How Statistics Work and Why They Don't Seem To - Warhammer 40K Fantasy
 

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  1. #1
    Senior Member Lemt's Avatar
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    How Statistics Work and Why They Don't Seem To

    OK, this is going to be short, simple, and as mathless as I can manage to do it. However I think a lot of players can benefit from learning how rolling dice REALLY works, hence I present this tactica.

    THE EXAMPLE

    You are playing a game, you play Space Marines versus Imperial Guard. You hace a unit of 5 Marines, no special weapons, and assault a Chimera that has moved the previous turn. 10 attacks, and you manage to get 6 hits.
    Now, a Chimera has rear armour 10, and your SMs have Strength 4, so you need a 6 to glance. You roll the six dice, and get a single 6.
    Now you think that it's a perfect roll, just the avergare result.
    Except it's not.
    That roll was slightly below average.

    THE REASON

    When you roll a single dice, you have 1 out of 6 chances of getting any result, in this case a 6. When you roll two dice, you have 1/6th of a chance of getting a six on either dice, but you also have a chance of getting a 6 on both dice, so you have to add that up.
    What this means is, the more dice you roll, the higher the chance you have of rolling at least one 6.

    The contrary is also true. If you roll several dice and need to get EXACTLY one result of 6, the odds of rolling that single 6 are lower than 1/6th.

    THE IMPLICATIONS

    I've seen many people think they were getting results that were really good or bad, because they didn't really know how statistics worked. Because of this, people often judge units and formations they are trying out incorrectly. This makes them think that certain options are better than they are, while they see some other options are worse than the truth.
    Now, you have the power of being able to avoid that mistake!

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  3. #2
    resident iconoclast Left of West's Avatar
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    159 (x3)

    Um, this is wrong.

    The "average" or expected return from rolling six dice and looking for sixes is exactly one. It is not slightly more than one. To find expected returns, one must simply multiply the number of trials (presuming they are identical) by the probability of success in each trial. That's commonly expressed as "p*n". In this case, P is 1/6 and N is 6, and 1/6*6 is exactly one.

    You haven't presented any math here at all. Though, as you say, the possibility of getting multiple successes needs to be accounted for, you haven't told us how to do so (and, in fact, you don't really have to since the formula above does account for it).

    You've basically asserted three things:

    That the "average" of six dice rolling for sixes is greater than one. This is incorrect.

    That the probability of getting at least one six increases as you roll more dice. This is obvious. You might have bothered to take the time to publish the formulas necessary to find out how likely it is that you will roll at least one six on a given number of dice, but without that your observation is puerile. As a side note, that formula has been presented before, on multiple occasions, including this one which was posted by me.

    That the probability of getting exactly one six on 6 dice is less than 1/6. This is also wrong. In fact, the probability of getting exactly one six on 6 dice is almost 40.2%, (about 2/5) something you'd have known if you actually knew how to do the math involved.


    This post fails to include any useful instruction and, in fact, includes two assertions which are false. It is, in short, a bad post. It misinforms and works contrary to the goal of helping others understand the math involved in the game.


    Also, for the record, you have about a 66.5% chance of getting at least one six (that is, one or more sixes) on six dice.
    Last edited by Left of West; September 16th, 2009 at 13:48.
    Once again, the conservative, sandwich-heavy portfolio pays off for the hungry investor!

  4. #3
    Son of LO
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    523 (x8)

  5. #4
    Senior Member Sancraer's Avatar
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    I agree, some actual maths is needed.

    Note: Space marines all have krak grenades which are S6 against vechiles so unless the player hadn't read the rules the space marines attcks would be S6.

  6. #5
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    The first bit is fine.

    'That roll was slightly below average.'
    Those rolls are actually slightly above the long-term average (which is 1/2 x 1/6 x 10=0.833)

    'When you roll a single dice, you have 1 out of 6 chances of getting any result, in this case a 6. When you roll two dice, you have 1/6th of a chance of getting a six on either dice, but you also have a chance of getting a 6 on both dice, so you have to add that up.
    What this means is, the more dice you roll, the higher the chance you have of rolling at least one 6.
    '
    This part looks alright as well.

    'The contrary is also true. If you roll several dice and need to get EXACTLY one result of 6, the odds of rolling that single 6 are lower than 1/6th.'
    I never saw this in any of my maths or stats subjects. Probably should have used Microsoft Excel (BINOMDIST()) or a graphics calculator for this.

    Also note that models attacking with grenades only get 1 Attack regardless of armament or charging bonuses (unless my brain's run away and abandoned me).

  7. #6
    Scourge Lord Krovin-Rezh's Avatar
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    204 (x6)

    Since this tactica is meant to teach (and not just point to other resources), I'm going to try and spell out how to determine your odds of rolling a 6 in as simple as I can manage.

    How often do you not roll a 6 on 1D6? 5 out of 6 times, or 5/6.

    How many dice are we rolling? 6 in this case.

    So our math needs to be 5/6 times itself for each die rolled:

    = 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6
    = 5*5*5*5*5*5/6*6*6*6*6*6
    = 15625/46656
    = 33.49%

    Subtract that from 100% and you have you percent chance of success instead of failure (66.5%).

    Now this math is not something most people are capable of doing quickly in their head, which is why we jump to simpler conclusions that are false. But it's possible with a calculator handy! Just plug it in like this:

    S = 1 - F^N

    S = odds of success we want to determine
    F = odds of failing on a single die
    N = number of dice to be rolled

    Remember that the ^ symbol means "to the power of", so you will never multiply by the number of dice, but instead by the odds of each die roll.

    This also works perfectly as is for needing any single result (like a 1), since your odds are the same for each die (5/6). One important thing to remember is that adding more dice when you need one result is an exercise in diminishing returns, so you'll never reach a point where rolling a 6 is guaranteed.
    Last edited by Krovin-Rezh; February 6th, 2010 at 10:28. Reason: Correction from LoW and Red

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  8. #7
    That Which Has No Time Red Archer's Avatar
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    531 (x8)

    Quote Originally Posted by Krovin-Rezh View Post
    "abs" means absolute value, which just means we disregard that the final answer is a negative number.
    Why is that?!

  9. #8
    resident iconoclast Left of West's Avatar
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    159 (x3)

    S = abs[1 - F^N]

    S = odds of success we want to determine
    F = odds of failing on a single die
    N = number of dice to be rolled
    This is close, but not quite right. F will never be greater than 1, so F^N will never be greater than one. Thus, 1-F^N will always be greater than zero, and the "abs" will not be necessary.

    All this formula will tell you is the probability that you fail all of the rolls. To find the probability that you succeed on at least one roll, you need 1-(1-F^N).

    What you were probably thinking, Krovin, was abs[F^N-1], which accomplishes the same thing as 1-(1-F^N).
    Once again, the conservative, sandwich-heavy portfolio pays off for the hungry investor!

  10. #9
    That Which Has No Time Red Archer's Avatar
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    531 (x8)

    I agree that the "abs" is not required. That's what I was asking Krovin: how the result could ever be negative.

    But I have to disagree with your version too, Left of West. If rolling at least a single six with six dice is a success, the formula 1-(5/6)^6 does indeed give us the chance of success.

    Your formula of 1-(1-(5/6)^6) is actually nothing else than 1-1+(5/6)^6, which in turn is nothing more than (5/6)^6 - the chance to roll no single six on any of the six dice and thus the chance of failure.

  11. #10
    resident iconoclast Left of West's Avatar
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    159 (x3)

    You're right. I was thinking 1-(1-P)^N where P is the probability of success. So, it's 1-F^N where F is the chance of failure (or 1-P).

    So, ignore my last post. It's just 1-F^N. No abs required. Sorry, Krovin.
    Once again, the conservative, sandwich-heavy portfolio pays off for the hungry investor!

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