🚀 go-pugleaf

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            -