5) = -1/[log2(1-p)] Dividing the median by the mean starts to converge on a number as p gets smaller. As p approaches 0, what is the limit of -1/[log2(1-p)] ÷ 1/p Fecking hell. Haven't seen that in a few years. I actually knew that. Good question Golden Duck. 1 Link to comment Share on other sites More sharing options...
danydandan Posted March 24, 2019 Author #255 Share Posted March 24, 2019 @Golden Duck, since we could not get your last question and you enlightened us with the answer. Would you please set a new question, if you'd be so kind? Link to comment Share on other sites More sharing options...
Golden Duck Posted March 24, 2019 #256 Share Posted March 24, 2019 1 hour ago, danydandan said: @Golden Duck, since we could not get your last question and you enlightened us with the answer. Would you please set a new question, if you'd be so kind? The actual answer is... Spoiler ln(2) I feel I got lucky finding the answer using a spreadsheet. I had to hit the textbooks to help work out the proof. I'm lacking a bit of inspiration at the moment. Here's one I remember seeing online. Two trains are 200km apart when they set off towards each other at 50km/h. At the same a bird sets of flying to and fro between the trains, at a speed of 60km/h, until it is squashed between the two trains when they meet. What is the total distance the bird flies? Link to comment Share on other sites More sharing options...
Ozymandias Posted March 25, 2019 #257 Share Posted March 25, 2019 1 hour ago, Golden Duck said: The actual answer is... Hide contents ln(2) I feel I got lucky finding the answer using a spreadsheet. I had to hit the textbooks to help work out the proof. I'm lacking a bit of inspiration at the moment. Here's one I remember seeing online. Two trains are 200km apart when they set off towards each other at 50km/h. At the same a bird sets of flying to and fro between the trains, at a speed of 60km/h, until it is squashed between the two trains when they meet. What is the total distance the bird flies? 120 km. The trains' closing speed is 100km/hr and the gap between the trains is 200km. They will hit each other after travelling for 2 hours. During that 2 hours the bird flying at 60km/hr will have travelled 120km. 1 Link to comment Share on other sites More sharing options...
Golden Duck Posted March 25, 2019 #258 Share Posted March 25, 2019 1 hour ago, Ozymandias said: 120 km. The trains' closing speed is 100km/hr and the gap between the trains is 200km. They will hit each other after travelling for 2 hours. During that 2 hours the bird flying at 60km/hr will have travelled 120km. Too easy! Your turn. Link to comment Share on other sites More sharing options...
danydandan Posted March 25, 2019 Author #259 Share Posted March 25, 2019 (edited) 9 hours ago, Golden Duck said: The actual answer is... Reveal hidden contents ln(2) I feel I got lucky finding the answer using a spreadsheet. I had to hit the textbooks to help work out the proof. I'm lacking a bit of inspiration at the moment. Here's one I remember seeing online. Two trains are 200km apart when they set off towards each other at 50km/h. At the same a bird sets of flying to and fro between the trains, at a speed of 60km/h, until it is squashed between the two trains when they meet. What is the total distance the bird flies? 120. Edit: Damn it, looks up.... Ozy already answered and you've given an answer. Edited March 25, 2019 by danydandan 1 Link to comment Share on other sites More sharing options...
Ozymandias Posted March 25, 2019 #260 Share Posted March 25, 2019 What do the following numbers represent? .... 11, 101, 1000, 1101, ..... Link to comment Share on other sites More sharing options...
danydandan Posted March 25, 2019 Author #261 Share Posted March 25, 2019 5 hours ago, Ozymandias said: What do the following numbers represent? .... 11, 101, 1000, 1101, ..... Probably not going to be correct. But.....3,5,8,13? Binary to Decimal Base. Or is that just the low hanging fruit? And I believe that 3,5,8,13 is apart of the very famous Fibonacci Sequence. 1 Link to comment Share on other sites More sharing options...
Ozymandias Posted March 25, 2019 #262 Share Posted March 25, 2019 (edited) 38 minutes ago, danydandan said: Probably not going to be correct. But.....3,5,8,13? Binary to Decimal Base. Or is that just the low hanging fruit? And I believe that 3,5,8,13 is apart of the very famous Fibonacci Sequence. That's it, Dan. You got it in one! You're turn to set a question. But you haven't given or laid out a solition method. Edited March 25, 2019 by Ozymandias 1 Link to comment Share on other sites More sharing options...
danydandan Posted March 25, 2019 Author #263 Share Posted March 25, 2019 Using only the number eight (8), as many times as you'd like, write an equation to equal one thousand (1000). You can use any operator, function or whatever but you can only use the number eight (8). Link to comment Share on other sites More sharing options...
Ozymandias Posted March 25, 2019 #264 Share Posted March 25, 2019 (edited) 1000 = 8 × 125 1000 = 8 × 8 × 15.675 1000 = (8 × 8)(8 + 8 - 0.375) 1000 = (8 × 8)(8 + 8 - 3/8) 1000 = (8 × 8)[8 + 8 - (1/8) - (1/8) - (1/8)] 1000 = (8 × 8)[8 + 8 - (8/64) - (8/64) - (8/64)] 1000 = (8 × 8){8 + 8 - [8/8(8)] - [8/8(8)] - [8/8(8)]} Alternatively (and prompted by Golden Duck's following post #269): 8 [(8 × 8) + (8 × 8) - 8/8 - 8/8 - 8/8] Edited March 25, 2019 by Ozymandias 1 Link to comment Share on other sites More sharing options...
Golden Duck Posted March 25, 2019 #265 Share Posted March 25, 2019 Is this cheating? >>> s = '8' >>> for i in range(int(8*8 + 8*8 - 8/8 - 8/8 - 8/8 - 8/8)): s = s + " + 8" >>> s '8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8' >>> eval(_) 1000 >>> 1 Link to comment Share on other sites More sharing options...
Ozymandias Posted March 25, 2019 #266 Share Posted March 25, 2019 (edited) 25 minutes ago, Golden Duck said: Is this cheating? >>> s = '8' >>> for i in range(int(8*8 + 8*8 - 8/8 - 8/8 - 8/8 - 8/8)): s = s + " + 8" >>> s '8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8' >>> eval(_) 1000 >>> Why do you think it might be cheating? Is it because you are using a computer? Edited March 25, 2019 by Ozymandias Link to comment Share on other sites More sharing options...
Golden Duck Posted March 25, 2019 #267 Share Posted March 25, 2019 20 minutes ago, Ozymandias said: Why do you think it might be cheating? Is it because you are using a computer? Perhaps because I'm using alphabetical characters. Thinking about this question a bit deeper... is the benefit of this type of question to cement in understanding of various identity laws? Link to comment Share on other sites More sharing options...
Harte Posted March 25, 2019 #268 Share Posted March 25, 2019 A man walks up to three boys playing in a vacant lot. All three boys have a dirt smudge on their forehead. The man says "I'll give a dollar to the first boy that can tell me if he has a smudge on his forehead." The boys think about it, then finally one raises his hand and says "I have a smudge on my forehead." The man paid him. How did the boy know? Harte 1 1 Link to comment Share on other sites More sharing options...
danydandan Posted March 26, 2019 Author #269 Share Posted March 26, 2019 (edited) Both answers are correct, @Ozymandias and @Golden Duck. I did say just use the number eight and any operator, function or whatever. A very simple solution is the following, hopefully that question was a good brain teaser. 888+88+8+8+8=1000. Yes I use eight eights to get the result, I thought that was eloquent. Take it away Ozy it's your turn you posted first. Edited March 26, 2019 by danydandan 1 1 Link to comment Share on other sites More sharing options...
Ozymandias Posted March 26, 2019 #270 Share Posted March 26, 2019 10 hours ago, Harte said: A man walks up to three boys playing in a vacant lot. All three boys have a dirt smudge on their forehead. The man says "I'll give a dollar to the first boy that can tell me if he has a smudge on his forehead." The boys think about it, then finally one raises his hand and says "I have a smudge on my forehead." The man paid him. How did the boy know? Harte Instead of setting a new problem, and if Harte doesn't mind, I choose to let his teaser (quoted above) to stand for mine. 1 Link to comment Share on other sites More sharing options...
Harte Posted March 26, 2019 #271 Share Posted March 26, 2019 1 hour ago, Ozymandias said: Instead of setting a new problem, and if Harte doesn't mind, I choose to let his teaser (quoted above) to stand for mine. I didn't mean to post out of turn. I just remembered that little logic problem so I posted it. Harte 1 1 Link to comment Share on other sites More sharing options...
Golden Duck Posted March 26, 2019 #272 Share Posted March 26, 2019 11 hours ago, Harte said: A man walks up to three boys playing in a vacant lot. All three boys have a dirt smudge on their forehead. The man says "I'll give a dollar to the first boy that can tell me if he has a smudge on his forehead." The boys think about it, then finally one raises his hand and says "I have a smudge on my forehead." The man paid him. How did the boy know? Harte 1 Each boy has the knowledge of what he sees - two boys with smudges; and, the words of the man - "I'll give a dollar to the first boy that can tell me if he has a smudge on his forehead." If you can infer "...first boy..." means there is more than must mean there is more than one boy with a smudge, then one boy must, based on the uncertainty of the others, can rule out the group of boys is made up of two smudges and one clean. Therefore, it means all boys must be smudged. Again I'm banking on "...first boy..." cannot mean only one. Link to comment Share on other sites More sharing options...
Ozymandias Posted March 26, 2019 #273 Share Posted March 26, 2019 (edited) My take is this: None of the boys can know that they have a smudge on their own forehead, only that the others do. That is why none can claim they have a smudge with certainty. The best bet, in your ignorance, is to take a punt and claim that you do before the other two, because 'if you're not in you can't win'! Relying on the man's use of 'first boy' is logically unsound because he could be saying that if only two boys had smudges. Edited March 26, 2019 by Ozymandias Link to comment Share on other sites More sharing options...
Golden Duck Posted March 26, 2019 #274 Share Posted March 26, 2019 40 minutes ago, Ozymandias said: My take is this: None of the boys can know that they have a smudge on their own forehead, only that the others do. That is why none can claim they have a smudge with certainty. The best bet, in your ignorance, is to take a punt and claim that you do before the other two, because 'if you're not in you can't win'! Relying on the man's use of 'first boy' is logically unsound because he could be saying that if only two boys had smudges. Corey sees two smudges. Corey knows Billy and Andy each see at least one smudge. If "first boy" means more that one and if there were only two smudges, then Corey also knows that Billy and Andy should know they have a smudge. Corey deduces that both Billy and Andy, because of their uncertainty, see two smudges meaning Corey himself must be smudged. 1 Link to comment Share on other sites More sharing options...
danydandan Posted March 26, 2019 Author #275 Share Posted March 26, 2019 (edited) 21 hours ago, Harte said: A man walks up to three boys playing in a vacant lot. All three boys have a dirt smudge on their forehead. The man says "I'll give a dollar to the first boy that can tell me if he has a smudge on his forehead." The boys think about it, then finally one raises his hand and says "I have a smudge on my forehead." The man paid him. How did the boy know? Harte He assumed he did because the others didn't raise their hands. Also he wiped his forehead. Edited March 26, 2019 by danydandan 1 1 Link to comment Share on other sites More sharing options...
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