Wikipedia:Reference desk/Archives/Mathematics/2007 August 10

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August 10

Superannuation Question

Hi guys, this isn't a homework question but it came up in a test a week ago and I can't figure out where I'm going wrong. Wonder if anyone can help...

Sally plans to retire tomorrow, she has $450,000 in her superannuation fund which compounds monthly at an interest rate of 15% p.a. If she intends to withdraw $6000 per month after the interest is paid such that the amount owing after the first month is A₁ = 450,000(1.0125) - 6000 find:

a) An expression for the amount owing after n months (ie A_n = ...)
b) How long the superannuation fund will last Sally if she continues to withdraw $6000 per month and the interest continues to be compounded monthly at the same rate
c) If Sally wants the fund to last 30 years what must the interest rate be?

I've got an expression for the first bit as being A_n = 480000 - 30000(1.0125ⁿ) which makes no sense because it means at a higher interest rate the fund will last less time. Using that formula I ended up with the original fund lasting 18.6 years and the interest rate required for it to last 30 years being 9.3%, which is lower than the original interest rate.

Can someone tell me where I've gone wrong and if possible how to get the right answer. Thanks

I think your problem's in the first step. Interest gets charged on the amount of money in the account, so as it says, after the 1st month we have A1 = 450000 * 1.0125 - 6000, then in the 2nd month we have A2 = A1 * 1.0125 - 6000 = 450000 * 1.0125 * 1.0125 - 6000 * 1.0125 - 6000. If you do that a few times, do you see a pattern? In more generic terms, if the principal is P, the interest rate r/100% per month, and the withdrawal W, you can write it as A1 = P(1+r) - W, A2 = (P(1+r) - W)(1+r) - W = P(1+r)^2 - W(1+r) - W, and so forth. Does that help? Confusing Manifestation 03:52, 10 August 2007 (UTC)

Thanks for the reply. My working for the first part of the question was:

A₂ = 1.0125(A₁) = 450000(1.0125²) - 6000(1.0125 + 1)

A₃ = 1.0125(A₂) = 450000(1.0125³) - 6000(1.0125² + 1.0125 + 1)

A₄ = 1.0125(A₃) = 450000(1.0125⁴ - 6000(1.0125³ + 1.0125² + 1.0125 + 1)

Therefore following the pattern...

A_n = 450000(1.0125ⁿ) - 6000(1.0125^n-1 + 1.0125^n-2 +...+ 1)

The last brackets contains a GP w/ a=1, r=1.0125, n=n

Sum_GP = a(rⁿ-1)/(r-1) = (1.0125ⁿ - 1)/(1.0125 - 1) = 80(1.0125ⁿ - 1)

Subbing back into original equation

A_n = 450000(1.0125ⁿ) - 6000(80(1.0125ⁿ - 1)

A_n = 450000(1.0125ⁿ) - 480000(1.0125ⁿ) + 480000

A_n = 480000 - 30000(1.0125ⁿ)

Which to me seems to say that if you increase the interest rate the fund is going to last a shorter time period, so I'm pretty sure my mistake is in those steps, but I can't seem to figure out what it is... Can anyone figure it out? Thanks

I think your mistake is in working with a specific interest rate, and then trying to guess how varying it would affect the result, rather than working with an unknown interest rate r to begin with. Doing that, I get:

A_n = 450000(1+r)ⁿ - 6000((1+r)^n-1 + (1+r)^n-2 +...+ 1)

A_n = 450000(1+r)ⁿ - 6000((1+r)ⁿ - 1)/((1+r) - 1)

A_n = 450000(1+r)ⁿ - (6000/r)((1+r)ⁿ - 1)

A_n = (450000 - 6000/r)(1+r)ⁿ + 6000/r

So you see that the dependence on r isn't that simple after all. Your mistake above was that you missed the factor of 1/r = 80, treating it as a constant instead. —Ilmari Karonen (talk) 11:54, 10 August 2007 (UTC)

Conceptualizing irrational numbers

I was reading about irrational numbers and the following question came to mind. Please can somebody help. Square root of 2 is irrational, that is it is 1.41 with the decimal digits going on for ever. However, suppose I have drawing/measuring instruments of infinite precision(is this point relevant ??), and if I draw a right angled triangle of sides 1 cm each, and if I measure the hypotenuse, it will be some FINITE PRECISE number (that is it will be a rational number) - Or will it not be? Because as per the formula, in this case, the hypotenuse should be square root of 2 which is irrational? How do I 'get' this / visualize this / understand this concept? Forml 11 08:14, 10 August 2007 (UTC)

The concept of "infinite precision" may not be possible to really define. I think the problem you're having is with an infinite decimal expansion rather than irrationality; consider a line of length 1/9 = 0.111... cm which you measure with a ruler. You measure this in steps: you measure the first millimeter, but this isn't accurate enough, so then you measure a little bit more, but that isn't accurate enough still, so you measure a bit more, etc. This process would not stop.

Actually, even I was not sure about the 'infinite precision' part - and so I had wondered if it was relevant.

The question though is, if I draw a right angled traingale of sides 1 cm and measure the hypotenuse, it will be sone FINITE PRECISE rational number. It will *not' be equal to square root of 2 (because it is irrational). That means what I measure is not what I get from the formula - right? Or is there some basic flaw in my reasoning here?

To ask the same question in another way, I want to draw a line of length square root of 2? Without using the ringht angled triangle, how do i draw the line having precisely that length?

Forml 11 09:15, 10 August 2007 (UTC)

The square root of two cannot be expressed as a ratio of integers. A physical ruler, even a very small one, has a finite unit of measurement. If the side is an integral multiple of that unit, perhaps some number of nanometres, then the ruler cannot give an exact integral measure for the hypotenuse. We could determine that when the side is exactly 70 units long then the hypotenuse is approximately 99 units long, and when the side is exactly 99 units long then the hypotenuse is approximately 140 units long; but ⁹⁹⁄₇₀ is too large and ¹⁴⁰⁄₉₉ is too small.

In other words, we cannot make a ruler whose unit can give an integral measure for both the side and the hypotenuse. These two lengths are incommensurable.

If you measure the hypotenuse exactly, the result will not be a rational number. For, if that number were ^p⁄_q, say, then we could simply make our unit q times smaller, so that the measure was the integer p. --KSmrq^T 10:14, 10 August 2007 (UTC)

Applying the infinitesimal to the physical is always problematic in that we can't physically measure things to an arbitrary measure of precision, while we could do this theoretically. Even if this were possible, any measurement of the hypotenuse is an approximation. Thus, it is only possible to draw a line of length sqrt 2 approximately.

Isn't this Zeno's Paradox ... or no? (Joseph A. Spadaro 16:57, 10 August 2007 (UTC))

The future cannot change the past

There is a urn with 5 white balls and 5 black balls. A blindfolded man draws a ball from the urn, gives it to his assistant who writes down the color of the ball in a notebook and then destroys the ball. Aka. drawing without replacement. This is done ten times until all the balls are drawn from the urn.

Now if the assistant wrote in the notebook "The tenth ball drawn from the urn is colored white", what is the probability that the first ball drawn from the urn is black?

I known that this is a trick question because the first ball drawn occurs when the urn has 5 white balls and 5 black balls, so the answer is

The probability the first ball drawn is black is 5/10 = 1/2

But my stupid Cow of a teacher marked my answer wrong.

She say its 5/9

But that is clearly wrong because it violates the LAWS OF PHYSICS. The future cannot change the past. Does anyone believe that such a stupid Cow can be a mathematics teacher!!

So how can I convince her that she is wrong? 211.28.119.203 10:30, 10 August 2007 (UTC)

I believe she's correct, not a stupid cow. You are given some extra information, namely that "the tenth ball drawn is white", and you are asked to calculate the probability given that information. Out of the nine unknown balls in the urn, five are black.

Think of it in reverse. Suppose the man drew a ball at the beginning and the assistant wrote down "the first ball drawn is white". What is the probability that the next ball is black?

(edit conflict)Actually... the question is more along the line of "how can we convince you she is right?" Because she is.

Look... Put a number on all the balls, 1-5 on the black and 1-5 on the white. Assume the assistant wrote down that the last ball was white-5. This means that that on the first draw, black 1-5 or white 1-4 could have been picked, all with equal probability, meaning a 5/9 chance of black.

Or, try a smaller example. 2 balls, one white, one black. Second ball is white. What's the chance the first is black? And then with four balls, 2 white, 2 black. Last ball is white, so the first ball has a 2/3 chance of being black, because you can't pick the white ball that was removed later.

Do either of those explanations help? Gscshoyru 10:39, 10 August 2007 (UTC)

How can an event in the future (10th drawn ball) reach back in time to change the event in the past when there is clearly 5 white balls and 5 black balls in the urn to have a probability of more than 5/10. MAthematics cannot violate physcial Causality! 211.28.119.203 10:42, 10 August 2007 (UTC)

Suppose the assistant wrote down all the drawings. The question "what is the probability that she wrote down that the first ball was black" is equivalent to "what is the probability that the first ball drawn was black", and is equivalent to the backwards scenario I posed to you above. The probabilities will be the same. Your teacher is correct.

The events in the future are not affecting the events in the past. The thing with probability is that it's really a measure of how likely things are, -based on our current knowledge-. Suppose person A flips a coin, but doesn't look at it, and person B then looks at the coin. It turns out it came up tails, but person A doesn't know this yet. We ask person A what the odds are that the coin came up heads. He says, "one in two". We ask person B what the odds are that it came up heads. He says, "zero". Person A doesn't have the information that person B had. The fact is that the coin flip happened in the past. Its result is already determined. But one person has more information about that result than the other. So, when we ask, "What's the chance that the first ball is black?" we're actually asking, "Given our knowledge of the outcome of this experiment, how likely is it that the first ball chosen was black?" I hope this helps clear it up. Maelin (Talk | Contribs) 10:50, 10 August 2007 (UTC)

Are the you saying that the probability of an event in the past can change upon gaining knowledge of subsequent events that occurs afterwards. That probability of a particular past event is NOT FIXED at the time of occurrence? 211.28.119.203 11:03, 10 August 2007 (UTC)

None of us are saying that. The act of writing each drawing down fixes each event. You do not know what the assistant has written except one drawing, it doesn't matter whether it's the last or the first. You have enough information to determine the probability of the other drawings that have been written down.

At the time the ball was drawn, the chance was 50/50. But, for instance, say that we know that the last five balls drawn are white. What is the probability that the first ball drawn is black? Answer that question. We're not talking about probability at time of occurrence, we're talking about probability that something did happen. Gscshoyru 11:12, 10 August 2007 (UTC)

You should also try to read Monty Hall problem, where a similar fallacy is discussed. Kusma (talk) 11:13, 10 August 2007 (UTC)

If the last five balls are white then the first ball must be black to preserve causality.

Hmmm... Does the subsequent events in the future do generate some kind of Quantum Mechanic probability waves that travel backwards in time and interact with past events? This is quite deep. No one has mention this kind of relationship between Mathematics and Quantum Mechanics to me before. 211.28.119.203 11:19, 10 August 2007 (UTC)

No, see, it doesn't have anything to do with that... though it would be interesting and funny if it did.

Even if the last five balls picked are black, at the time of the first pick. The probability that the first ball picked is black is 50/50. This will always be true (assuming there are 5 white and 5 black, etc) But that's not what's being asked, you see. What's being asked is the probability of an event based on events that happened later. What's the chance that you're going to win the lottery? Near zero. However, if you do later win the lottery, what is the chance that you did win the lottery? 100%. We're asking a question of past events based on information from future events. That's where your misunderstanding lies. Gscshoyru 11:24, 10 August 2007 (UTC)

(edit conflict) Take a simpler example. You are sitting in your room and wondering what is your mother currently doing. Half an hour later you hear her announcing, "Dinner is ready!". You think, "Oh, half an hour ago she must have been preparing dinner". Does her announcement travel back in time to affect what she was doing? Of course not; Her announcement is effected by her earlier activities, and knowledge about her announcement in the future allows you to make deductions about her activity in the past.

Another simple example. You only have one black ball and white ball in the urn. The blind-folded man takes out the balls one after the other. You are not told what was the first ball he picked, but you are told the second was white. With the information you currently have, what is the probability that the first ball is black? Of course it is 1, since if the second is white then the first must be black.

As was mentioned, probability depends on what you know. If a ball has not yet been taken out, then any person in the world will tell you that the probability for first being black is 1/2. But that changes after the balls are taken out and you have some knowledge about the result.

But this isn't the important issue. The important issue is that everyone makes mistakes - your teacher does, of course, but so do you. It is reasonable to assume that your teacher knows much more than you and is far less likely to make mistakes than you. So, regardless of whether you are right or not, your first reaction should not be "how to prove her wrong", but rather to check yourself to see if you have made a mistake. If that fails, you should ask the teacher to explain her reasoning. If, after all that, you are still convinced that you are correct - which you very well may be - then it is still unhelpful to call her either "stupid" or a "cow". -- Meni Rosenfeld (talk) 11:31, 10 August 2007 (UTC)

Oh, and one more thing - mathematics is independent of physics. While the nature of our physical universe certainly inspires the mathematical development, mathematical observations do not have to answer to physical laws. -- Meni Rosenfeld (talk) 11:35, 10 August 2007 (UTC)

If you had not been told that the last ball was white then you would be correct, the chances of the first being black is 5/10. However since you are given this information, you know that the white ball that was picked last cannot possibly have a chnace of being the first ball that was picked - since the two contradict each other. Now that you know the last ball's colour, you could treat this as a problem where there are are 5 black balls and 4 white balls, and you want to find the probability that the first one taken from the urn is black. In this case, it is obviously 5/9. asyndeton 11:38, 10 August 2007 (UTC)

I have to sit down and think about this slowly. 211.28.119.203 11:46, 10 August 2007 (UTC)

By all means, do so. If you have any problems or whatever that you need to clear up while thinking, don't hesitate to ask. Gscshoyru 11:49, 10 August 2007 (UTC)

And I will add that this is a much better approach. People make many more mistakes when they are emotional or in a hurry. -- Meni Rosenfeld (talk) 11:51, 10 August 2007 (UTC)

The future may not predict the past, but it may impact our assessment of what may have happened in the past, when we don't know exactly what did happen. This indeed is the essence of Bayes' theorem. I think Gscshoyru's example with only one ball of each color gives the feel better (you learn someting about the first ball by what happens later), but let's see what it says...

Let X₁ be the first ball being black and X₂ it being white. The probabilities of each is 1/2. Since one and exactly one of these things occurs, we can use the theorem with any other event. Let Y be the 10th ball being white; this is our other event. If X₁ happens, the probability of Y is 5/9 (=Pr(Y|X₁)); conversely, if X₂ happens, the probability of Y is 4/9 (=Pr(Y|X₂)). What is the probability of X₁ happening, the first ball being black, if we know Y happened (the 10th was white)? This is Pr(X₁|Y).

${\begin{aligned}\Pr {(X_{1}|Y)}&={\frac {\Pr(Y|X_{1})\Pr(X_{1})}{\Pr(Y|X_{1})\Pr(X_{1})+\Pr(Y|X_{2})\Pr(X_{2})}}\\&={\frac {{\frac {5}{9}}\times {\frac {1}{2}}}{{\frac {5}{9}}\times {\frac {1}{2}}+{\frac {4}{9}}\times {\frac {1}{2}}}}\\&={\frac {\frac {5}{18}}{\frac {9}{18}}}\\&={\frac {5}{9}}\end{aligned}}$

A heuristic I like is that the last ball could have been black. But it wasn't. This is evidence that the earlier balls were individually more likely to be white. How much so is given by the formula above. Baccyak4H (Yak!) 14:00, 10 August 2007 (UTC)

The entire conversation / thread above has been most interesting and, actually, fascinating. Not only from a mathematical standpoint, but from a psychological one. On some level, I can understand the original poster's question and confusion over whether or not the future event "changes" the past event. Above, everyone's mathematical explanations have been helpful and enlightening (i.e., the substantive mathematical content of probability). However, Meni Rosenfeld nits the nail on the head with his/her comments. The actual math / probability here is secondary. The real issue is how to approach all problems (including, but not limited to, math problems) ... and dealing with the possibility of making mistakes. In my opinion, calling your teacher a "stupid cow" is not only immature, but arrogant. The base line of enlightened learning should be: "OK, in all probability, she is right ... let me see if I can understand why, even though it is confusing." However, your approach was: "Damn, this lady is so stupid and I certainly know more about math than her ... I'd love to know how the hell she ever got a job as a math teacher." Substantive mathematics and probability aside, that second statement speaks volumes about your approach to learning. And to dealing with other human beings / people (who may or may not make mistakes in their lives). Not to mention, the potential that you yourself might be wrong and might have made a mistake (or might not have). In the future ... the substantive math is essentially irrelevent and trivial ... the other issue will have far greater impact on your life and your future (and your academic career). Think about it. I am sure that this "stupid cow" is a nice person who, by the way, is only trying to help you learn. And the thanks she gets is that you call her a "stupid cow" ( by the way, are you, like, 5 years old ? ) ... but, more importantly, that you think she is a stupid cow (whether you speak it or not). Shame on you. You yourself may ultimately become a math (or whatever field) genius ... but I certainly wouldn't want you as a math teacher explaining math to my kids. Learning (and teaching) ... and, for that matter, relating to other human beings ... goes far beyond the substantive concepts of math and probability. Now, there's a lesson for you. (Joseph A. Spadaro 17:24, 10 August 2007 (UTC))

Follow Up to the Original Poster -- After reading the above mathematical answers and explanations, you probably now see that your answer was wrong and that your teacher's answer was right (i.e., that you made a mistake and that she did not). So, if she is "stupid" ... what exactly does that make you? And, how does it feel to be called -- or to be thought of as -- "stupid" just because you made a simple and reasonable mistake? Please think about these questions and, with introspection, your honest answers to them. Thanks. (Joseph A. Spadaro 17:39, 10 August 2007 (UTC))

A reference to conditional probability and posterior probability would have been helpful in the initial reply.…81.154.108.76 17:50, 10 August 2007 (UTC)

Just for the record I agree with the above; to assume you know more maths than someone who is paid to teach it is extremely arrogant and to insult her by calling her a cow is a sad reflction on your personal opinions on the world; you are not better than other people and you ought to treat others with the respect with which you would like to be treated. Even if you had been correct, which you weren't, that would have made your teacher neither stupid nor a cow. Humans by nature make mistakes and you, as proven by the above, are not an exception to that rule. Also do you now consider yourself a stupid cow for being the one who was wrong? asyndeton 17:54, 10 August 2007 (UTC)

Amen to that. (Joseph A. Spadaro 18:03, 10 August 2007 (UTC))

Many sayings reflect this wisdom.

• People in glass houses shouldn't throw rocks.
• The pot calling the kettle black.
• When I was very young, my teacher was an ignorant fool; years later I was amazed at how much wiser she became.
• What goes around, comes around.
• When you become a teacher, may you have students as arrogant as you.
• We're all bozos on this bus.

Some teachers can be quite wicked with a student like you. They will take your bold mistake, encourage you to dig yourself a deeper and deeper hole, and then confront you with an unescapable consequence of your flawed reasoning, inviting public ridicule. --KSmrq^T 19:53, 10 August 2007 (UTC)

Eventhough this thread has generated a lot of useful remarks both on mathematics and the psychology of being a student there may still be something left to say which the OP might find useful. (This has to do with the different probability interpretations.)

What does it mean that the probability that the first ball drawn from the urn is black given that the tenth ball drawn from the urn is colored white is 5/9? One way of looking at it is the following. Consider that the experiment the OP describes is carried on simultaneously in 900 different rooms and that there is an overseer which can see what happens in each of the rooms. When the first ball has been drawn in each room, the overseer goes and checks which colour it was and paints the door of the room in the same colour as the ball. How many rooms does the overseer paint white? Well, the probability of the first ball being white is 1/2 so the number of doors painted white is very close to 900 * 1/2 = 450. The rest of the doors will be painted black, and again that will be very close to 450 doors. Now the rest of the experiment is carried out in each room and when all of the balls have been drawn, the overseer goes and checks the colour of the last ball drawn. If the last ball drawn was white the overseer puts a white flag by the door, if the last ball drawn was black the overseer puts a black flag by the door. How many white flags will the overseer put down? Well, again, given that the balls come out in random order, the last ball will be black very close to half of the time and it will be white very close to half of the time. So there will very close to 450 white flags and very close to 450 black flags. Now the question you are asked is the following: How many doors with a white flag are painted black? The answer to that, as explained above, is very close to 5/9 of the doors, that is very close to 5/9 * 450 = 250 doors. What happened? Did somebody travel back in time and repaint the doors? Of course not, there are still very close to 450 doors painted black, so still very close to 1/2 of the total of 900 doors is black, but you are not being asked to look at all of the doors, only those which have a white flag by them, and of the very close to 450 doors which have a white flag by them there are very close to 250 doors which are painted black. So where is the rest of the black-painted doors? We have not accounted for very close to 200 black doors. These doors have a black flag by them and the question asks you to ignore the doors which have a black flag by them. Stefán 21:34, 10 August 2007 (UTC)

The question you are answering is "Given that the last ball drawn will be white, what is the probability that the first ball will be black?" You are quite correct in answering this as "1/2", since the future cannot change the past.

But this is not the question being asked. The question being asked is "Given that the last ball drawn was white, what is the probability that the first ball drawn was black?" Here, the course of history is known, and there are only nine balls that can have been drawn first: any of the five black balls, and four out of the five white balls. The answer is "5/9". --Carnildo 21:47, 10 August 2007 (UTC)

The teacher has set up a logical impossibility in the question - the assistant cannot know what colour the last ball picked will be. No wonder they are angry/confused. How does this represent good teaching - it's just an abuse.87.102.74.130 21:56, 10 August 2007 (UTC)

Can't any of you mathematicians see that this sentence

Now if the assistant wrote in the notebook "The tenth ball drawn from the urn is colored white", what is the probability that the first ball drawn from the urn is black?

is not acceptable english, one alternative would be

Now if the assistant wrote in the notebook "The tenth ball drawn from the urn is colored white", what was the probability that the first ball drawn from the urn was black?

- in which case the probability is unchanged.

I'll write it out to make it simple "the first sentence is grammatically unacceptable english".87.102.74.130 22:05, 10 August 2007 (UTC)

How about: Now if the assistant has written in the notebook "The tenth ball drawn from the urn is colored white", what is the probability that the first ball drawn from the urn is black? Is that grammar acceptable to you? Stefán 22:11, 10 August 2007 (UTC)

No it's not acceptable - because if the "assistant has written the tenth ball drawn from the urn is colored white" then the previous nine balls will have already been picked and the question is not one of probability - since by that time their colours will be known.

I've assumed here that the assistant is not capable of time travel or predicting the future - does that make sense?87.102.5.144 09:30, 11 August 2007 (UTC)

Incorrect or not, we all understood the problem perfectly. This is the maths reference desk and we are willing to forgive, even ignore, an incorrectly used tense. Of all the things the poster deserves criticism for, I don't feel his grammar schould be top of the list. asyndeton 09:44, 11 August 2007 (UTC)

I'm not criticising the poster - I'm criticising the teacher's question and those who answered it - that's why I say that the poster has every right to be "pissed off".

It's not a case of an incorrectly used term - it's a matter of logic - as written the question is nonsense. Myabe thats why the question poster was pissed off - did you think of that?87.102.5.144 10:01, 11 August 2007 (UTC)

It is safe to assume that the poster was pissed off because he thought, incorrectly, that his maths teacher was making an elementary mistake, one which he thought he had detected and she hadn't. He did not call her a stupid cow becuase she used the present indicative where she should have used past historic indicative or the imperfect indicative, or some other indicative past tense. Indeed, he was the one who brought the question here and I severely doubt he used exactly the same wording as his teacher. asyndeton 13:10, 11 August 2007 (UTC)

It's still a question of logic - if the tenth ball has been drawn the previous balls colours are already known - so the question of probability is meaningless - (again ignoring the possibility of time travel) - I think that the poster has at least partially recognised this. No amount of tense changing makes the question logically consistent unless time travel is invoked.87.102.5.144 13:33, 11 August 2007 (UTC)

That is as pedantic as you can get - saying that since the tenth ball has been drawn we obviously know the colour of the first ball - and you are and the original poster are both being unnecessarily difficult by talking about time travel. Do you really need everything flawlessly explained to you as otherwise you will not be able to work out every little ambiguity for yourself? If so, I will adapt the question so that you can fully understand. The person taking the balls from the urn was blindfolded on the first nine turns and when the balls were removed, they were thrown randomly in a corner, making it impossible to tell the order in which they were removed. Then on the tenth go, the blindfold was removed and the person was allowed to the see the bal's colour. Happy? asyndeton 13:42, 11 August 2007 (UTC)

Not really. Simply rewriting the question to help it make sense is one option - however it doesn't help the original poster - to be helpful you should help the poster - not just show that "I can do this" - the poster had issues with the timeline of the events - entirely justified as I see it, that was due to the wording of the question. The poster's point of view was that the colour of the tenth ball does not affect the probability of the colour of the first ball - since the picking of the first ball happens before the tenth ball is picked - I was able to pick that up from comments such as "mathematics cannot violate causality", as it stands the questions is a "trick question" because "The tenth ball drawn from the urn is colored white" does not affect "what is the probability that the first ball drawn from the urn is black?" - despite being constructed in a sentence suggesting that it will.

Clearly the poster can answer questions of the type "there are 5 black and 5 white balls in a bag, a ball is removed (it is white), what then is the probabilty of the next ball being picked being black/white"

In terms of a timeframe it would be obvious to assume that "The tenth ball drawn from the urn is colored white" happens after 'the first ball is picked' - however the question as written is misleading in attempting to suggest that it affects the probability of an event that occurs before. ie the first ball picked is an event that happens before the tenth ball being picked.

There is a problem with the grammar of the sentence - not the maths.87.102.38.115 14:33, 11 August 2007 (UTC)

Stepping back a little bit, 87.102.5.144 said: if the tenth ball has been drawn the previous balls colours are already known - so the question of probability is meaningless. This is not correct. It is meaningful do ask about the probability of events that have already happened. To understand this, take a look at Probability interpretations. Bayesian probability allows you to assign a probability to almost any statement but if you are sceptical about that, frequency probability allows you only to assign probabilites to events that can be repeated. The question the teacher asked the OP has a definite meaning in both of these interpretations. Stefán 20:12, 11 August 2007 (UTC)

The grammar is irrelevant, what is important is the scenario and how it clarifies the absence/presence of information and how it affects the probabilities involved.

Here's a simplified version of the problem: You take a marble out of an urn without looking. After you take it out, you look. It's black. What color was the marble?

You could have taken out any color of marble. You didn't see the marble until after you took it out, so that can't affect what color it was. Would you still agree that the marble was black?

As a question closer to the original, but still with an obvious answer, suppose that there were only two marbles: one black, one white. If the last marble was black, what color was the first? — Daniel 23:47, 10 August 2007 (UTC)

I fully agree with Asyndeton's posts above. I am sure that the original poster's grievances were with the mathematical answers on probability, and not with the (perceived) ambiguous verb tense. If the original poster's grievances were focused on "what is the proper verb tense to correctly articulate this math question?" ... then, I am sure that he would have sought assistance at the Grammar/Language help desk and not the Math help desk. (Joseph A. Spadaro 16:24, 11 August 2007 (UTC))

Even if the problem, as formulated by the OP, is grammatically incorrect, there is only one reasonable interpretation which is even close to it, and that is "what is the probability that the first ball drawn was black, given that the last ball drawn was white?". If the formulation of the problem was too confusing for this meaning to be understood, the correct course of action would have been to ask for clarification (at the time of the exam). -- Meni Rosenfeld (talk) 18:17, 11 August 2007 (UTC)

The issue here is that the poster is 'a child at school' with 'a problem with a question' - I fail to see how a bunch of anal-retentive wogs demonstrating their own superior intelligence and understanding have helped the poster.87.102.8.162 21:46, 11 August 2007 (UTC)

<rhetoric>Were we supposed to demonstrate our stupidity and bewilderment?</rhetoric>

Seriously, the OP has come here with what is essentially a problem in probability; Various people have tried to shed some light on the mathematical issues; some of them have commented on the way the OP has handled the situation. Hopefully, the OP's probabilistic intuition and human-interaction skills have improved as a result. What exactly is wrong? -- Meni Rosenfeld (talk) 21:55, 11 August 2007 (UTC)

Being rude will not help anyone. The OP came here for help and received it; we were also prompted by what he wrote to point out that his tone and manner were not appropriate - neither are yours. No-one here [demonstrated] their own superior intelligence and understanding; they did their best to explain the problem in his logic, and remind him that human beings all make mistakes and all deserve respect. The OP, like you, was extremely sarcastic, pedantic and generally rude when given the help he asked for. That is not a mature way to deal with this issue. asyndeton 22:01, 11 August 2007 (UTC)

Well it's great to see that a stuck up wog like you gets the chance to judge other people - that's what this is all about anyway isn't it.87.102.1.234 10:43, 12 August 2007 (UTC)

I repeat the first sentence of my previous comment. Also, I, along with the rest of the people who watch this page I'm sure, am getting bored of this tiresome argument. asyndeton 11:57, 12 August 2007 (UTC)

Yeah. Seriously. He came in here with an incorrect idea about two things -- the way a certain mathematical question worked, and how to be civil to others, including his teacher. We tried to be helpful and correct him on both counts. What's wrong with that? Nothing. There's no reason for you to be arguing about this at all, and especially no reason for you to be uncivil and nasty about it (stuck up wog?) Please stop arguing over this. Gscshoyru 12:25, 12 August 2007 (UTC)

This is nothing other than Newcomb's paradox, if only in a different form. Unfortunately, that is a paradox, so both sides are kind of correct. 76.19.240.87 02:31, 15 August 2007 (UTC)

Probability vector

How does a Probability vector work? --Savedthat 18:59, 10 August 2007 (UTC)

Suppose you have a suspected loaded die. You throw it a zillion times and find that 1 comes up 14% of the time, 2 through 5 each 17% of the time, while you get a 6 a whopping 18% of the time. This you then can represent as a probability vector [0.14, 0.17, ..., 0.17, 0.18], which I've written sideways because I'm too lazy to make it vertical – but in fact mathematically row vectors are often more convenient here. The fun is when the outcome of a throw is input to a further probabilistic process. If the possible inputs are numbered 1 through m (where m would be 6 for a die roll as input), the possible outputs 1 through n, and P_ij is the probability of outcome j given input i, then the matrix-times-vector expression P^Tv gives the probability vector of the output if v is the probability vector of the input. Here the superscript T denotes matrix transpose. If you work with row vectors, you can use vP. Such row-vector-times-matrix expressions occur when dealing with Markov chains.  --Lambiam 19:33, 10 August 2007 (UTC)

What is meant by outcome of a throw is input to a further probabilistic process.? --Savedthat 20:35, 10 August 2007 (UTC)

For example, if you throw for instance the value of 3 with the die, you may then throw 3 coins and count the number of heads. So what is the chance you will get 4 heads or more, assuming the coins are fair? With a fair die, it is 19/192, or slightly less than 9.9%. With the loaded die as above, it is slightly above 10.4%.  --Lambiam 21:55, 10 August 2007 (UTC)

What does n stand for? --Savedthat 03:11, 11 August 2007 (UTC)

Above it stands for a natural number, in particular the number of possible different outcomes of some process. For the specific example of throwing as many coins as the value coming up on the die and counting the number of heads up, n would be equal to 7.  --Lambiam 22:47, 11 August 2007 (UTC)

How did you get the number 7, I mean what did you do to get it? --Savedthat 05:27, 13 August 2007 (UTC)

Big E?

What does the Big E mean? Is there an article about this big E?

$\sum _{i=1}^{n}p_{i}=1$

--Savedthat 20:33, 10 August 2007 (UTC)

Sigma notation Dlong 20:35, 10 August 2007 (UTC)

edit conflict Great question. The letter is a Greek capital Sigma, which is used to denote a sum. See Sum#Capital_sigma_notation. In this case, the notation means

$\sum _{i=1}^{n}p_{i}=p_{1}+p_{2}+p_{3}+\dots +p_{n}$

Hope this helps, --TeaDrinker 20:37, 10 August 2007 (UTC)

Note that this is said aloud as "the summation for all values of p sub i for i from 1 to n". StuRat 00:10, 13 August 2007 (UTC)

Why do people use Sigma notation in mathematics? --Savedthat 20:40, 10 August 2007 (UTC)

In many cases it is not feasible to write out every term one wishes to sum. An alternative to sigma notation is to use an ellipsis (three dots: ...), as I did above. However using ellipes requires the reader to recognize the a pattern (since the ellipsis just means "continue this pattern"). Sometimes this is easy, such as

$\sum _{i=1}^{n}i^{2}=1^{2}+2^{2}+3^{2}+\dots +n^{2}$

Other sums are not so easy to recognize the pattern, leading to difficulty and potential ambiguity for the reader. --TeaDrinker 21:10, 10 August 2007 (UTC)

Because it's much more concise than the alternative, and it's also set up that way so there can be no confusion. 3 + 5 + 7 ... 17 could mean the sum of odd numbers, or odd primes, or something else totally different. By stating specifically what pattern is being used in a formula or otherwise, everyone understands what is meant. Gscshoyru 21:05, 10 August 2007 (UTC)

To actually answer the question, the "big E" is expectation.

Thanks. --Savedthat 05:26, 13 August 2007 (UTC)

Greedy algorithms and making change

As whoever answers this question will know, the term greedy algorithm refers to an algorithm under which one proceeds in a step-by-step making locally optimum choices without regard to longer-term consequences. For example, if a shopkeeper makes change using a greedy procedure, he or she will use the largest coin that is less than the amount of change to be made, then the largest coin that is less than the amount left, and so forth. E.g., the United States has (commonly used) coin denominations of 25 cents, 10 cents, 5 cents, 1 cent. If 40 cents' change is owed, the shopkeeper will use a 25-cent piece, then a 10-cent piece, then a 5-cent piece.

It has been shown that with this system of coinage, the greedy algorithm will always satisfy the criterion of making the correct change with the smallest number of coins. For other coin sets, however, this is not the case. For example, consider a country that has 25-cent, 20-cent, 10-cent, 5-cent, and 1-cent pieces (as the U.S. did for a time in the 19th century). Under this scenario, the shopkeeper would still make change for 40 cents with 25 cents plus 10 cents plus 5 cents (3 coins), even though the optimal solution is two 20-cent coins (2 coins).

Is there a general formula or test that establishes which coin sets will, and which will not, have the making of change optimized using the greedy procedure? Newyorkbrad 20:40, 10 August 2007 (UTC)

So can "Greedy algorithms" predict future using probabilities? --Savedthat 20:56, 10 August 2007 (UTC)

No... this a completely different question that yours, sorry if you're confused. And a very interesting question at that. Now I have to think about this more... Gscshoyru 21:00, 10 August 2007 (UTC)

Well, an obvious sufficient condition is that the denominations, listed from lowest to highest, are such that each value divides the next. This is the case for the US example above. It is not, however, necessary, as shown (for instance) by British coinage, with its denominations of 1, 2, 5, 10, 20, 50, 100 and 200 pence. Algebraist 22:13, 10 August 2007 (UTC)

Thanks for the response, but I think your second sentence is not true (U.S. coins are 1, 5, 10, 25; 10 doesn't divide 25). Newyorkbrad 22:56, 10 August 2007 (UTC)

<sigh>I thought by now I'f learnt not to edit when both half-asleep and half-drunk... Algebraist 23:42, 10 August 2007 (UTC)

Consider a three coin set a,b,c with a>b>c. The gready algorithm will fail if a+2c=2b, or indeed if a+nc=2b for n>=2 and nc<a. This suggests that it may fail if 2b>a. This is probably the simplest set of cases for failure no doubt there are others. Its also worth observing that not all coin sets will always be able to give the correct change. --Salix alba (talk) 22:34, 10 August 2007 (UTC).

Good point... Newyorkbrad, may we assume that the smallest coin is 1, or, that is, you could pay for anything exactly with enough of smallest coin?

I wonder if it makes a difference... Gscshoyru 00:03, 11 August 2007 (UTC)

This seems to be solved in the following papers:

• Chang, S. K.; Gill, A. (1970), "Algorithmic Solution of the Change-Making Problem", J. ACM, 17 (1): 113–122, doi:10.1145/321556.321567
• Chang, Lena; Korsh, James F. (1976), "Canonical Coin Changing and Greedy Solutions", J. ACM, 23 (3): 418–422, doi:10.1145/321958.321961.
• Kozen, Dexter; Zaks, Shmuel (1994), "Optimal Bounds for the Change-Making Problem", Theor. Comput. Sci., 123 (2): 377–388, doi:10.1016/0304-3975(94)90134-1

—David Eppstein 00:28, 11 August 2007 (UTC)

Thanks! I'll take a look at these. Regards, Newyorkbrad 00:33, 11 August 2007 (UTC)

Interestingly, the abstracts (I'll look up the papers themselves tomorrow) seem to suggest that the problem actually still has not been solved—they speak in terms of bounds for testing a given coin set to see whether a counterexample to greedy-optimality exists, rather than looking at the coin combination itself and determining it. This appears to be a far more complicated problem than I would have assumed. Newyorkbrad 00:41, 11 August 2007 (UTC)

When you do look, do you think you could sorta summarize some of the more interesting parts, here? We seem to need to pay to see the papers, unless you have a subscription or whatever, and this question has me seriously interested. Gscshoyru 00:49, 11 August 2007 (UTC)

I think the abstracts, which are available without charge, probably give as good a summary as I could. Maybe one of the more mathematically advanced readers here could do better. Regards, Newyorkbrad 01:16, 11 August 2007 (UTC)

I remember looking at this algorithm in a computer science elective subject I took last year. As I recall, the coin sets can fail the algorithm if there are any two coins of values a and b such that a < b < 2a, or to put it another way, if any coin is smaller than another coin but larger than half of that other coin. I'm not sure if this is necessary and sufficient for failure; we only looked at it briefly. Maelin (Talk | Contribs) 01:26, 11 August 2007 (UTC)

a < b < 2a is unnecessary for failure: consider the set 1, 10, 25 (try to make 30). It is also not sufficient: the greedy algorithm works for 1,2,3. Algebraist 19:01, 11 August 2007 (UTC)

One more reference:

• Pearson, David (2005), "A polynomial-time algorithm for the change-making problem", Oper. Res. Lett., 33 (3): 231–234, doi:10.1016/j.orl.2004.06.001

—David Eppstein 01:29, 11 August 2007 (UTC)

greedy algorithm predicting stock price

Can this formula work:

C_t - LL_t-_n Divided by HH_t-_n - LL_t-_n

Multiplied by 100, where LL_t and HH_t mean lowest low and highest high in the last t days, respectively? --Savedthat 21:12, 10 August 2007 (UTC)

I have no idea what you mean. What are you trying to do? What do you mean by "work?" Gscshoyru 21:15, 10 August 2007 (UTC)

For the above algorithm, I mean. --Savedthat 21:17, 10 August 2007 (UTC)

Wait... you mean the greedy algorithm one? Then... Highest high and lowest low? Days? I'm very confused... Gscshoyru 21:19, 10 August 2007 (UTC)

Well consider the greedy algorithm formula listed above for a stock? --Savedthat 21:29, 10 August 2007 (UTC)

That's not an algorithm. It's barely a formula. And you've not defined with $C_{t}$ and $n$ mean. In any event, any attempt to predict stock prices is bound to failure. One that could be expressed with a line of rudimentary arithmetic, if it worked, would be in such widespread use that it would probably cause itself to fail over time. Donald Hosek 22:57, 10 August 2007 (UTC)

C stands for the stock price. t stands for time. n stands for the number of rows in the matrix I believe. --Savedthat 03:10, 11 August 2007 (UTC)

You believe?? Didn't you come up with it yourself?

I have a feeling that you have no idea what a greedy algorithm is. It has absolutely nothing to do with probability, or predicting the future. I'd suggest reading the article first, so you have a better understanding, rather than asking about everything that you don't understand. Gscshoyru 03:31, 11 August 2007 (UTC)

Doubly stochastic matrix - n

What does n stand for in the article Doubly stochastic matrix? --Savedthat 21:38, 10 August 2007 (UTC)

I think its just the number of rows or columns in the matrix. The class of n × n doubly stochastic matrices. --Salix alba (talk) 21:58, 10 August 2007 (UTC)

Thanks. --Savedthat 05:26, 13 August 2007 (UTC)