• If 4 submarines attack a battleship what is the probability that at there is at least 2 hits


  • With 4 subs, you may score anywhere from 0 to 4 hits. Each sub has a 2/6 (or 1/3) chance of hitting, and therefore a 4/6 (or 2/3) chance of missing. You can figure the total probability by multiplying the chance of each sub hitting, along with the chance of each sub missing. So for instance:

    0 hits: 2/3 * 2/3 * 2/3 * 2/3 (all 4 subs miss) = 0.1975, about 20% chance
    4 hits: 1/3 * 1/3 * 1/3 * 1/3 (all 4 subs hit) = 0.0123, about 1% chance

    Things get a little more complicated with some hits and some misses involved. You might have:

    • First sub hits (hmmm)
    • Second sub hits (mhmm)
    • Third sub hits (mmhm)
    • Fourth sub hits (mmmh)

    The probability of each of those four things is the same:

    1 hit: 1/3 * 2/3 * 2/3 * 2/3 (1 sub hits, the rest miss) = 0.0987, about 10% chance

    But since there are four ways it can happen, there’s a 40% chance of getting exactly one hit. Similar for 2 hits–there are 6 different ways of getting exactly 2 hits (hhmm, hmhm, hmmh, mhhm, mhmh, mmhh) and the probability of each is:

    2 hits: 1/3 * 1/3 * 2/3 * 2/3 (2 hits, 2 misses) = 0.0494, about 5% chance

    So 6 * 5% is about 30% for exactly 2 hits. For exactly 3 hits, there are again 4 ways, and probability of:

    3 hits: 1/3 * 1/3 * 1/3 * 2/3 = 0.0247, about 2% chance

    About 4 * 0.0247 = 10% for 3 hits.

    The summary then:

    • 0 hit: 20%
    • 1 hit: 40%
    • 2 hit: 30%
    • 3 hit: 10%
    • 4 hit: 1%

    Which totals to just a little over 100% because of my rounding to whole percentages, but you get the idea. So to your original question, the chance of getting at least 2 hits is 30 + 10 + 1 = about 41%.

    Now, if you want to take into consideration rerolling–that is, your subs got 0 or 1 hit, and need to go into a second round of combat, you’d have to figure the probability of the battleship hitting one of your subs (so you know whether you’ll have 4 or 3 subs in the next round), and then figure the probability of each contingency again, recursing forever… I’ll leave that as an exercise to the reader.


  • And for those who are mathematically challenged, you can also use an online calculator: http://aacalc.nfshost.com/


  • I just wrote a series of articles in which, among other things, I mention how to handle the aforementioned problem of infinite recursion.

    I am Such a Genieus.   :wink:

    Maybe if Posters put up a lot of Demand for my Articles, there will be More Of Them.  :-D

    (I submitted the article series a couple weeks ago, I think - they may be released sometime in the next couple months)


  • Thanks

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