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# Re: [OT] Math

From: "Nathan" <Nathan_at_Spring_Grove_UK@e...>
Date: Sun, 21 Jan 2001 17:06:52 -0000
Subject: Re: [OT] Math
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> Laserlight wrote:

>
>> 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.
>> <snip> I want to know what the probability is for getting
>> four hits?  I could work out all
> the 46000+ possible results and track them....

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,
> or AT LEAST 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:
>
> P = (1/6)^4*(5/6)^2 (4 missiles hit, 2 miss).

Which works out at 0.0536%, or less than the 0.0772%
chance of every missile hitting from a salvo of four. If I
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
>
> 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.

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...

Nathan
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