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Thread: Practical Survival

  1. #41
    Join Date
    May 2009
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    46
    Phaze - I will only respond briefly, since I think this is going somewhat off-topic.

    No, probability of a given string of occurrances is completely independent of previous / following occurrances.
    ...
    Take each probability individually. For a two-hit chain in the above example: p =.1114. The probability of NOT having two hits in two attacks (1 hit then 1 miss, 1 miss then 1 hit, 2 misses) is .8886. Simple as that.
    The original example uses a tank with 55% avoidance. If the question is, what is the probability that two out of two swings connect?, the answer is trivial to obtain... it's also not .1114. It's .55 * .55 or .3025. Of course the outcomes of other swings have no bearing on this - they're independent. My understanding of the examples, and the discussion, was rather that it was an attempt to determine the relative frequency of two-hit chains, as opposed to three-hit chains, one-hit-one-miss-chains and so on. From this perspective it's clear that the swings need some context to determine whether you're looking at a chain of exactly two hits (which must by definition be bookended by misses, otherwise it would just be part of a longer chain of hits) versus a chain of at least two hits (which is just two hits occurring in a row regardless of context). Notice Hypatia makes the distinction between a chain of exactly two hits and a chain of at least two hits, and furthermore includes the additional a factor in determining the probability in the exact case, which I believe is an acknowledgement of what I said above (although I may be wrong, since I can't see the derivation).

    As a final point in response, I don't quite understand the random variable for which outcomes and their probabilities are being calculated, which is responsible for my confusion. The number of hits taken in a 2 minute fight is a perfectly valid random variable (with outcomes ranging from 0 to 60, the total number of swings, each of which has an associated probability). The number of times you will be hit exactly five times in a row is also valid, although harder to reason about, because the worst case would be if you were hit five times in a row, then avoided, then hit fives times, then avoided, etc. Without the avoided hit between 5 chains, it's not a string of exactly five hits, it's more hits than that (i.e. a 5+ chain). Hypatia's table shows various events along with probabilities associated with them, but it's not clear whether these are all possible outcomes of the same random variable (in which case they should sum to one) or possible outcomes of lots of different random variables, in which case what are the other possible values for these variables?

    Also notice I took care to state I don't disagree with any of the analysis where expected damage is concerned, and I have taken great pains to read all the relevant material. The point I was trying to make is that in considering only the expected benefit of avoidance for the purposes of comparison, rather than the whole distribution of possible benefits and their associated probabilities, you're effectively obscuring its random nature. This is why a lot of recent applied statistical analysis refrains from using single-statistic summaries of intermediate variables in favour of retaining the whole distribution to reason about the final thing of interest (in this case, something like what's the probability of death in this fight?). But this is a technical point and perhaps not appropriate for discussion in this forum.

    And Satorri, you're quite right that not taking damage for periods of time is a wonderful thing... provided your healers are able to heal you through the periods where you do take damage, because these are going to happen sooner or later. There are all kinds of ways to ensure you're able to deal with the damage which might be incoming and still conserve your mana if it turns out not to, cancel casting being the most obvious. What this means is that once your mitigation and healer throughput are sufficient to sustain you in the worst case scenario, adding avoidance reduces the burden on healer mana and generally makes everyone happier.

    I suppose the true question we want to ask is, what effect does adding x amount of avoidance versus y mitigation have on my probability to survive this encounter? Depressingly, I suspect the answer is, very little in either case, since you're extremely unlikely to die from normal attacks anyway. The overwhelmingly more frequent cause of death is some screw up in the mechanics of the encounter. But the discussion is stimulating at least

  2. #42
    Join Date
    Feb 2008
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    615
    Quote Originally Posted by bashef View Post
    Hypatia: the minor point first - something is nagging me about those probabilities, and it has to do with how the two, three, four chains and so on all interact with one another.
    I thought about that a while when I made my original post before concluding that it's not necessary to worry about that. I may have been mistaken, however.

    Let's consider the following events, where "o" is a miss, "x"is a hit, and "." is a "don't care".

    Code:
    1 ...oxo
    2 ..oxxo
    3 .oxxxo
    4 oxxxxo
    To calculate the probability of any one of those happening is (1.0) for every ".", (a) for every "o", and (1-a) for every "x". Now here's the thing--if you take the probability of each of those events *given* that you end with "xo", you can divide by (a)(1-a) to remove that term. Now you have the probability that a chain will be of the appropriate length. (Note: That a *chain* will have that length. This is the probability of the length given that you *already have a chain*.)

    The probability that a chain of hits will be length 1 is (a). The probability that a chain of hits will be length 2 is (a)(1-a). The probability of that it will be of length 3 is (a)(1-a)(1-a). And in general, the probability will be (a)(1-a)^(n-1).

    If we sum up over all chain lengths, we see that the total is 1.0.

    Now, there are two ways to take this. I believe that what my original probabilities describe is "the probability that a given hit is part of a chain of length n". And in fact, we likely would rather have "the probability that a given hit is the end of a chain of length n". Which means you are correct. Let's see what the numbers look like then:

    Code:
    Original Numbers (probability of a hit/miss being in a chain of length n)
     1.84% 5+hit chains   5.03% 5+miss chains
     2.26% 4-hit chains   4.12% 4-miss chains
     5.01% 3-hit chains   7.49% 3-miss chains
    11.14% 2-hit chains  13.61% 2-miss chains
    24.75% 1-hit chains  24.75% 1-miss chains
    
    New Numbers (probability of a chain being of length n)
     1.01% 5+hit chains   2.26% 4+miss chains
     1.24% 4-hit chains   1.85% 4-miss chains
     2.76% 3-hit chains   3.37% 3-miss chains
     6.13% 2-hit chains   6.13% 2-miss chains
    13.61% 1-hit chains  11.14% 1-miss chains
    
    New Numbers (expected number of chains of length n in a 600-swing fight)
      6.09 5+hit chains   13.59 5+miss chains
      7.44 4-hit chains   11.12 4-miss chains
     16.54 3-hit chains   20.21 3-miss chains
     36.75 2-hit chains   36.75 2-miss chains
     81.67 1-hit chains   66.83 1-miss chains
    
    148.5  hit chains    148.5  miss chains
    It's interesting that there'll always be the same number of expected hit and miss chains, but it makes sense: every hit chain must be followed by a miss chain, and vice-versa.

    So, this is the correct way to count the number of expected chains. Thanks for getting me to re-visit that.

    Quote Originally Posted by bashef View Post
    what you mean of course is that the expected frequency of the event is as you show.
    Correct. Apologies for using poor terminology. The intent of using expected frequency instead of probability is that it becomes much easier (in my opinion) to interpret as the raw probability becomes very small. (i.e. very few people can eyeball "1% chance" vs "0.07% chance" and really understand the difference in magnitude. It's much easier to compare "happens on average once every 1m40s" vs "happens on average once every 23m48s".)
    Learn to science and stop theorycrapping in its tracks.

  3. #43
    Join Date
    Oct 2008
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    4,930
    bumping for easy access
    The (Old) Book on Death Knight Tanking
    The New Testament on Death Knight Tanking
    -----------------------------------------
    Quote Originally Posted by Horacio View Post
    Who f-ing divided by zero?!?

  4. #44
    Join Date
    Jul 2009
    Posts
    135
    Great thread, it started out as a simple answer for "everytank" and turned into an hour long reread just to allow me to understand that...
    its still a chance if you are gonna be missed or hit and do everything you can to build you toon to be able to take a second or third hit.
    I think we can all agree that there is always a "one shot" death but as an experienced tank we know when to make our EH or avoidance become MORE!

    Thanks for writing this.

  5. #45
    Join Date
    Oct 2008
    Posts
    4,930
    Bump for access.
    The (Old) Book on Death Knight Tanking
    The New Testament on Death Knight Tanking
    -----------------------------------------
    Quote Originally Posted by Horacio View Post
    Who f-ing divided by zero?!?

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