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  1. #1
    Brother Sergeant Bubba's Avatar
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    Question Arkham Horror probability problem

    Hello everyone!

    So yesterday my friends and I were playing a good game of arkham horror. As we were playing, I pointed out that one of my favorite weapons to use (in the base game alone) was the shotgun. Always seem to blow apart my foes in ease with a blessing on hand, passing the Will checks, with +1 Fight skill, and 6's keep coming out for the double successes.

    One of the other guys who's never played before last night felt the urge to say/argue that the chances of killing the same said monster with a blessing, passing the will check, with the same combat rating, the +1 Fight skill, but with a pair of enchanted blades is the better option in an equal setting. So, since I was not able to sleep tonight and being a former 40k player, I decided to "mathhammer" it out between the two controlled scenarios against a really tough monster to determine the better combo. Problem is, I crunched the math (which is all over the place and I am starting to get tired) and I think I made errors. If someone could give me a hand that would be great, so I can check my work and share with him my results (cause there's no worse thing than arkham players with bad blood between them playing again) so this can put my mind at ease. Here's what I have:

    Target Monster: Has a combat rating of -3 and 3 toughness.

    Character: Fight rating base is 3, +1 fight from skill (no clue tokens for rerolls), and a blessing (4+ to succeed).

    Note: We assume Will check was passed.

    Here's my math work:

    Scenario #1: Dual Enchanted Blades (+4 combat each).
    -> Total combat dice= (3+1+4+4)-3= 9d6
    -> Combat chances minus toughness:
    =>(3/6)^9 -(3/6)^3= .00195-.125= -0.1269

    Scenario #2: Single Shotgun (+4 combat and each 6 rolled is a double success)
    -> Total combat dice= (3+1+4)-3= 5d6
    -> Combat chances minus toughness:
    =>[(3/6)^5+ (1/6)^5] - (3/6)^3= (.07776+.00013)- 0.125= 0.07788- 0.125= -0.04712

    So I don't think my math makes much sense at this point (negative values?) and I bet I'm missing steps or forgetting things I need to do (like accounting for the 6's for the shotgun correctly). It's been a while since I've done things like this in my gaming days, plus I haven't taken probability in so long, but it would be nice to know. Sorry if I make anyone else's brain hurt too, cause now I need sleep.

    What am I missing or not understanding? :confused:

  2. #2
    Brother Sergeant Bubba's Avatar
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    Re: Arkham Horror probability problem

    Any help is greatly appreciated!! Any at all!!

  3. #3
    Chapter Master Bloodknight's Avatar
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    Re: Arkham Horror probability problem

    It would probably help if you could describe how combat in Arkham Horror works. I've been trying to make sense of your formulas, but I don't really understand what the numbers mean.
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  4. #4
    Brother Sergeant Bubba's Avatar
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    Re: Arkham Horror probability problem

    Quote Originally Posted by Bloodknight View Post
    It would probably help if you could describe how combat in Arkham Horror works. I've been trying to make sense of your formulas, but I don't really understand what the numbers mean.
    That makes a difference in getting help lol. Basically, with the number of combat dice you would have available to throw against the monster, you have to be equal or greater to the number of toughness to kill the monster outright. Partially wounding has no effect. I don't know how the extra dice thrown affect scenario #1 and how the double success condition on a roll of a natural 6 on scenario #2 affects the probability.

    So to give you an example of how many dice is used in scenario #1: 9d6 is thrown with needing a 4+ to succeed. After your roll, you need a total 3 successes for the kill.

  5. #5
    Chapter Master de Selby's Avatar
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    Re: Arkham Horror probability problem

    So you want to know the probability of getting three or more rolls of 4+ on nine dice?

    This is easier stated as 1 minus the probability of only getting 0,1 or 2 such successes. The binomial theorem gives us
    1 way to get 0 successes
    9 ways to get only 1 success
    9*8/2=36 ways to get only 2 successes
    p=1-((1*(3/6)^9)+(9*(3/6)*(3/6)^8)+(36*(3/6)^2*(3/6)^7))
    =1-(46*(1/2)^9)
    that's a 91 percent chance of success if I've understood the question.

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