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TheDeath
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posted November 08, 2008 02:08 PM |
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So we have Corribus and Binabik with 2/3, dimis and TA with 1/2, which is right? 
I agree with Corribus however. When you look at a side, and it is a H, you have 1/2 chances so the other is a H, right? While when you look and it is a T, then you are 100% sure the other is a H. So the probability must be higher than 1/2, although I didn't do the math myself.
@Keldorn: The problem is that you're mixing up how much they actually PAID and the money the serviceman kept (which is similar to the "cost" since it's not theirs).
Thus, in that scenario, they initially paid 30, right?
Then, the owner decides to give them back 5. This means they only (theoretically) should have paid 25.
The serviceman then only gives them 3 out of 5, so it's 25+2 = 27, how much they paid (normally). HOWEVER, there is a problem because you MIX the amount of how much they PAID with how much the serviceman/owner gets. The operation is like this:
30 - 5 = 25 (should be)
30 - 3 = 27 (how the serviceman decided)
but then, if you want to find out "the sum", you don't add up the amount the serviceman kept to find out the total number of coins they PAID:
27 + 2 = 29
more like:
27 - 2 = 25 (to get the total amount of how much they should have been given back)
If we look at it like this, and count the TOTAL number of coins, we end up with:
25 + 3 + 2 = 30
25 is the number of coins they should theoretically pay
3 is the number of coins they were actually given back
2 is the number of coinsthe serviceman kept for himself
the sum is 30, no missing coin
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JollyJoker
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posted November 08, 2008 04:11 PM |
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Quote:
On the other hand, you might want to Aim for Luck + Logistics + Light instead, which will open up for Nature's Luck which, when things are taken into acount, is superior even to Imbued War Machines.
I doubt that, because for Elven Luck you sacrifice Resistance, which is a downer, plus it may depend on whether you get a 50% bonus artifact. The other game I had Phoenixfeather Cape and with High Druids the Ballista banged out a triple 600 points Fireball shich is something even a Warlock will gulp. 
@TheDeath
We have 2 problems, haven't we?
Corribus' problem all agree upon 2/3.
The problem with the 5 coins most(?) agree on 1/2, right?
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Corribus
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posted November 08, 2008 05:02 PM |
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Quote:
So we have Corribus and Binabik with 2/3, dimis and TA with 1/2, which is right?
I agree with Corribus however. When you look at a side, and it is a H, you have 1/2 chances so the other is a H, right? While when you look and it is a T, then you are 100% sure the other is a H. So the probability must be higher than 1/2, although I didn't do the math myself.
Trust me, it's 2/3. Do the experiment yourself if you don't believe me. Use two pieces of paper with H/T written on them instead of coins. Do the experiment 100 times, and see what the result is. (I've done it.)
The problem is most people get stuck on the idea of having 2 coins. You can't think like that. There are 3 total possible scenarios of drawing a heads. Of those 3, 2 of them have heads on the other side. It doesn't matter that some of those options involve the same coin.
In my experience, no matter how many time you try to convince some people, they won't believe that the answer is not 1/2. Same thing with the Monty Hall Problem.
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I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. -Mitch Hedberg
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TheDeath
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posted November 08, 2008 06:18 PM |
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Quote:
We have 2 problems, haven't we?
Corribus' problem all agree upon 2/3.
The problem with the 5 coins most(?) agree on 1/2, right?
Oh yes sorry 
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Asheera
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posted November 08, 2008 06:21 PM |
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Quote:
1a) I sketched another possible magic system which would work the following way: if you build a level of the guild it would SHOW 4 random spells, from which you would PICK the allowed number (3 for levels 1-3, 2 for levels 4-5, plus 1 for the Library), with the picked ones being the ones that could actually be learned in the guild.
Question: what probability do you have with THAT system to get ANY specific spell starting with 2 towns?
I take it it's not with TotE spells which would make it very difficult?
With one town we have 4/8 (1/2) chances to get a specific spell for a level. So that makes it 75% with two towns.
Or did I miss something?
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friendofgunnar
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posted November 08, 2008 06:22 PM |
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Quote:
In my experience, no matter how many time you try to convince some people, they won't believe that the answer is not 1/2. Same thing with the Monty Hall Problem.
lol, I actually had to do the monty hall problem a couple dozen times to convince myself the math wasn't fruity 
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TheDeath
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posted November 08, 2008 06:36 PM |
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Edited by TheDeath at 18:37, 08 Nov 2008.
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Quote:
2) For my suggested system, what probability do you have in ONE town, to get at least ONE spell of a specific magic school on all 5 mage guild levels (as opposed to the 100% actually guaranteed for 2 schools in each town)?
Let's take an example with the Dark magic school.
So for a level, you get 4 random spells from which you can pick (that means, at least 1 dark to show up in those 4) from a total of 8. So we need combinations.
The full number of combinations for 8 elements, taking 4 at a time, is:
C(8,4) = 8! / ((8-4)! * 4!) = 70 (total combinations)
Now, how many of those combinations contain at LEAST one dark spell (and there are 2 in total)? First, we make all combinations with exactly one dark spell, let's call it A (same for the other B). That means, we take all possible combinations with 3 at a time (because the fourth 'selection' is the dark spell) from 6, not from 8, because we excluded the two dark spells.
C(6,3) = 6! / ((6-3)! * 3!) = 20
We double this, because we do the same for B (when one of the four selections is FIXED on B, and the other 3 make all the possible combinations from the remaining 6 non-dark spells).
So we have 40 such combinations. Note that we can also have when we get TWO (BOTH) dark spells with the same selection. This means, we need to take all possible combinations from those 6 non-dark spells, taking 2 at a time (the other 2 are for the two dark spells).
C(6,2) = 6! / ((6-2)! * 2!) = 15
So in total, the number of combinations that include at least one dark spell is 20*2 + 15 = 55. The probability to get these is simply the number above, divided by the total amount of combinations, which is 70 (see above).
55/70 = 11/14 = 78.5% (approximately)
This is to get if you need only ONE level (guild level). To get it for ALL levels, the chances lower considerably.
(11/14)^5 = 29.9% (approximately)
Did I make any mistakes?
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JollyJoker
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posted November 08, 2008 06:53 PM |
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@Ash and Death
Both right.
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TheDeath
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posted November 08, 2008 07:00 PM |
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Now that I look at it, the explanation (written down, without just formulas) is longer than I expected (in my head, it's a lot "shorter") 
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dimis
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posted November 09, 2008 08:58 AM |
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Edited by dimis at 10:03, 09 Nov 2008.
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Towards infinity ...
If anyone still has questions about the 1 T/T + 2 T/H coin-problem (which is a simplification of Celfious' 5 coin-problem), just post.
Since we recently talked about the 0.999... = 1, I think it's a good opportunity to "review" some stuff about infinity. So, how do people prove that two sets have the same size? Well ... for finite sets, it is rather straightforward. One can simply count the elements of one and the other and then decide. But this fails if the sets are both infinite (simply because you 'll never stop counting). So, the idea is to match the elements of one, with the elements of the other and it's very natural since it avoids counting and does what it intuitively has to do for equality among these two sets. For example, if one wants to prove that the set A = {1, 2} is "of the same size" (from now on "has the same cardinality") as B = {3,4}, he can simply exhibit a way of "pairing" elements between the two sets. Hence, f : A >--->> B, with f(x) = x + 2, does the trick, since f(1) = 3 and f(2) = 4.
That f matches elements, means that every element of one set will be paired with exactly one element of the other set, and vice versa (in math language this is called bijection or "one-to-one and onto").
Note that g : B >--->> A, with g(x) = x - 2 also does the same trick.
So, here are some problems, which although simple (may be not all of them ...), indicate that strange things happen at infinity.
1] Show that the set of natural numbers N = {0,1,2,3,4,...} has the same cardinality (denoted as =_c from now on) as the set of even natural numbers E = {0,2,4,6,8,...}.
Once you show that note that you proved that a proper infinite subset of an infinite set can have the same cardinality as the "bigger" one.
2] Continue the shock by proving that N =_c Z, where N again the natural numbers {0,1,2,3,4,...} and Z the set of integers {...,-3,-2,-1,0,1,2,3,...}.
3] What about P = (0,+oo) and the Reals R = (-oo,+oo)?
4] What about U = (0,1) and the Reals R?
5] What about P = (0,+oo) and U = (0,1)?
Have fun!
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keldorn
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posted November 09, 2008 09:41 AM |
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To react to the solutions of my problem:
This is a math paradox. The solution is only wrong in the way I've shown you. All other ways give good solutions (I've seen 2 good ones, well done ) By the way, scientists don't really understand what is the problem with this exercise and how can this be true.
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TitaniumAlloy
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posted November 09, 2008 09:53 AM |
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the monty hall problem is cool
i forgot about that one
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dimis
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posted November 09, 2008 09:57 AM |
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Edited by dimis at 09:59, 09 Nov 2008.
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Just to note here something. When "paradox" doesn't mean "mistake/wrong" it means "a fact that contradicts our intuition". And yours is a paradox in the first sense. We 'll see a paradox of the second kind soon.
So, any solutions to the "cardinality" problems?
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JollyJoker
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posted November 09, 2008 10:18 AM |
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I love that.
You can easily prove that the fractional or rational numbers have the same cardinality than natutals and integers, by painting a coordinate plane with x-axis being the numerator and the y-axis the denominator. By starting at 0/0 and then going in a spiral round like {1/0, 1/1, 0/1, -1/1, -1/0, -1/-1, 0/-1, 1/-1, 2/-1, 2/0, 2/1, 2/2, 1/2...} you can count them all and pair them with naturals.
You cannot do that for the irrationals. Irrationals have a higher cardinality than rationals, integers and so on.
You might say, up to rationals every number is a point, the points themselves, while their distance to wach other is infinitely low, are still separate points.
This changes with the irrationals, which create "space" by what I would call the mathematical uncertainty relation. You can take any irrational and try to pinpoint it, but if you stop at, say, the 20th figure behind the decimal point it becomes undefined from that figure: let's say you compute square root of 2 and reach at 1.4142. At this point, you cannot say anything about the next figure - therefore sqrt2 "fills up" the whole space between 1.4142 and 1.4143
It's interesting to note that the simplest geometric constructions already involve irrational numbers.
So that's how the irrationals make a straight line out of a string of points, creating a higher dimension.
Speaking about interesting mathematics - anyone interested in fractal mathematics?
http://en.wikipedia.org/wiki/Image:Mandel_zoom_00_mandelbrot_set.jpg#file
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william
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posted November 09, 2008 10:19 AM |
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I'm interested in Fractals and learning how to make them and all that, but I have never really been any good at it since I'm really not that good at Maths. I love Fractal Images, some of them are fantastic.
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Ecoris
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posted November 09, 2008 03:50 PM |
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Edited by Ecoris at 15:52, 09 Nov 2008.
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Quote:
How can you determine whether any given natural number (so no fractions or decimals or such) is divisible by 2, 3, 4, 5, 6, 9, 10 and 11?
Why is 7 skipped? It's not too difficuly. If you have an integer abcd..xyz you "remove" the last digit multiply it by two and subtract from the rest. I.e. 7 divides abcd...xyz if and only if 7 divides abcd...xy - 2z.
E.g. 7 divides 1092 since 7 divides 105 = 109 - 2*2. (If you don't know that 7 divides 105 you can perform the iteration once more: 7 divides 105 since 7 divides 0 = 10 - 2*5).
11 divides a number if and only if 11 divides the alternating checksum a - b + c - d + ... .
Quote:
3] What about P = (0,+oo) and the Reals R = (-oo,+oo)?
4] What about U = (0,1) and the Reals R?
5] What about P = (0,+oo) and U = (0,1)?
3]: The natural logarithm ln is a bijection (0,+oo) -> (-oo,+oo) (the inverse is the exponetial function).
5]: The map given by x -> x/(1-x) is a bijection (0,1) -> (0,+oo) (the inverse is given by x -> x/(1+x).
4] follows from 3] and 5].
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TheDeath
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posted November 09, 2008 05:31 PM |
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Quote:
I'm interested in Fractals and learning how to make them and all that, but I have never really been any good at it since I'm really not that good at Maths. I love Fractal Images, some of them are fantastic.
Well that technically requires not only math but also knowledge of designing an algorithm and using math (sometimes even finite state machines are used), usually trigonometry (if you want "rotating" or other weird effects) is used for movement or rotation, otherwise you'll end up just with splashes. I'm not too familiar but suffice to say it's not just plugging in a formula to do the job, at least if you want to make your OWN you have to use your own formula.
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dimis
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posted November 09, 2008 05:56 PM |
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Edited by dimis at 18:00, 09 Nov 2008.
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@william: Why don't you try to participate here? My goal was to see some very interesting problems which can be easily stated nevertheless. If you want we can see some examples on fractals too. But first we need some foundation on tedious programming tasks. Then we can move on... TheDeath is right. You need math and programming for generating your own fractals.
@Ecoris: Thanks for participating. This gives me the opportunity for the following problem:
@all who haven't seen this before: We have a new problem:
So far the participants of this thread are:
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dimis, Corribus, Celfious, william, TitaniumAlloy, Binabik, JollyJoker, Asheera, TheDeath, Gnoll_Mage, Domzilla, broadstrong, JoonasTo, homm3megejas, NiKitA, PhoenixReborn, Vlaad, OmegaDestroyer, mamgaeater, Keldorn, alcibiades, FriendOfGunnar, Ecoris
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I know nothing about their birthdays (of course I know mine! ) but I claim that at least two people among the participants have birthday on the same calendar day (NOT including the year...). Would you bet for or against that statement? Why?
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TheDeath
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posted November 09, 2008 06:02 PM |
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There are 23 right? (or did I count wrong)
Question: does the month matter, or just the "day" of the month?
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dimis
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posted November 09, 2008 06:06 PM |
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Edited by dimis at 19:06, 09 Nov 2008.
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Yes, the people are 23.
Yes, of course the month matters (I excluded only the year).
"Same date" means day X of month Y.
edit: Probably it is obvious, but X is just the number; e.g. 1. NOT for example Tuesday 1, or Tuesday.
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This thread is pages long: 
