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ohforfsake
Promising
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Initiate
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posted October 12, 2009 01:33 PM |
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Quote:
I think we should take into consideration the difference between attack and defense. Assuming this is HOMM3 Vampire Lord 10att 10def, Wraith 7att 7def. Vampire Lord damage is 5-8. The question to all: Is it correct that vampire lord will do 6-9 damage with same probabilities? The same for the wraiths. I don't want to search algorithms of calculation of damage now. 
Can somebody say exactly which damage could do those creatures and with which exact probabilities?
Until we have those, can't we just assume my values are correct, and that all outcomes (of damage) is random and thereby equal likely? 
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ihor
Supreme Hero
Accidental Hero
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posted October 13, 2009 08:06 AM |
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Edited by ihor at 08:06, 13 Oct 2009.
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So assuming Vampire Lord can do 5-9 damage with same probabilities we have - he can do 9 damage with 1/5 prob. or he can do 9 damage two times in a row with 1/5 * 1/5 = 1/25.
That mean that probability of killing wraiths for one particular round is 1/25. -> 24/25 probability that wraith will survive.
Now wraith do 1-3 damage. We could take mathematical expectation = 2 damage per round.(??? I'm not sure about this step, have to think later about it when have time)
That means that wraith will kill vampire in 20 rounds(vampire lord = 40hp).
So the probability of winning of wraith is (24/25)^20 -> 44.2%
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dimis
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posted January 06, 2010 04:09 AM |
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Edited by dimis at 04:12, 06 Jan 2010.
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(0,1) =_c [0,1]
Back with a difficult problem this time.
Show that (0,1) has the same cardinality as [0,1].
Initially it might seem easier to show that (0,1) has the same cardinality as (0,1].
Good luck guys!
EDIT: Ecoris had answered similar problems on page 7 - in other words that's the technique/idea for the proof. On the other hand ... 
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The empty set
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ihor
Supreme Hero
Accidental Hero
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posted January 07, 2010 12:23 PM |
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I know the solution and will not post it to leave for others. Even more easy problem which could help is building bijection between the set of natural numbers and all non-negative numbers. Later on it is easy to prove that if A - infinite set, B - finite set, then union of A and B have the same cardinality that A.
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dimis
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posted January 07, 2010 06:37 PM |
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I didn't get your hint.
You can not mean a bijection between N = {0, 1, 2, 3, 4, 5, ...} and R_+ = [0, +oo), right ? And I don't see any finite set there.
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The empty set
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gnollking
Supreme Hero
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posted January 07, 2010 07:03 PM |
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TheDeath
Responsible
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with serious business
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posted January 07, 2010 07:07 PM |
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Is it even allowed to count up to infinity?
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The above post is subject to SIRIOUSness.
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zamfir
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posted January 07, 2010 07:13 PM |
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Quote:
Is it even allowed to count up to infinity?
Yes, but only if you are Chuck Norris.
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dimis
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posted January 12, 2010 03:02 AM |
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hint
gnollking:  nice!
Ok, a hint for those who do not know the problem.
We want to prove that (0,1) has the same cardinality as (0,1]. (We write (0,1) =_c (0,1].)
The way it is proven is by showing a bijection between the two sets.
Moreover, the domain is one of the sets and the image is the other one; whichever makes you feel more comfortable. I think the most intuitive bijection comes when we define a function from (0,1] to (0,1).
Now, the "obvious" candidate is
f(x) = x with x \in (0,1] .
But of course the problem is that f(1) = 1 and we don't want to include 1 in the image; i.e. the image has to be (0,1). So, the question now becomes:
What value should f(1) take ?
Or in other words, how can we "compress" that line ? Because clearly, f(1) should belong in (0,1) and the way f is defined so far all the "candidate" values are already taken. Is there a way out ?
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The empty set
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ihor
Supreme Hero
Accidental Hero
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posted January 12, 2010 02:16 PM |
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@dimis
I meant that building bijection between (0,1) and [0,1] is similar to building bijection between N = {1,2,3,...} and Z+ = {0,1,2,3,...} because [0,1] = union of (0,1) and {0,1} ,where {0,1} is finite set here (contain 2 elements).
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dimis
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posted January 12, 2010 03:35 PM |
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I see ihor. Sorry for the misunderstanding.
However 0 belongs to N. Check!
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AlexSpl
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posted January 12, 2010 03:41 PM |
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0 is natural number? how do you prove it?
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alcibiades
Honorable
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posted January 12, 2010 04:14 PM |
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Last I learned, 0 did not belong to natural numbers.
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dimis
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posted January 12, 2010 05:05 PM |
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Edited by dimis at 17:08, 12 Jan 2010.
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Alex:
Alci: Incorrect whoever said so.
Well, 0 is a natural number by definition. In set theoretic notions it represents the empty set (which has no members). And I am referring to Set Theory, because this is where the foundations of Mathematics are. If you open a book on Set Theory, you will see that they rely on the definitions for Peano axioms. For example check the 5th axiom here.
Perhaps another intuitive approach is to consider what we really want to do with numbers ==> add and multiply.
For instance, we want to have a magic number M such that
a + M = a
where a is a natural number. And that M is what we write down 0. I hope this helps.
Historical Remark: If I remember correctly 0 was "invented" around 700 AD. I will have to check again, but it will take some time. I think I first read about that in The Parrots Theorem (which by the way is highly recommended).
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ihor
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posted January 12, 2010 08:18 PM |
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Edited by ihor at 20:28, 12 Jan 2010.
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Certainly I was taught that natural numbers set is different in different countries. You can check it. It's definition and can't be proved, someone said so we have it.
I think it is logical to denote N={1,2,3,...} cause for the set {0,1,2,3,...} we have Z+, similar to R+ = all real numbers >= 0.
Edit:
Also, the main historical purpose of natural numbers is counting. Counting from zero is not a very natural thing to do.
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alcibiades
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posted January 12, 2010 08:47 PM |
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Wikipedia knows everything:
"In mathematics, there are two conventions for the set of natural numbers: it is either the set of positive integers {1, 2, 3, ...} according to the traditional definition or the set of non-negative integers {0, 1, 2, ...} according to a definition first appearing in the nineteenth century."
So, I stand with my previous statement:
N = {1, 2, 3, ...}
Z+ = {0, 1, 2, 3 ...}
Edit > Just to emphasize total confusion, I ment to write N_0 and not Z+, in case anybody goes back reading through this again.
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dimis
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posted January 12, 2010 11:58 PM |
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Edited by dimis at 00:09, 13 Jan 2010.
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Guys, no.
Ihor, you can't use your argument of Z_+ because as you already know we also have the other symbolism which is Z_+^* = {1,2,3,4, ...}; i.e. the * excludes zero. So, now what ? Either way N will be the same as either Z_+ or Z_+^*.
Alci, you can believe to that, but if you are consistent with your beliefs, you should never include as a basis of any proof that involves induction the case n=0. Do you really exclude zero always ? And by the way, wikipedia has mistakes here and there in maths. It is not the best source; it is just a handy source.
Finally, are you sure guys that you remember the definition from your high school books correctly ? I challenge you to find them and see what is actually written.
As of starting counting from 1, this is again not true ihor. You never do that. You always have some "scratch" area in your head which is initialized to zero, and as objects appear in front of you, you simply increment. Think about it. How much is the empty sum ?
For example, compute the sum
S = sum_{i=2}^1 i [i.e. i starts from 2 and ends up to 1]
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Ecoris
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posted January 13, 2010 01:00 AM |
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This is a silly discussion. Where I come from, 0 is not a natural number. As you know I study mathematics at the university and the definition here is that N = {1,2,3,...}. You can take my word for that, or I can look it up.
Also, Z_+ and R_+ (and similar sets) do not include 0. In Denmark the notation N_0 is quite common for the set of non-negative integers, {0,1,2,3,...}.
I dislike the notation Z_+^* since ^* (superscript *) usually means the multiplicative subgroup.
And I don't buy the idea of having a "scratch" area in my mind which is initialized to zero when I am to count something. The number 0 is a more abstract concept than a positive integer.
Do you have a "scratch" area in your mind which you initialize to 1 when you are to multiply something?
But all this is just about definitions. One can use whichever one pleases, there is no right and wrong.
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dimis
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posted January 13, 2010 01:48 AM |
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Edited by dimis at 03:59, 13 Jan 2010.
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I know that it is a matter of definition Ecoris, but it is a rather "counter-intuitive" definition and therefore not ok. Moreover, although kind of silly discussion, I think it is fun. Anyway, regarding the product, it really depends on the context. If I have to multiply 2 or 3 numbers I will not initialize something to 1. However, if I have a product like the sum above, I do initialize to 1; e.g. replace in my example above the \sum with \product. That's the intuition of initialization. You have to do it if you want to be consistent.
And now talking about multiplication, here is a small "paradox" if N does not include 0. Ok, so, what do we really need for natural numbers ? A minimum number (say 0 or 1 - we agree on that one), and a successor function (which gives us the next number after the one where we apply the function). Basically this is it. Now, multiplication is a convenient extension of addition.
Hence, the natural numbers have an identity element under multiplication, since a*1 = a.
However, since multiplication is nothing more than addition in that set, you should also have an identity element for addition; i.e. a + something = a. But the way you prefer N to be defined ... Where can I find such a magical element in N ?
Anyway, it is a matter of definition, but you don't give me any argument for not including it. In other words, I can agree with anything, but it is neither convenient, nor convincing, nor intuitive when we do not include 0 in the natural numbers.
Another example on defining emptiness.
Let's see conjunctions for a little while. The more variables we introduce in a conjunction, the less entries are actually true in a truth table. Hence it makes sense to say that the empty conjunction is "true". Another definition. I would also accept "false" but it is counter-intuitive.
On the other hand, if a formula is formed by disjunctions of variables, then, the more variables one introduces, the more frequently a formula is satisfied. So, under this setup, it makes sense to say that the empty disjunction is "false". Another definition; and of course anyone can define it to be "true".
Some references (jump to natural numbers if I didn't do it on the link; there are tables of contents):
1 It has an existence and uniqueness theorem (follows from the axioms) for natural numbers.
2, 3, 4
Finally, when we are in high school, we do not really care about the axioms. We have the intuition and we want to do maths or simply have fun. Who cares about axioms ? By the way guys, have you never encountered the other symbolism of N^* ? What is that supposed to mean ?
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ihor
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posted January 13, 2010 09:58 AM |
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Firstly, yes I'm sure that we defined it without zero. You can google "natural numbers" and see how many of links on the first page indicates that 0 is not natural number.
Yes, its definition, and you produced no more and no less arguments for 0 to be natural number than we.
We assume:
N >= 1
Z_+ >= 0
R_+ >= 0
R_++ > 0
God made natural numbers, all else is the work of man. - Leopold Kronecker
Historically natural numbers are not supposed to be some good set with good properties, its the set which used for counting. By evolution humanity wanted to give to set nice properties, so then we got group Z_+, ring Z, fields R and C. There is no purpose to redefine set of natural numbers just to have nice properties. Definition is definition.
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