🚀 go-pugleaf

In article <00967347.5659DB00@uinpla.npl.uiuc.edu> tayloe@uinpla.npl.uiuc.edu (Rex Tayloe) writes:
>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            -
>
Yeah, it's the Geometric Distribution. The word "average" can be explained
in the following sense. You ask 1000 volunteers for the game. Ask each
of them to repeat steps a-c until he/she kills himself/herself. 
Then you yourself (the sole survivor of the game) records the data.
e.g. Mary kills herself after repeating steps a-c  4 times
     John kills himself after repeating steps a-c  6 times
          :       :      :       :         :          :

       ... etc.

The average is the mean of these numbers. Suppose your sample only includes
Mary and John, the (sample) average will be 5. For this particular problem the
answer is 6 (this is the mean for the distrubution, not for the sample). 
It means a person who plays the game is expected to die after
repeating steps a-c 6 times. (If you find the mentioning of sample and 
distribution mean confusing, just forget about this paragraph, okay?)

Dicky Yan.