Probability question

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)