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March 6

How many edges on a cone?

Ive typed the phrase "how many edges on a cone?" into google and gotten moderate success. I am wondering if anyone knew of a definitive answer and could they explain? —The preceding unsigned comment was added by 71.162.95.164 (talk) 01:16, 6 March 2007 (UTC).[reply]

What do you mean edges? If you mean the edge as in where the slanted part tapers off to the circle, then it's one. If you mean how many tangent lines there are, it's infinite. --Wirbelwindヴィルヴェルヴィント (talk) 03:19, 6 March 2007 (UTC)[reply]

Wouldn't there also be one at the peak? − Twas Now ( talk • contribs • e-mail ) 16:57, 6 March 2007 (UTC)[reply]

If you mean sharp angle changes on an infinite cone, the answer is zero. StuRat 12:37, 6 March 2007 (UTC)[reply]

But if you mean faces, wouldn't there be two? One on the bottom and one wrapped around the cone? --Theunicyclegirl 19:25, 12 March 2007 (UTC)[reply]

coordinate geometry and permutation and combination

i'm studying permutations and combinatorics for this entrance exam and i just thought of a really interesting(not sure if its feasible) problem:In a 3d orthogonal coordinate system, in how many ways could u move from a point(2,3,5) to a point(9,11,14)[(or in general from(x,y,z) to (x',y',z')]assuming that u could only move parallel to the coordinate axes and on integral points.Is there some sort of generalized algorithm to find the solution??59.183.42.170 18:24, 6 March 2007 (UTC)[reply]

I will assume we are only allowing steps in positive directions, otherwise there are clearly infinitely many possible routes. In this case, we have to make a total of 7+8+9=24 steps, 7 in the x-direction, 8 in the y-direction and 9 in the z-direction. Hopefully it is obvious how to proceed; I won't say more in case this is homework. Algebraist 20:02, 6 March 2007 (UTC)[reply]

An inductive approach may prove fruitful. Consider a grid with vertices at integer points, label each vertex with the number of ways in which to reach the point (0,0) when moving down/left along grid lines. I've translated and reflected your grid to make things easier.

|   |   |
1 - 3 - 6 -
|   |   |
1 - 2 - 3 -
|   |   |
1 - 1 - 1 -

Continuing the numbers for a little bit and you will find a well know set of coefficients. (hint the guy who gave his name to pressure). The three dimensional case continues in a similar fashion and there is a less well know sequence which gives the numbers. --Salix alba (talk) 22:50, 6 March 2007 (UTC)[reply]

maybe Multinomial theorem will help you --Spoon! 00:57, 7 March 2007 (UTC)[reply]

Best tactic in a gambling game

There is this game on line that my mother likes to play called Limbo. The idea is that everyone that plays picks a number between 1 and 99999 and whoever picks the lowest unique number (i.e. the the lowest number that only one person picked). What is the optimum strategy to play here? I haven't studied probability theory in years and years, but my gut tells me that you would have to use some sort of density function that plots what numbers people are expected to pick based on how many people are playing (so like f₁₀₀₀(x) would plot what numbers a 1000 people are expected to play and its definite integral would plot how many people have played below a certain number), and from that you could figure out the expected value. Or something. As I said, years and years, I'm probably completely off-base :) Can someone give my mother some pointers? Oskar 21:30, 6 March 2007 (UTC)[reply]

Hey pal, I'm no mathematician but I believe the larger the number of participants, the larger you should pick your number. Perhaps the number that would give you the best probability of winning is a function of the number of participants. —The preceding unsigned comment was added by 209.53.180.38 (talk) 22:00, 6 March 2007 (UTC).[reply]

f (2)=1 When there are only two players, pick the number 1.

f (3)=1 When there are 3 players, pick the number 1 because the other two players will pick the number 2. 202.168.50.40 22:17, 6 March 2007 (UTC)[reply]

For two players, as you say, picking 1 is a symmetric Nash equilibrium. If this leads to the players sharing the prize, the equilibrium is strict and also Pareto optimal, and thus can be expected to be stable against collusion. If nobody gets the prize, however, it is neither, and the players might be tempted to make a deal to let each other win on alternate turns, since they can only gain from doing so. As soon as either player deviates from always picking 1, however, the other player can gain a relative (though not absolute) advantage from reverting to the equilibrium strategy. Thus the strategy of picking 1 satisfies Maynard-Smith's second condition, and is therefore an evolutionarily stable strategy. In fact, the game is equivalent to a degenerate version of the Prisoner's dilemma with S = P.

For three players, however, picking 1 is not an equilibrium. Neither is picking 2 nor randomly picking 1 or 2. In fact, I haven't been able to find a symmetric Nash equilibrium for the three-player game, and I have a hunch that there might not be any, though I haven't managed to prove that so far. Of course, per Nash's theorem, there must be a (possibly asymmetric) equilibrium. For the case where prizes are not shared, one player picking 1 and the others picking any other number is indeed a Nash equilibrium; so is one player picking 2 and the others 1. Both are, however, not strict, and therefore unstable. —Ilmari Karonen (talk) 22:00, 7 March 2007 (UTC)[reply]

What if there are 5000 players? Oskar 23:38, 6 March 2007 (UTC)[reply]

I can't imagine you could figure it out without knowing what the particular people you're up against usually do. For instance, does anyone take 1, or are they too scared? Hard to say which it is without looking. If they don't, and you stepped in and started winning that way, would anyone sacrifice themselves to make your vote irrelevant by taking 1 as well? If they did, but paused first, could you time it right to sidestep the rush? Hard to guess at. One thing that makes it easier is that you don't mention anything about runners-up, so it doesn't matter how far down the distribution you are, only if you're at the very bottom of the pile. So, playing an extreme game, near the very very bottom, would be the most practical system, surely. If you play up near the top, or even the middle, you're gambling that every single other person in the game will play above that, which is almost certainly not going to happen with 5000 players. If you play low, you're only betting you won't land on anyone else, which has better (though still not very good) chances. Black Carrot 07:08, 7 March 2007 (UTC)[reply]

You know, the more I look at it, the worse a game it sounds. Since you suggest you might go to the work of setting working out a density function of some kind (which could only be found through empirical observation), consider keeping track of who wins. Is there a small group of people who win consistently whenever they're playing, or is it closer to random? If there is, ignore everyone else and focus on emulating and then beating that small group. If not, it's quite possible there's no way to significantly beat the odds. Keeping in mind, BTW, that the odds are roughly 5000 to 1. Black Carrot 07:12, 7 March 2007 (UTC)[reply]

This game is played on The Hits Television channel, they'll offer up a huge prize for the price that someone txts in that is the lowest unique price they recieved, and they would run it for a week or so. it'll be something like a £15000 car, and the person who wins will normally have guessed around the £2-10 mark. They ust get a huge number of entrants. In this kind of system, plenty of people will guess very low numbers. Capuchin 12:04, 7 March 2007 (UTC)[reply]

I'd look at the winning numbers from the previous runs, and guess somewhat near (within 10%), but not extremely near (within 1%), those. If the prize is a bit more valuable than average, assume more entrants and guess higher, if the prize is a bit less valuable, guess lower. However, I'm not confident that this will increase your chances enough so it's actually worth the risk, this would require a large number of poor players and/or a very generous contest that isn't very profitable. StuRat 14:26, 7 March 2007 (UTC)[reply]

If this is a TV show, I'd assume that it's completely crooked if the prize has any real value. They'll not admit anyone has won until enough mugs have placed ripoff phone calls to generate the desired amount (this is clear from the "week or so" comment above), then just pick someone with a with a lowish bid at random.—86.132.234.51 18:40, 7 March 2007 (UTC)[reply]

I wouldn't say that this is a pure math question. It looks much like an for mathematics untreatable game like poker, where knowledge about people is more important than statistics. Mr.K. (talk) 16:52, 8 March 2007 (UTC)[reply]

I'm not so sure about that. While individual human behavior can't be predicted mathematically, collective human behavior can be, using probability and statistics and things like the Black–Scholes formula. StuRat 17:10, 8 March 2007 (UTC)[reply]

Yes, but this people are part of the statistics and the result of it. Considering that all the players have access to the stats you still don't know in what way to go. It look like a traffic jam situation. If you try to avoid the traffic jam that builds up direct after work, you can wait a couple of hours to drive home. But then you will stuck in the traffic jam of those that wait a couple of hours after work to drive home. Furthermore, modelling a process doesn't mean that you can predict it. The underlying math beneath Black–Scholes, the Brownian motion, is a way of modelling a random movement (=> non-deterministic, non-predictable) Mr.K. (talk) 21:31, 8 March 2007 (UTC)[reply]

I would think that, if a mathematical study were done, we could predict the most likely winner for a given number of entries, just as we can predict whether a given poker hand is likely to be a winner, even considering that each of the other player's hands is unpredictable. The guesses for this game should fall into some general distribution (bell-shaped ?). Then there might be preferences for certain types of numbers, such as primes. Once we determine those type of things, we could come up with some rule, like finding the next prime number after the square root of the number of entries. StuRat 02:17, 11 March 2007 (UTC)[reply]

Debt question

I'm in debt a total of $9400 to all my credit cards. I have an offer from a debt consolodator that he will pay off all my credit cards and I would pay him back $200/month for 5 years for a total of $12,000 (however it could be more if i dont make my payments on time, but i dont think that will be a problem).
Consider this: my credit cards all have annual interest rates of about 17%. I'm not very good at math, but I'm able to do this much: 17% of $9400 devided by 12 months means about $133 of interest a month to start. Obviously, as the balance gets smaller, I will be paying less interest and more of my $200 would be going towards chipping away at the balance. Can someone show me how to do:
a) how long would it take me to pay it off on my own if i do $200/month and how much would i have paid by the end?
b) to what annual percentage rate is the $12,000 offer equivalent to?
c) and just out of curiosity, if i don't take the offer, how do i know how much of my $200 is going to interest and how much to paying off the balance after X number of months? there should be a simple formula for that? Answer as many as you can and please don't just give me the answers, show me how its done with the formulas and stuff.
P.S. I might take the offer even if it's $12,000 is a little more than I'd pay otherwise, just because its less hassle than 8 credit cards. —The preceding unsigned comment was added by 209.53.180.38 (talk) 22:31, 6 March 2007 (UTC).[reply]

For (a), I wrote a simple computer program to simulate a series of monthly payments. You could also do it with a spreadsheet. The core of the calculation is:

interest paid this month = principal remaining * (rate/12)

principal paid this month = payment - interest paid this month

principal remaining = principal remaining - principal paid this month

which represents one month's payment. Repeat the calculation until the principal drops below 0. Multiply the number of months by the amount you're paying each month to get approximately the total amount you're paying. (It won't be exact, because of rounding and because the final month's payment will be less than $200, but it gives a rough idea.)

For (b), you calculate the APR by dividing the interest paid by the amount you owe to get the total interest rate, then divide that by the number of years to get the annual percentage rate. Because of the wonders of compound interest and amortization, calculating the actual interest rate is much harder, but the actual interest rate is largely irrelevant.

For (c), see my answer to (a).

Hope this helps. --Carnildo 23:21, 6 March 2007 (UTC)[reply]

For (a), you can use Time value of money#Present value of an annuity. You want to pay off a present value of PVA = $9400, with an annuity of A = $200, and a rate of

r=(1+.17)^{1/12}-1

per month, and you solve for the number of periods n you have to pay. You get

n=-\log _{1+r}\left(1-{\frac {PVA\cdot r}{A}}\right)

—The preceding unsigned comment was added by Spoon! (talk • contribs) 01:29, 7 March 2007 (UTC).[reply]

I don't understand, why can't you do this using an excel spreadsheet? The monthly interest rate is either 0.17/12 or ((1.17)^(1/12) - 1 ). Thus either 1.416% per month or 1.31696% per month depending on how the monthly interest is calculated. Oh yes, you have to decided whether you want to pay at the start of the month (before the monthly interest is calculated) or pay at the end of the month (after the monthly interest is calculated).

Using 1.416% and excel spreadsheet, it shows you will finished paying it off yourself at the end of the 76th month. That's 6 years and four months at $200 per month. with $15200 (76 payments). 220.239.107.247 06:59, 7 March 2007 (UTC)[reply]

Some other ways to pay less:

1) Get a low interest rate loan, such as a house mortgage, and use that to pay off the debt. You might also have a friend or relative who is willing to give you a low interest (or even no interest) loan.

2) Do the credit-card shuffle game, continuously moving the balance to new cards with low or no interest introductory periods. Be careful, though, as many have "gotchas", and will charge you huge interest rates retroactively if you are a day late on a payment, etc. Be sure to cancel the cards after the introductory period ends.

And, to keep from getting into trouble again, you need to avoid getting into debt for most items. There are only a few things that justify debt, in my opinion: education, because of it's high return on investment, a house, due to avoiding flushing rent down the toilet, a reliable car, if needed to get to work, and health care, if needed. And, in those cases, you should buy the minimum house and car you actually need, not a mansion and a Mercedes. Other things, like vacations, clothes, entertainment, appliances, electronics, and gifts, should never be financed. You can make due with cheaper versions of those things, skip them altogether, or wait until you can actually pay for them. For gifts, for example, instead of buying expensive presents, offer to clean their gutters for them. For vacation, drive to the zoo, museums, aquariums, etc., in your area instead of taking a cruise. For entertainment, watch free TV instead of cable or paying for concerts. Dine at home instead of at expensive restaurants. Buy only the clothes you actually need (no designer labels). Skip the cell phone. Play free online or downloadable video games instead of buying games. Then, once you've paid off your debt and can actually afford a few of life's luxuries, that's the time to buy them (while still saving some money for the future and retirement). I'm sorry if you're a financially responsible person who got into debt for a legit reason, like medical bills, but I went with the odds and assumed you're as financially irresponsible as most people. StuRat 14:11, 7 March 2007 (UTC)[reply]

I think too many people are treating this as a math question and not a finance question. There are a number of places you can turn to get the rates reduced on your credit cards. I believe the law states that the people who negotiate with the companies must be non profit but you still should be careful.

I had a very poor, very old friend who gave me a power of attorney so I could tell the credit card companies to leave her alone. I basically told them they were not going to get a single dime from her and it wasn't worth forcing her into bankruptcy because there was no money. They might have been able to reposess a few dresses but she didn't even have enough social security money to pay her rent. I wasn't going to negotiate because there was nothing to negotiate about. However, a number of the companies were willing to cut the rates significantly and some of them would even cut the amount that she owed. I don't remember the details.

My point is that you should talk to various professionals and get their advice. It should be free.