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/
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