[OT] Math

A long time ago, in a university far, far away....I took statistics. And have
since forgotten most of it.

If I fire six missiles, each of which hits with a 1 in 6 chance, I
know the chance of getting six hits is (1/6)^6 and the chance of
getting six misses is (5/6)^6.  Let's say, though, that I want to know
what the probability is for getting four hits? I could work out all

> If I fire six missiles, each of which hits with a 1 in 6 chance, I

Four hits is (1/6)^4, IIRC

> Laserlight wrote:

That depends on whether you want EXACTLY 4 hits, or AT LEAST 4 hits. For
EXACTLY 4, your probability is:

P = (1/6)^4*(5/6)^2 (4 missiles hit, 2 miss). For AT LEAST 4, your
probability is

P = (1/6)^4*(5/6)^2 + (1/6)^5*(5/6) + (1/6)^6 (4, 5, or 6 hits). That
may simplify down to (1/6)^4, but I'm not sure.

> Laserlight wrote:

> If I fire six missiles, each of which hits with a 1 in 6 chance,

Fortunately there are only 64 cases, not 46,656. This is because you are only
interested in two outcomes per
missile:  p(hit)=1/6, p(miss)=5/6, two possibilities
repeated six times =2^6 outcomes in total.

From: Brian Quirt <baqrt@mta.ca>
Date: Sunday, January 21, 2001 02:33
Subject: Re: [OT] Math

> That depends on whether you want EXACTLY 4 hits,

In reply to both this and Schoon's posting, I would say LL wants to know the
chance of exactly four out of his salvo of six missiles hitting. The chance of
four or more of the six hitting would also be interesting...

> For EXACTLY 4, your probability is:

Which works out at 0.0536%, or less than the 0.0772% chance of every missile
hitting from a salvo of four. If I read your formula correctly, you've stated
the chance of the first four missiles hitting and the fifth and sixth missing;
one exact case out of our 64.

> For AT LEAST 4, your probability is

This correctly sums three specific outcomes, not the general "any four (or
more) hit" scenario. Common sense dictates that if every missile has the same
chance of hitting, six must have a better chance of achieving the objective
than four.

Unfortunately I have the same problem as Laserlight; my last education in
statistics was 16 years ago and I can only remember the basic operators. My
item "Chance of Normal Space Drive survival" posted
5/1/2001 was created <looks nervously around the
list for fear of derisive laughter> using an elementary probability tree. That
was OK for 8 outcomes but is more trouble than it's worth for 64.

I would be certainly be keen to know what the correct method is for working
this out. Fortunately, computers also give us the option to bludgeon
mathematical problems to death. I will get back to the list if I get any
worthwhile results...

Here we go.

Granted that my initial answer of p=(1/6)^4=7.716*10^(-4) only takes
into account four hits and totally disregards the other two dice.

The chance of the other two dice missing is p=(5/6)^2.

Thus EXACTLY 4 dice hitting is p=(1/6)^4*(5/6)^2=5.358*10^(-4)

...or .05358%

From: Laserlight <laserlight@quixnet.net>
Date: Sunday, January 21, 2001 04:49
Subject: [OT] Math

> If I fire six missiles, each of which hits with a 1 in 6 chance,

p(exactly four hits)=0.008037551, or 0.8% p(more than three hits)=0.008701989,
or 0.9%

Having built my bludgeon, I thought I'd better get my money's worth from it:

p(exactly three)=0.053583676, or 5.4% p(more than two)=0.062285665, or 6.2%

p(exactly two)=0.200938786 or 20.1% p(more than one)=0.263224451 or 26.3%

And for those of you fighting in a Teske field:

p(exactly one)=0.401877572, or 40.1% p(more than none)=0.665102023 or 66.5%

Now, does that earn me the title Malleus Munchkorum, or qualify me for burning
at the stake for the same offence?

> Nathan wrote:

Oops. I forgot about that. You have to add in the combinations possible. I
guess that would mean something slightly different. Note that I will be using
6C4 as my abbreviation of "6 choose 4" (combination notation)

        Thus 6C4*(1/6)^4*(5/6)^2  = 0.8% (which is the same as you got).
Note that my own statistical ability is (currently) based on one class in
high school ("Finite math" - covered probability, statistics, induction,
combinations/permutations, binomial theorem, etc. etc. but all briefly)
(although in getting my degree I'm very likely to be taking more at some
point - I suspect that either Physics or Computer Science will require
it).

> > For AT LEAST 4, your probability is

Again, here we should have:

P = 6C4*(1/6)^4*(5/6)^2 + 6C5*(1/6)^5*(5/6) + 6C6*(1/6)^6 = 0.87% = 0.9%
(again as you got).

In general, when you want to know whether or not exactly n events, with P=a
will occur out of a set of m events, m>n,

P = mCn*(a)^n*(1-a)^(m-n)

Hope that helps,

[quoted original message omitted]

> Nathan wrote:

I didn't see anything about how to calculate combinatorials (sp?). Maybe I
just missed it, so here goes:

mCn = (m)! / (n! (m-n)!)

So in the above case you've got 6C4 = (6!) / (4! (6-4)!)
= 6 * 5 * 4 * 3 * 2 * 1 / 4 * 3 * 2 * 2 * 1 * 1
= 15 There are 15 ways that 4 items can be arranged inside of 6 items.

Hope that helps,

OK, now hold yer horses:

I want to know the equation for finding the answer to the original question:
what is the probability of either exactly four, or four or more, hits given
each shot has a 1 in 6 chance of hitting.

I don't want the answer; I want the general equation from which the answer is
derived.

If an explanation would take some great length, then feel free to
contact me off-list.

Thanx,

> Sean Bayan Schoonmaker wrote:

For exactly 4:

P(4) = 6C4*(1/6)^4*(5/6)^2.

In general, for exactly N out of M hitting, with the probability of a hit
being P(hit) and the probability of a miss being P(miss):

P(N) = MCN*(P(hit))^N*(P(miss))^M

For 4 or more:

P(4 or more) = P(4) + P(5) + P(6)

or

P(4 or more) = 1 - P(3 or less)

That gets a bit complex for larger numbers, but it's at least the only general
equation that I can remember.

MCN here is M choose N (combination notation) or M!/((M-N)!*N!)

Remember "if I have six shots at 1/6 chance to hit, what's the chance
of getting 4 hits"?

> MCN here is M choose N (combination notation) or M!/((M-N)!*N!)

Turns out MS Excel has a function COMBIN--putting    =COMBIN(6,4)
into a cell yields the result 15.