Expected Value

Date: 11/07/2001 at 01:49:09
From: Hermann
Subject: Expected Value question

I am playing a game where I toss a fair coin. I get a bead if I flip 
a head and I lose a bead if I flip a tail. If I flip a tail and I 
have no beads, nothing happens. If I want to get a total of 5 beads, 
what is the expected number of coin tosses I must make? 

I have no idea where to begin. Is this a geometric distribution 
problem? 

I appreciate your help.

Date: 11/07/2001 at 10:52:34
From: Doctor Anthony
Subject: Re: Expected Value question

Let a = expected number of throws when you have no beads.
    b = expected number of throws when you already have 1 bead.
    c =    "        "        "                "         2 beads.
    d =    "        "        "                "         3 beads.
    e =    "        "        "                "         4 beads.

  a = (1/2)(1+a) + (1/2)(1+b) ...........(1)
  b = (1/2)(1+a) + (1/2)(1+c) ...........(2)
  c = (1/2)(1+b) + (1/2)(1+d) ...........(3)
  d = (1/2)(1+c) + (1/2)(1+e) ...........(4)
  e = (1/2)(1+d) + (1/2)(1)   ...........(5)

we must use this chain of equations to find a.

  From equation (1) 

   a/2 = 1 + b/2
     a = 2 + b        so  b = a-2  and  from (2)

   a-2 = 1 + a/2 + c/2
   a/2 - 3 = c/2      so   c = a-6  and from (3)
   
   a-6 = 1 + b/2 + d/2
   a-7 = b/2 + d/2
   2a-14 = b + d
   2a-14 = a-2 + d
    a-12 = d      so  d = a-12  and from (4)

  a-12 = 1 + c/2 + e/2
  a-13 = c/2 + e/2
  2a-26 = c + e 
  2a-26 = a-6 + e
   a-20 = e        so  e = a-20  and from (5)

  a-20 = 1 + d/2 
  a-21 = d/2
  2a-42 = d
  2a-42 = a-12
     a = 30

Therefore the expected number of throws is 30 to get 5 beads. 

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/