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  1. #1
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    76 (x1)

    probability help...

    ok, I'm working on some maths here...
    A model with BS 2 has a 33% chance of hitting at range. What are the chances of hitting if it fires twice (for rapid fire, or Assault 2) ?

    now, before you yell '66%', it's not correct, because a model with BS 3 (50% hit rate) firing twice is DEFINITELY NOT 100% hit rate.

    So, the BS 2 model will always have a 33% chance to hit with each shot, because each result has no bearing on the next... but surely there has to be a way to work out the success rate for a set amount of attempts (firing 5 shots, for example, should have a higher chance of hitting than firing once)- if there is a formula for this, then what is it?

    any help would be great.


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  3. #2
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    523 (x8)

    P(at least one hit) = 1 - P(no hits)

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    Senior Member Royal_Marine_Machine's Avatar
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    38 (x1)

    33% = 0.33

    0.33 x 0.33 = 0.1089

    Oh the joys of having just finished a statistics exam.

    so that's 10.89%

    You just times all the probability's together, hope you get it

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    523 (x8)

    Quote Originally Posted by Royal_Marine_Machine View Post
    33% = 0.33

    0.33 x 0.33 = 0.1089

    Oh the joys of having just finished a statistics exam.

    so that's 10.89%

    You just times all the probability's together, hope you get it
    That is the chance of both hitting, not the probability of at least one hit, which is what I think the OP was asking

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    Senior Member Royal_Marine_Machine's Avatar
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    Oooh Ok.

    Then iiits *works out fervently"

    ookkk
    possibilities:
    0.33 x 0.66 = 0.2178
    0.66 x 0.33 = 0.2178

    0.2178 x 2 = 0.4356

    Possibility = 43.56%

    That better?

    My sources tell me that you are a Maths God... Muwahahhaha
    Last edited by Royal_Marine_Machine; June 16th, 2009 at 18:05.

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    523 (x8)

    I think the OP was interested in what the probability of hitting at least once is when you fire two shots with a 33% chance of hitting on each shot.

    if the chance of hitting is .33 then the chance of missing is .67. So the chance of missing with both shots is .67 x .67 which is .4489

    So the chance of hitting with at least one shot is 1- .4489 which is .5511

    So with two shots at 33% the probability of getting a hit (or better) is roughly 55%

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    Senior Member carrotman50's Avatar
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    You need binomial distribution:

    (pq)^2 = p^2+2pq+2qp+q^2

    p = hit (0.33)
    q = no hit (0.67)

    to hit at least once you need p^2 (hits twice) and 2pq + 2pq (hits once each)

    according to my calculations is 0.9933, (0.33^2+(2x0.33x0.67)+(2x0.33x0.67)), which when you think about it, you only have to hit with ONE dice, while the probability for TWO hits would be 0.1089 (0.33^2).
    Wow, I never thought I would find a use for Stats, but I find one barely 6 hours after my exam!

    Although, im not sure if the actual binomial distribution equation is right...although it should be...
    Last edited by carrotman50; June 16th, 2009 at 18:38.
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    Senior Member Royal_Marine_Machine's Avatar
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    38 (x1)

    This P and Q stuff isn't what I was taught in my Stats, (exam was 10 hours or so ago, 8 O'clock)

    I was taught that the probability of hitting is 0.33, the probability of missing is 0.67 (my mistake earlier), so the possibilities are HIT then MISS or MISS then HIT.

    So that's 0.33 x 0.67 and 0.67 x 0.33, which are both 0.2211, times it by two (two possibilities) = 0.4422 or 44.22%.

    hmm... You guys are probably right, seeing as by my calculations there's a strange possibility, maybe if it's Orks - the remaining 13% is the chance of the gun blowing up.

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    523 (x8)

    Quote Originally Posted by carrotman50 View Post
    You need binomial distribution:

    (pq)^2 = p^2+2pq+2qp+q^2

    p = hit (0.33)
    q = no hit (0.67)

    to hit at least once you need p^2 (hits twice) and 2pq + 2pq (hits once each)

    according to my calculations is 0.9933, (0.33^2+(2x0.33x0.67)+(2x0.33x0.67)), which when you think about it, you only have to hit with ONE dice, while the probability for TWO hits would be 0.1089 (0.33^2).
    Wow, I never thought I would find a use for Stats, but I find one barely 6 hours after my exam!

    Although, im not sure if the actual binomial distribution equation is right...although it should be...
    If this is what you did in your exam then start worrying, think about what you have said for 2 seconds, Over a 99% chance of hitting?

    This is what you mean

    (p+q)^2 = p^2 + 2pq + q^2

    discarding q^2 and letting p = 0.33 and q=0.67 gives the same result as I gave earlier in the thread.

    Please, if you don't know what you're talking about or have only a rudimentary grasp of maths then don't post, it's not helpful.

    Question answered, thread closed

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