Upgrade Disable Compromise

[MisterWinter]

I''m sorry, but this does not work out this way IMO.

Hey Gump, if you can't convince yourself with math, why don't you try simulation?

For instance this very nice line of Bash

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w="0"; for i in {1..100000}; do w=$((w + (RANDOM%6 + RANDOM%6 + RANDOM%6 + RANDOM%6 + RANDOM%6 + RANDOM%6 + RANDOM%6 + RANDOM%6 + 8 < RANDOM%6 + RANDOM%6 + RANDOM%6 + RANDOM%6 + RANDOM%6 + RANDOM%6 + RANDOM%6 + RANDOM%6 + 8))); done; echo ${w:0:2}.${w:2:1}%;

simulates 100'000 8v8s and prints the win/loss ratio of the attacking player. (It's about 47%, Bash confirmed...)

[techgump]

Ok, 47%... But this is not factoring a few things still.
1. it is two rolls which determine the outcome. Not one.
2. values rolled are potentally greater than 6, much greater.
3. I think neither 1 or 2 affect the % outcome, but not knowing Lopdo's alog may very well change it.

Either way, I can accept 47%.

That said, here, this explains my point precisely:
http://www.mathsrevision.net/gcse/pages.php?page=32

You are acting like the second roll's outcome is dependant on the first. Other than the fact that you do not need a second roll if you win the first one (as attacker), nothing else mathematically is dependant. It is not like you are taking 1 blue ball away, leaving two reds and one blue... which WOULD create a dependant variable. In this case, the first roll is made and lost, the second roll has nothing to do statically with what happened on the first roll. It is entirely Independent.

What Darkas is saying would be no different than Darkas saying each time I play the lottery my odds of WINNING increase. This is NOT true however. You odds are the same, each time, every time. They are completely Independent.
http://mathforum.org/library/drmath/view/56585.html

EDIT: OR like saying if I flip a coin twice my chances of getting heads is 75%, or is it 25%. LOL. This is where Drakas is changing his formula so that heads is 75% and tails is 25%. But in reality, it does not matter if you are heads or tails. As I state, using his methods, both heads and tails would have a 25% chance, leaving 50% in the dark void.

I am telling you, WINNING chances do not change when the probably remains the same each time. lol

[techgump]

Ok, after all my laughing and "I'm telling you"... I think I must concede to Darkas ... boy I hate being wrong.

I also asked myself:
If someone bet me 500 dollars that I could not flip heads once over 5 coin tosses, I'd take them up on it, betting that I could
If they asked me to do this over one coin toss, I would decline, as my odds are definitely 50/50.
Obviously, as Lopdo first pointed out, the more tosses you have to hit it just once, the better your odds. And as Darkas has stated, it does only take one win as an attacker to win the battle.

So, the odds of winning one over many does logically appear to increase, while the odds of winning each one does not. That said, the formula used still did not makes sense to me in a practical sense what-so ever, as I surely do not expereince this type of result using the boosts with equals... I still lose more often than not on both rolls. However, assuming a 47% win on one roll and being an attacker, in order to lose with a bonus, the math is in fact:

.53 x .53 = 28%, and hence the attacker is 72% likely to win on the FIRST roll, and 47% likely to win on the second, if the first roll was lost. The value does change based on the number of roles you have left. This makes me want to attack equals more with boosts, however, I have such awful luck attacking -1 properties (properties with 1 man less than mine), that I use my bonus attacks mostly on the first turn to assure I win 4s over 3s, and 3s over 2s to connect up and build strong from the start. So 1/2 my boosts are wasted to assure a strong start and win, and hence considering usage, does degrade the overall % gain in men killed by using them.

For those interested in a good reference on Darkar's math:
http://en.wikipedia.org/wiki/Gambler%27 ... _fair_coin

[MisterWinter]

Ok, 47%... But this is not factoring a few things still.
1. it is two rolls which determine the outcome. Not one.
2. values rolled are potentially greater than 6, much greater.
3. I think neither 1 or 2 affect the % outcome, but not knowing Lopdo's alog may very well change it.

1)2) On my code RANDOM%6 + 1 simulate a dice roll: it will return with equal probabilities a number in { 1, 2, 3, 4, 5, 6 }. By summing 8 of this you get a simulation of rolling 8 dices : RANDOM%6 + RANDOM%6 + RANDOM%6 + RANDOM%6 + RANDOM%6 + RANDOM%6 + RANDOM%6 + RANDOM%6 + 8. Then you can do this simulation once per player, and compare the results. If(f) the attacker's roll is strictly greater that the defender one, the attacker wins.

3) Of course we don't know the "algorithm" used in-game to resolve the fight. But I think that we could very reasonably guess that the two number displayed durring fights are actually linked to the internal "algorithm". So far the numbers I got were matching the hypothesis that the game used one dice 6 per supply for both the attacker and the defender. But ofc the game could be implemented with biased dices or something like that.

What Darkas is saying would be no different than Darkas saying each time I play the lottery my odds of WINNING increase. This is NOT true however. You odds are the same, each time, every time. They are completely Independent.

Man, really? Let me try to make things clear and formal. Let's denote by A the event of "the attacker wins a fight" and D the event of "the attacker loses a fight".

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P(A) = p
P(D) = 1 - P(A) = 1 - p

Now consider that we do the fight experiment (sounds great xD) twice, and that this experiments are independent. Here are the 4 possible outcomes with they probabilities:

Code: Select all

{ AA, AD, DA, DD }
P(AA) = P(A) * P(A) = p * p
P(AD) = P(A) * P(D) = p * (1 - p)
P(DA) = P(D) * P(A) = (1 - p) * p
P(DD) = P(D) * P(D) = (1 - p) * (1 - p)

(verification)

As you probably know from playing the game, if you attack using the attack boost you win if you are in AA AD DA, and you lose when you are in DD. So

Code: Select all

P(Attacker a wins with boost) = P(AA or AD or DA) = P(AA) + P(AD) + P(DA) = 1 - P(DD)
                              = 1 - (1 - p) * (1 - p)

You should stop stating things you obviously don't know using some random wiki links, no offense.

:-/

[Secondwind]

Is it me or do people on the internet almost always have no idea what probability means and assume it's the same as chance?

Ultimately no matter how you look at it, it's a significant advantage for those that have it versus those that don't, nothing else in the game provides such a massive advantage. It doesn't belong.

he is too lazy to do the math but he dares to call anyone that questions him idiot, go figure

I foolishly assumed the readers would be intelligent enough to do it themselves if they couldn't understand what I was saying without having it written explicitly. I'm quite certain the majority of people that actually understand math at a moderate level knew exactly what I was talking about without needing to do all of the math, they just know within a certain margin of error what the outcomes will be. Just like you know if you roll 1d2 (Flip a coin) it's a 50/50 chance of 1 or 2. I can simply take that further without needing to do the math for it. I imagine the math was done simply to try and explain it to others rather than for themselves, I simply don't care enough to explain it to people so thoroughly like the others in this thread. They're simply far nicer and more patient than me, but it doesn't really change that I was right. Just now everyone knows I was rather than just myself knowing it. And well as mentioned, I don't care what other people know, it's not my job to convince people who can't figure out flipping a coin twice in a row changes the probability of Heads or Tails and then insist the chance is the same when of bloody course it is, but we're talking about probability not chance.

Another simple example: What is 10x10, do you know it's 100 or do you need to do 10+10+10+10+10+10+10+10+10+10 in order to arrive at the fact that it's 100? Almost everyone just knows the answer without needing to do the math, right? I simply go much further with that and I don't care to do the math for others when I already know within a reasonable margin of error myself what the outcome will be.

This is all incredibly basic math and I did not think it would need to be spelled out, I don't need to write out a model or bruteforce because the difference between 25/75 and 28/72 is irrelevant in the grand scheme of things.


And you still didn't mention if the rolls are 1d(Nx6) or Nd6. Rerolls are even more overpowered if it's the former.

[techgump]

.53 x .53 = 28%, and hence the attacker is 72% likely to win on the FIRST roll, and 47% likely to win on the second, if the first roll was lost. The value does change based on the number of roles you have left. This makes me want to attack equals more with boosts, however, I have such awful luck attacking -1 properties (properties with 1 man less than mine), that I use my bonus attacks mostly on the first turn to assure I win 4s over 3s, and 3s over 2s to connect up and build strong from the start. So 1/2 my boosts are wasted to assure a strong start and win, and hence considering usage, does degrade the overall % gain in men killed by using them.

Could someone still explain to me why the math does not add up if we took this same approach from a defensive standpoint?
The odds for a single roll win for a defender is 53% (again assuming 47% attacker). So using the same formula for two roll odds we did for attacker, but instead inserting defender ratio:
.47 x .47 = 22%, and hence the DEFENDER is 78% likely to win on the FIRST roll, and 53% likely on the second.

I guess I keep coming back to this and asking... How can a defender be 78% likely to win on First roll, and attacker be 72% likely to win on First roll. Man, my head is spinning over this quandary, so I guess I still believe our formula is wrong.

[techgump]

Secondwind wrote:

Is it me or do people on the internet almost always have no idea what probability means and assume it's the same as chance?

Um, it's not just people on the Internet. It is people. Considering they are both Synonyms for each other by any dictionary, they are interchangeable English words. If they are not, do explain the difference in this statement:

I have a 1/6 chance of rolling a 6 on a six sided die.
I have a 1/6 probability of rolling a 6 on a six sided die.

If they are not the same, one of these statements is incorrect. Which one Secondwind?

[Secondwind]

Rofl, you really have no idea what they mean do you? I'm talking about usage in statistics, it's called context. They're decidedly not the same in math as they might be in common usage.

http://dictionary.reference.com/browse/probability

"the relative possibility that an event will occur, as expressed by the ratio of the number of actual occurrences to the total number of possible occurrences."


http://dictionary.reference.com/browse/Chance

"a possibility or probability of anything happening: a fifty-percent chance of success."


I went ahead and highlighted the important parts for you.

It's a very clear and distinct difference.

If you flip a coin and get heads you have a 50% chance of getting heads again. Because chance is not effected by previous results, you always have a 50% chance.

However your probability of flipping a coin, getting heads then flipping the coin and getting heads again is 25%.

If they are not, do explain the difference in this statement:

I have a 1/6 chance of rolling a 6 on a six sided die.
I have a 1/6 probability of rolling a 6 on a six sided die.

If they are not the same, one of these statements is incorrect. Which one Secondwind?

The difference is you not understanding english... The fact that the probability and chance are equal is because you're only rolling it ONCE. How the hell does that make them mean the same thing?


This has a chart of the probability for 2d6. The forums screw with the formatting.
edit: better chart
http://www.sarna.net/wiki/2d6_to_1d20_E ... robability


Notice how the distribution is different from 1d6? Seriously I don't know how to explain this any simpler.

[techgump]

WEll, I think you are amusing Secondwind... You yourself pointed out my point

http://dictionary.reference.com/browse/Chance
"a possibility or probability of anything happening: a fifty-percent chance of success."

Let me repeat that. CHANCE: "a possibility or probability of anything happening: a fifty-percent chance of success."

EDIT/Addition:
And you never answered my question:

Considering they are both Synonyms for each other by any dictionary, they are interchangeable English words. If they are not, do explain the difference in this statement:
I have a 1/6 chance of rolling a 6 on a six sided die.
I have a 1/6 probability of rolling a 6 on a six sided die.
If they are not the same, one of these statements is incorrect. Which one Secondwind?

You act as if it is stupid of Internet users to use chance and probability as one in the same... and that is because they are, as you yourself reference above. I find it so very funny when people feel the need to insult and express disdain for those that are not as superior to one's skills and understandings to make one feel better about himself; being righteous... especially when those with perhaps a less of an understanding, are not necessarily incorrect.

The difference is you not understanding english...

Yes, that is my problem.

Hey, I am no exception and am guilty too, probably by posting this, but I'll make the point... It's funny

I think your stupid, a complete idiot I call you, if you don't have over one million dollars in your bank. Stupid people, you don't understand the world. Wow, I feel better already.

[techgump]

And still, no one is answering my question, as to why the % formula is not equating correctly when looking at it from both an attacked and defender standpoint. (Darkas, some help?)