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Heroes Community > Tavern of the Rising Sun > Thread: Let's talk about Maths!!!
Thread: Let's talk about Maths!!! This thread is 55 pages long: 1 2 3 4 5 ... 10 20 30 40 ... 51 52 53 54 55 · «PREV / NEXT»
Nikita
Nikita


Famous Hero
Meepo is underrated
posted November 07, 2008 11:26 PM

Can i post my homework in here so u guys can do it?

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dimis
dimis


Responsible
Supreme Hero
Digitally signed by FoG
posted November 07, 2008 11:36 PM

I can't follow the problem on H5. The reason is that I've played only once or twice the game, I don't know how many spells there are, what the weights are, and the CONFUSION spell is ... simply casted on me! There are so many interesting problems, waaay easier formulated than that one related to H5 (or something very-very specific). To my opinion you should be able to state a problem without implying anything extra, or assume any prior "implied" knowledge by others.
Best
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The empty set

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JollyJoker
JollyJoker


Honorable
Undefeatable Hero
posted November 08, 2008 12:04 AM
Edited by JollyJoker at 00:08, 08 Nov 2008.

@Gnoll_Mage
What you have computed is the following:
If I play TWO random town games with ONE town each (two basic Homm games where I switch town to random and have one town only), what probability do I have to get Confusion in at least one of those two games.
Ok?

@dimis
Well, sorry, but this is a HoMM forum, so I thought a Homm-reöated math problem might hit some nerves. Math is best when used to solve problems asociated with what you know, and here I thought we'd know Homm.

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phoenixreborn
phoenixreborn


Promising
Legendary Hero
Unicorn
posted November 08, 2008 12:29 AM
Edited by phoenixreborn at 01:49, 08 Nov 2008.

I solved Corribus' problem the way Binabik did.

Corribus' explanation was overly complex but did make sense.  However, the question didn't specify the details of how the father and son were working together...so it's imprecisely worded...the example you gave about them each painting their own half is valid...how can you say a different answer is wrong?

Edit: Ok so I didn't read a couple posts of explanation.

Maths are a weak spot for me so I'm glad the thread has been made.  I think a heroes math question is very relevant to math and to this forum so I'm glad it came up.

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Vlaad
Vlaad


Admirable
Legendary Hero
ghost of the past
posted November 08, 2008 12:34 AM
Edited by Vlaad at 05:43, 08 Nov 2008.

I've always wanted to know what the Ranger's chances of getting Basic War Machines are if he already has 2, 3, 4 etc skills...?

Taken from the Manual:
Quote:
Note that the values below only apply to new skills. Chances of improving various skills, or being offered various abilities are uniform. Once a skill is learnt, the chances below are scaled among the remaining unknown skills.

Skill           %

Avenger        10
Attack          6
Defense        10
Leadership      8
Logistics      15
Luck           15
War Machines    2
Enlightenment  10
Sorcery         4
Dark Magic      2
Destructive     8
Light           8
Summoning       2

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dimis
dimis


Responsible
Supreme Hero
Digitally signed by FoG
posted November 08, 2008 12:46 AM

Yes, guys. I agree with you and I am glad that there are Heroes-related problems. However, I think that a problem of more than 3-4 lines involves many different subproblems, and usually you want to stress one of them; not many. To put it simply, divide and conquer your "main" problem! Then, just post a solution to that bigger problem glueing all the necessary pieces together. That way, with some guesses someone else can follow too.
I 'll interfere again when you are through with the problem.
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The empty set

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TitaniumAlloy
TitaniumAlloy


Honorable
Legendary Hero
Professional
posted November 08, 2008 01:35 AM

Hey, Corribus, are you a maths teacher?
At what level do you teach?

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OmegaDestroyer
OmegaDestroyer

Hero of Order
Fox or Chicken?
posted November 08, 2008 01:50 AM

Corribus
____________
The giant has awakened
You drink my blood and drown
Wrath and raving I will not stop
You'll never take me down

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TitaniumAlloy
TitaniumAlloy


Honorable
Legendary Hero
Professional
posted November 08, 2008 01:56 AM

Ah

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Corribus
Corribus

Hero of Order
The Abyss Staring Back at You
posted November 08, 2008 02:48 AM
Edited by Corribus at 02:58, 08 Nov 2008.

Quote:
Hey, Corribus, are you a maths teacher?
At what level do you teach?

No, but I've taught a lot of physical chemistry courses which require a lot of math.

Do we have a current problem?  

Here's another simple one that most people nevertheless get incorrect.

You have a bag that contains two coins. One coin has heads on both sides and the other has heads on one side and tails on the other. You select one coin from the bag at random and observe one face of the coin. If the face is heads, what is the probability that the other side is heads?
____________
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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Mamgaeater
Mamgaeater


Legendary Hero
Shroud, Flying, Trample, Haste
posted November 08, 2008 03:01 AM
Edited by Mamgaeater at 03:08, 08 Nov 2008.

you have 4 bags of marbles with 5 1 ounce marbles in each. But another bag contains 5 marbles that weigh 1.1 ounces each.
How can you figure out which bag has the marbles that weigh 1.1 using a single load scale once?



edit: this is actually more logical than math related sorry.
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Protection From Everything.
dota

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dimis
dimis


Responsible
Supreme Hero
Digitally signed by FoG
posted November 08, 2008 05:36 AM

Very nice one Corribus. It deserves a Math-QP: . My instinct gave the wrong answer initially.
Should I reveal what I did wrong?

@mamgaeater: there is another thread (Riddles) which is more suitable for your problem. Although, someone should have mentioned it by now (or a variation with 10 marbles instead of 5...)
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The empty set

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Celfious
Celfious


Responsible
Legendary Hero
Mickey cult member
posted November 08, 2008 05:40 AM
Edited by Celfious at 05:42, 08 Nov 2008.

Corribus I have to say 2/3 chances it is heads. 66.(6)%
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TitaniumAlloy
TitaniumAlloy


Honorable
Legendary Hero
Professional
posted November 08, 2008 05:54 AM

here's what I think

You have the two coins, HH and HT, call them x and y respectively.
You need the probability that the coin is h, given that the side is H.

That's the probability of it being both H and x, divided by the probability of it being H.

The probability that it's both H and x is the same as the probability of it being x, because if it's x it has to be H. Since there are two coins, that is 1/2.

The probability that it is H is 3/4 (four options, three of them are heads).

Hence it is (1/2)/(3/4) = 2/3 ?

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Corribus
Corribus

Hero of Order
The Abyss Staring Back at You
posted November 08, 2008 06:10 AM

Yes, the answer is 2/3.  You'd be surprised how many people refuse to believe the answer isn't 1/2.
____________
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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TitaniumAlloy
TitaniumAlloy


Honorable
Legendary Hero
Professional
posted November 08, 2008 06:13 AM

Now that you said that, 1/2 does sound like a very appealing option

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Celfious
Celfious


Responsible
Legendary Hero
Mickey cult member
posted November 08, 2008 06:23 AM
Edited by Celfious at 06:23, 08 Nov 2008.

goes about 5 steps up into the extra credit math problem

Ok theres 5 coins in a bag, 1 = t/t 2=h/h and the remaining 2 = h/t

You pull a coin and look and see a T

If you want, tell me the probability of the other side being a heads..
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TitaniumAlloy
TitaniumAlloy


Honorable
Legendary Hero
Professional
posted November 08, 2008 06:36 AM
Edited by TitaniumAlloy at 06:54, 08 Nov 2008.

edit:
nevermind, I thought there were 3 coins i misread the question
dimis got it

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dimis
dimis


Responsible
Supreme Hero
Digitally signed by FoG
posted November 08, 2008 06:48 AM

@ Celfious: This time it is half (1/2).

In the first problem by Corribus I made the following blunder when I first read it:
"The probability is a weighted sum. So, it is the sum of two probabilities, one for each coin. I get the first coin (H/H) with probability 1/2, and hence on the second step with prob. 1 I have H; hence, this part contributes 1/2 * 1 = 1/2. The other coin if it is H, it will be H if I turn it with prob. 0. So, 1/2 (for this second coin) * 0 = 0.
Hence, in total 1/2 + 0 = 1/2"
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The empty set

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Celfious
Celfious


Responsible
Legendary Hero
Mickey cult member
posted November 08, 2008 06:51 AM

1/2 is correct.. I think at least lol

Ok you have 85 coins... jk but it probably wouldn't be as hard as it seems it would be.
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