Ozymandias Posted March 20, 2019 #226 Share Posted March 20, 2019 1 hour ago, Golden Duck said: V_0 (r = 1, h = 1) == π(r^2)(h) == π(1)(1) == π V_1 (r = 0.9, h = 1.15) == π(r^2)(h) == π(0.9^2)(1.15) == π(0.81)(1.15) = π(0.9315) V_1 is 93.15 per cent of V_0. A difference of 6.85. Pretty good estimation by @danydandan. I would've just averaged 0.88 and 0.96 to get 0.92. Since you have given the necessary solution and the correct answer, viz. a 6.85% reduction in volume, you should set the next problem. 1 Link to comment Share on other sites More sharing options...
Golden Duck Posted March 20, 2019 #227 Share Posted March 20, 2019 12 minutes ago, Ozymandias said: Since you have given the necessary solution and the correct answer, viz. a 6.85% reduction in volume, you should set the next problem. For the upcoming 2019 Rugby World Cup, a betting agency has set the following market: New Zealand 2.20 England 5.50 Ireland 6.00 Wales 8.00 South Africa 9.00 Australia 12.00 France 26.00 Argentina 34.00 Scotland 34.00 Japan 251.00 Canada 501.00 Fiji 501.00 Italy 501.00 Samoa 501.00 Tonga 501.00 Georgia 1001.00 Namibia 1001.00 Russia 1001.00 United States 1001.00 Uruguay 1001.00 What is their margin? PS. I would've preferred to the RWC but it doesn't look like What is their margin? Link to comment Share on other sites More sharing options...
Habitat Posted March 20, 2019 #228 Share Posted March 20, 2019 (edited) You can work out "market percentage" by dividing 100 by each of the prices allotted to each team. Just doing it in my head, it appears to be about 122%. So, in theory, if each team was backed in proportion to its odds, to take out $100, it would cost $122 to do so, to be guaranteed to be on the winner of the comp, for the same collect of $100. Which is clearly bad business for the investor who was silly enough to do it. The bookie "cost" of $100 is the payout, but "sales" of $122. Profit was $22. Of course this is theory, and in practice it is much harder for the bookie to make a book where every contingency is a winner. The 22% gives a margin for error and operating expenses, and if the bookie's market assessments are as good as the punters, over time they will win, but not necessarily on any one event. Looked at another way, in theory they retain about $18 for every $100 invested, when betting 122%, but there are other factors that make it somewhat less lucrative, including that the margin at the shorter odds end is usually less, and that is where the bulk of money is bet. In a typical bookmakers market where a contingency is priced At $101, the real odds would be more like $201. I recall attending a greyhound track where the bookmakers were betting around 200%, which was obviously to provide a margin against bettors who were better informed about the chances of the dogs, than they were. Simple market % tells only part of the story. Edited March 20, 2019 by Habitat Link to comment Share on other sites More sharing options...
Ozymandias Posted March 20, 2019 #229 Share Posted March 20, 2019 (edited) 14 hours ago, Golden Duck said: For the upcoming 2019 Rugby World Cup, a betting agency has set the following market: New Zealand 2.20 England 5.50 Ireland 6.00 Wales 8.00 South Africa 9.00 Australia 12.00 France 26.00 Argentina 34.00 Scotland 34.00 Japan 251.00 Canada 501.00 Fiji 501.00 Italy 501.00 Samoa 501.00 Tonga 501.00 Georgia 1001.00 Namibia 1001.00 Russia 1001.00 United States 1001.00 Uruguay 1001.00 What is their margin? Habitat gave an answer which is not far off the mark, I think. Although having revisited this thread you did not respond to hom so I am assumimng you are wanting the solution and the exact answer, so here goes …. I am not a gambler or a bookie but I think that to find the margin as a percentage you must take the sum of the reciprocals of the list of numbers you gave above multiplied by 100 ; i.e. [1/2.2 + 1/5.5 + 1/6 + 1/8 + ….. 2/34 + …. 5/501 + ….5/1001] x 100 = 1.23871 … x 100 = 123.87% Therefore, the margin is 23.87% Edited March 20, 2019 by Ozymandias 2 Link to comment Share on other sites More sharing options...
danydandan Posted March 20, 2019 Author #230 Share Posted March 20, 2019 (edited) I've been trying to do this in my head for a few hours and I'm not sure still but anyways here I go. I think we need to take into account all the odds in the match. In this case you need to get the probability of each team winning then add them all together. And get that percentage from the sum of these probabilities! So 21% margin. 1.21 - 1= .21 or 21%. Edited: Apparently typing out the sum results in not being able to post for some odd reason. But the 1.21 is the sum of the probabilities. Edited March 20, 2019 by danydandan Link to comment Share on other sites More sharing options...
Golden Duck Posted March 20, 2019 #231 Share Posted March 20, 2019 3 hours ago, Ozymandias said: Habitat gave an answer which is not far off the mark, I think. Although having revisited this thread you did not respond to hom so I am assumimng you are wanting the solution and the exact answer, so here goes …. I am not a gambler or a bookie but I think that to find the margin as a percentage you must take the sum of the reciprocals of the list of numbers you gave above multiplied by 100 ; i.e. [1/2.2 + 1/5.5 + 1/6 + 1/8 + ….. 2/34 + …. 5/501 + ….5/1001] x 100 = 1.23871 … x 100 = 123.87% Therefore, the margin is 23.87% Diving 100 by each price works too. Should the exact answer be found each time? That's been the practice so far, hasn't it? You can set a new problem. 1 Link to comment Share on other sites More sharing options...
Ozymandias Posted March 20, 2019 #232 Share Posted March 20, 2019 54 minutes ago, Golden Duck said: Should the exact answer be found each time? That's been the practice so far, hasn't it? These are maths questions. In my view you need to give the solution method and the correct answer for two reasons: (1) It is important to give a solution method for the benefit of those who might like to know how the problem was solved. (2) Maths is black and white.There is no point giving approximate answers when the question asks for THE answer. Just sayin'. 1 Link to comment Share on other sites More sharing options...
Ozymandias Posted March 20, 2019 #233 Share Posted March 20, 2019 (edited) 1 hour ago, Golden Duck said: You can set a new problem. A boring machine drills a road tunnel with a diameter of 5 metres through a mountainside. A road surface having a maximum width of 8 metres is then laid horizontally within the tunnel. What is the maximum vertical height between the road surface and tunnel roof? Edited March 20, 2019 by Ozymandias Link to comment Share on other sites More sharing options...
danydandan Posted March 20, 2019 Author #234 Share Posted March 20, 2019 3 minutes ago, Ozymandias said: These are maths questions. In my view you need to give the solution method and the correct answer for two reasons: (1) It is important to give a solution method for the benefit of those who might like to know how the problem was solved. (2) Maths is black and white.There is no point giving approximate answers when the question asks for THE answer. Just sayin'. Yeah I agree. I know I haven't been exact in my answers (as I like to work out the solution in my head as much as I can) in the future I'll be more exacting in my responses. 2 Link to comment Share on other sites More sharing options...
danydandan Posted March 20, 2019 Author #235 Share Posted March 20, 2019 7 minutes ago, Ozymandias said: A boring machine drills a road tunnel with a diameter of 5 metres through a mountainside. A road surface having a maximum width of 8 metres is then laid horizontally within the tunnel . What is the maximum vertical height between the road surface and tunnel roof? Do you mean 5m radius? 1 Link to comment Share on other sites More sharing options...
Ozymandias Posted March 20, 2019 #236 Share Posted March 20, 2019 3 hours ago, danydandan said: Do you mean 5m radius? Thanks, Dan. Another hurried post. I meant a RADIUS of 5m, yes!!! Link to comment Share on other sites More sharing options...
Habitat Posted March 20, 2019 #237 Share Posted March 20, 2019 Actually, the theoretical margin for the bookmaker, isn't 22%, it is around 18%, because for every $100 "held" he should retain about $18 of it, betting 122% markets. But that is just theoretical, in actual practice they make around 10% at best, over time. Which is still a handsome result. Link to comment Share on other sites More sharing options...
Harte Posted March 21, 2019 #238 Share Posted March 21, 2019 5 hours ago, Ozymandias said: A boring machine drills a road tunnel with a diameter of 5 metres through a mountainside. A road surface having a maximum width of 8 metres is then laid horizontally within the tunnel. What is the maximum vertical height between the road surface and tunnel roof? x=3 so clearance is 5+3=8 meters. You asked the same question on page one. Harte Link to comment Share on other sites More sharing options...
Harte Posted March 21, 2019 #239 Share Posted March 21, 2019 Too lazy to make one up - so here's a problem I gave my Geometry Honors students: Given the circle with equation x2 + y2 = 26 and the line with equation y = -5x – 26, determine: 1) whether the line intersects the circle, 2) whether the line is tangent to the circle. If the line is tangent, find the coordinates of the point of tangency. Harte Link to comment Share on other sites More sharing options...
Harte Posted March 21, 2019 #240 Share Posted March 21, 2019 These aren't really "brain teasers" you know. Harte 1 Link to comment Share on other sites More sharing options...
Golden Duck Posted March 21, 2019 #241 Share Posted March 21, 2019 (edited) 8 hours ago, Harte said: Too lazy to make one up - so here's a problem I gave my Geometry Honors students: Given the circle with equation x2 + y2 = 26 and the line with equation y = -5x – 26, determine: 1) whether the line intersects the circle, 2) whether the line is tangent to the circle. If the line is tangent, find the coordinates of the point of tangency. Harte System of equations: x^2 + y^2 == 26 ## Equation (1) y == -5x - 26 ## Equation (2) Substitute (2) into (1): x^2 + (-5x - 26)^2 == 26 x^2 + 25x^2 + 260x + 676 == 26 26x^2 + 260x == -650 x^2 + 10x == -25 ## Divide by 26 x^2 + 10x + 25 == 25 - 25 ## Complete the square; ie, Add (10/2)^2 to both sides (x^2 + 5)^2 == 0 x + 5 == 0 x == -5 Only one solution for x, so (2) is tangential to (1). Substitute x = -5 into (2): y == -5(-5x) - 26 == 25 - 26 == -1 The point of tangency is (-5, -1) Edited March 21, 2019 by Golden Duck 1 Link to comment Share on other sites More sharing options...
Ozymandias Posted March 21, 2019 #242 Share Posted March 21, 2019 (edited) 9 hours ago, Harte said: x=3 so clearance is 5+3=8 meters. You asked the same question on page one. Harte Did I? God, I must be getting old and stuck in my ways. Anyway, as you well know, you are correct. The answer requires the application of Pythagoras' Theorem to the right-angled triangle (which is a standard 3 : 4 : 5 triangle with X corresponding to side 3). It would be nice if you stuck around and contributed to the thread. Edited March 21, 2019 by Ozymandias Link to comment Share on other sites More sharing options...
Ozymandias Posted March 21, 2019 #243 Share Posted March 21, 2019 8 hours ago, Harte said: These aren't really "brain teasers" you know. Harte Well that Coordinate Geometry problem you set could not be considered one, I suppose. Link to comment Share on other sites More sharing options...
danydandan Posted March 21, 2019 Author #244 Share Posted March 21, 2019 16 hours ago, Harte said: These aren't really "brain teasers" you know. Harte They are if you use just your brain to try figure them out. Link to comment Share on other sites More sharing options...
Ozymandias Posted March 21, 2019 #245 Share Posted March 21, 2019 10 hours ago, Golden Duck said: The point of tangency is (-5, -1) Good show, Golden Duck. You can also demonstrate that the line is a tangent to the circle using the perpendicular distance formula which turns out to be the same as the circle radius. The circle has its centre on the origin (0,0). For aX + bY + c = 0 to be a tangent to X^2 + Y^2 = R^2 there is also the condition that c = +/- R[rt(1 + m^2)] but you still have to find the coordinates of the point of tangency, as you did, using simultaneous equations. Link to comment Share on other sites More sharing options...
Harte Posted March 21, 2019 #246 Share Posted March 21, 2019 12 hours ago, Ozymandias said: Well that Coordinate Geometry problem you set could not be considered one, I suppose. My students solved it so fast it p***ed me off. Should have made it harder. LOL Harte Link to comment Share on other sites More sharing options...
Harte Posted March 21, 2019 #247 Share Posted March 21, 2019 12 hours ago, Ozymandias said: Did I? God, I must be getting old and stuck in my ways. Anyway, as you well know, you are correct. The answer requires the application of Pythagoras' Theorem to the right-angled triangle (which is a standard 3 : 4 : 5 triangle with X corresponding to side 3). It would be nice if you stuck around and contributed to the thread. Wasn't exactly the same. But it did involve the intersection of radius and chord, only the triangle formed was 30-60-90 instead of 3-4-5. Harte 1 Link to comment Share on other sites More sharing options...
Golden Duck Posted March 22, 2019 #248 Share Posted March 22, 2019 (edited) Alright, Im stuck for inspiration. But here goes... Over the last Australian summer I was pondering the main Cricket batting metric - the Batting Average or runs per dismissal. If I use Geometric Distribution as a model, by what should I multiply the batting average to derive a batting median? Edited March 22, 2019 by Golden Duck I mixed up divide and multiply. Link to comment Share on other sites More sharing options...
danydandan Posted March 23, 2019 Author #249 Share Posted March 23, 2019 (edited) On 22/3/2019 at 12:27 AM, Golden Duck said: Alright, Im stuck for inspiration. But here goes... Over the last Australian summer I was pondering the main Cricket batting metric - the Batting Average or runs per dismissal. If I use Geometric Distribution as a model, by what should I multiply the batting average to derive a batting median? I don't think you can derive a median from a mean or average, unless it's a set of averages. Edited March 23, 2019 by danydandan Link to comment Share on other sites More sharing options...
Golden Duck Posted March 23, 2019 #250 Share Posted March 23, 2019 1 hour ago, danydandan said: I don't think you can derive a median from a mean or average, unless it's a set of averages. It took me a few months to figure out; ie, the entire season. It's the question on limits I was alluding to earlier. Link to comment Share on other sites More sharing options...
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