🚀 go-pugleaf

An average is not an absolute: it must be taken over a domain. Do you 
define the solution as being the number of times played, when 1/2
of all people playing have dies?xxx sorry, died?

Assuming that you mean by the "average" person, the point at which half
of the people have diesxxx (sorry, I did it again!) diesxsasdc
died, the following progression applies:
Turn number    Cumulative proportion of people dead
1                .1667
2                .3056
3                 .4213
4                .5177  <-- Oops! Half of 'em are dead! Yuck!
  There may be a way to get an exact figure, like 3.9something something,
but it escapes me at the moment...
Any more questions?
By the way a Monte Carlo simulation, or a Las Vegas simulation might
help here...

Yes, a quick and dirty Monte Carlo (actually it was a Las Vegas simulation-
they're not the same, you know!) simlation has borne out my original
numbers. You can expect to die some time arounfd the 3rd or 4th try.
The curve vaguely resembles a log curve. I'm sure there's a way to get
an exact figure through clxxx calculus, I believe using a definite integral
and dividing to get an average is right, byxx but I always hated 
integration- differentiation is so much easier!

In article <1993Jan26.183234.20186@mksol.dseg.ti.com>, blanton@mksol.dseg.ti.com (John F Blanton) writes:
>THIS IS A HOMEWORK PROBLEM
>
>The game of Russian roulette consists of the following steps:
>
>a.  Load a 6-shot revolver with 1 bullet and 5 empty chambers.
>b.  Spin the cylinder.
>c.  Point the gun at your head and pull the trigger.
>
>On the average, how many times can one person play this game
>(steps a, b, c)?
>
>This assignment is due at 5 PM today.  Our class started out with
>a mix of theorists and experimentalists.  Now all of the     
>experimentalists are dead, and the theorists are confused.
>
>We need help.
>
>(solution tomorrow)
>
>John Blanton
>blanton@lobby.ti.com

    Look up the "negative binomial distribution".  The answer is 6.


    -        Rex Tayloe              TAYLOE@uinpla.npl.uiuc.edu       -
    -        U of Illinois           TAYLOE@uiucnpl.bitnet            -

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?

In article <1993Jan26.183234.20186@mksol.dseg.ti.com>, blanton@mksol.dseg.ti.com (John F Blanton) writes:
> THIS IS A HOMEWORK PROBLEM
> 
> The game of Russian roulette consists of the following steps:
> 
> a.  Load a 6-shot revolver with 1 bullet and 5 empty chambers.
> b.  Spin the cylinder.
> c.  Point the gun at your head and pull the trigger.
> 
> On the average, how many times can one person play this game
> (steps a, b, c)?
> 

An interesting (mathematically) variant is a,b,c,c,c, . . .

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            -

As promised, here is the solution:

The probability of surviving one round (interesting choice of
words there) is 5/6.  The probability of surviving N rounds
is           N
    p = (5/6)

The probability of surviving N-1 rounds and being killed on the
Nth round is
                N-1
   p = 1/6 (5/6)


The average lifespan of someone who plays Russian roulette is

                      N           i
   L = lim            Sum [i (5/6) ]/6
       N-->Infinity   i=1

                               N
     = lim            1 - (5/6)
       N-->Infinity (------------) / 6
                               2
                      (1 - 5/6)

     = 6.

My thanks to all who submitted solutions.  As you can see, I have
left out some steps, and the last limit expression presumes that
the limit of a derivative is the same as the derivative of the
limit.  I didn't verify that this is always true, but it turned
out to be true in this case, so I didn't type in the more complicated
limit expression that would have resulted by taking the derivative
first, then computing the limit of the result.

John Blanton
blanton@lobby.ti.com