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- From: tayloe@uinpla.npl.uiuc.edu (Rex Tayloe)
- Subject: Re: Desperate situation
- References: <1993Jan26.183234.20186@mksol.dseg.ti.com> <00967313.5DA0F1C0@uinpla.npl.uiuc.edu>,<2581@blue.cis.pitt.edu>
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- Sender: usenet@news.cso.uiuc.edu (Net Noise owner)
- Reply-To: tayloe@uinpla.npl.uiuc.edu (Rex Tayloe)
- Organization: Nuclear Physics Laboratory, University of Illinois, Urbana-Champaign
- Date: Wed, 27 Jan 1993 02:14:33 GMT
- Lines: 29
-
- In article <2581@blue.cis.pitt.edu>, wbdst+@pitt.edu (William B Dwinnell) writes:
- >
- >Rex: How are you defining the "average"? Do you mean the mean number of tries
- >successfully past by all people trying this? Do you mean the number of
- >tries at which half the people are dead? Possibly you mean something
- >else altogether?
-
- The problem was to determine the average number of times you pull the
- trigger before you get shot. = sum of number of triggers pulled in game
- i/number of games. This process would be described by the negative binomial
- distribution (or, more specifically, the geometric distribution). This
- distribution describes the probability for obtaining the kth success in the rth
- trial. In this case k=1 (the geometric distribution). The n.b.dist is given by:
-
- P(r;p) = p(1-p)**r-1 (for k=1)
- where p = probability of success in each trial (= 1/6 in this example)
-
- This distribution has a most probable value of 1 and an mean value of
- 1/p = 6. So by "average" I mean "mean". = (
- /rP(r;1/6)dr
- )
- 6 does seem high doesn't it? I think that it is because of the long tail on
- this distribution. It also just occurs to me that a most probable of 1 seems
- strange also. Oh well, I hope somebody yells if this is wrong.
-
-
- - Rex Tayloe TAYLOE@uinpla.npl.uiuc.edu -
- - U of Illinois TAYLOE@uiucnpl.bitnet -
-
-