• Does anyone know how to calculate probability for multiple dice rolls?

    For instance, two different rolls for an attacking sub, is it still 33% chance of a hit (still 4 out of 12 chances), 66% (two chances of 33%), or 44% (the original third plus a slightly chance of both hitting, for example).

    My Google-Fu has led me to all kinds of not helpful pages (they want sum totals, like playing craps or something) and I haven’t done this kind of math in decades.


  • Well, I happened to find it with a LOT more digging… so I thought I’d share.

    Apparently its easier to do it based on missing.  So…

    Example is for, let’s say, submarines attacking…

    Take your odds of one dice missing (2/3rd chance you WON’T get a 1 or a 2)

    Then you take 2/3 to the Nth power, N being however many rolls you get, so…

    Attacking with two subs gets you 2/3 to the 2nd power, or 5/9th chance you’ll miss OR, more helpful…
    two attacking subs have a 44.4% chance of hitting once and an 11.1% chance of both hitting.

    If you got 3 rolls, its 2/3 times 2/3 times 2/3rd, or 8/27th chance of missing or…
    a 70.4% chance of hitting once.

    Link to the solution in case I need it again https://math.stackexchange.com/questions/2087463/multiple-dice-rolling-probabilities?rq=1

  • '19 '18

    I think you have it right for finding at least one hit.  That case is relatively simple because it’s the probability of not hitting zero times.

    It gets a bit trickier if you want to know other values but generally looks like:

    Prob(# of hits) * Prob(# of misses) * # of ways it can happen

    Mathematically it looks like:

    Where, w = probability of a hit
    x = number of units hitting
    y = probability of a miss
    z = number of units missing

    = w^x * y^z * NCR(x+z , x)

    In your example you’re finding the probability of zero hits which is nice because the first and last term equal 1 and can be ignored:

    (2/6)^0 * (4/6)^3 * NCR(3,0)
    = 1 * (64/216) * 1
    = 29.6%

    So the probability of not hitting zero times is 70.4% like you said.

    Let’s say you want to know the probability of getting 2 hits exactly:
    (2/6)^2 * (4/6)^1 * NCR(3, 2)
    = (4/36) * (4/6) * 3
    = 22.22%

    If you’re unfamiliar with the NCR function, these are called “combinations”.  The function exists on some calculators and you can do it in excel using =combin(x,y).  With 3 subs there are 3 possible ways of getting 2 hits, sub A+B, sub B+C and sub A+C, hence 3 in my example.

    If you plan on going very far down this rabbit hole you’re likely going to want an excel spreadsheet.

    Hope that helps.


  • It does, thank you!

    The sub scenario is what kicked off this pursuit for me.  Specifically, at the opening of Global 1940, What are the odds of two German subs getting a single hit on Britain’s cruiser.

    I mean, its easy enough to figure out how many hits total you might get, on average, from a stack of infantry, but there are plenty of smaller strikes of trying to take down one or two units I was trying to figure out my odds for.

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