tag:blogger.com,1999:blog-4167684119478977136.post9214265751305722236..comments2024-03-22T08:23:38.715+00:00Comments on Trouble At T’Mill - a wargaming blog: Probability for Wargamers 11 - combinations and Chain of CommandMike Whitakerhttp://www.blogger.com/profile/02165272678144625943noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-4167684119478977136.post-31753992865080427472013-05-02T23:41:08.533+01:002013-05-02T23:41:08.533+01:00Blimey.
Thanks Mike, I was having trouble getting...Blimey.<br />Thanks Mike, I was having trouble getting to sleep so thought I'd catch up on your blog. Now my brain is melting. <br />Cheers anyway. Stephen Haranhttps://www.blogger.com/profile/07052611372084200629noreply@blogger.comtag:blogger.com,1999:blog-4167684119478977136.post-34997453829683485742013-04-30T03:45:23.568+01:002013-04-30T03:45:23.568+01:00This is where we get into the binomial theorem:
(A...This is where we get into the binomial theorem:<br />(A+B)^n = nCn.A^n + nCn-1.A^(n-1).B + nC(n-2).A^(n-2).B^2 + ... + nCr.A^(r).B^(n-r) + ... + nC1.A.B^(n-1) + nCo.B^n<br /><br />This is a very useful equation when determining multiples of either/or or hit/miss outcomes. (A+B)^n = 1, simply means that you will get a bunch of 'hits' and/or a bunch of 'misses'; no other result is possible. The expamsion might look like a waste of time, but they give you the number (n) of hits, say.<br /><br />(1/6 + 5/6)^5 expands to <br />(1/6)^5 + 5.(1/6)^4.(5.6) + 10.(1/6)^3.(5/6)^2 <br />+ 10.(1/6)^2.(5/6)^3 + 5.(1/6).(5/6)^4 + (5/6)^5.<br />Each of those terms (0-5) equals the probability of rolling 0-5 '1's (or any given number) on a D6.<br /><br />5 '1's = 1/7776<br />4 '1's = 25/7776<br />3 '1's = 250/7776<br />2 '1's = 1250/7776<br />1 '1' = 3125/7776<br />0 '1's = 3125/7776<br /><br />To check my maffs you simply add up the numerators, which should come out at 7776. They do. So you are more likely to roll at least one '1' than not; you are as likely to roll exactly 1 '1' as none at all; the probability of rolling more than 1 '1' is very nearly 1 in 5 (4 to 1 against).<br /><br />It is sometimes useful to determine the 'expected score' which involves term by term multiplying each probability by the actual number of 'hits' it represents. In the above example, your 'expected score' (ES) is about 0.73. It has to be admitted that it's not a particularly useful finding for this particular example, but it can be in other circumstances. <br /><br />For example, suppose you had several units - lets say it's 4 - firing, each receiving 5 dice to roll, '1's to 'hit'. Then as the ES is about 73%, we can reckon we can expect overall to score 73% x 4 'hits' - call it 3. We might score more, slightly more likely less (the miniscule chance of scoring 20 hits tends to skew the results). But we can feel unlucky if we score less than 2 hits, and very fortunate indeed if we roll more than 4.<br /><br />Cheers,<br />IonArchduke Piccolohttps://www.blogger.com/profile/15533325665451889661noreply@blogger.com