Penny-moving GameDate: 06/24/99 at 20:59:07 From: Kathy Subject: Probability The numbers 0 1 2 3 4 5 6 7 are on a number line. A penny is placed on the 3. A fair coin is flipped. If the flip of the coin shows heads, the penny is moved one space to the left. If the flip of the coin shows tails, the penny is moved one space to the right. What is the probability of the penny landing on 0 before it lands on 7? Date: 06/25/99 at 10:52:34 From: Doctor Anthony Subject: Re: Probability Since there are equal probabilities of going left or right, the probabilities are almost certainly in inverse proportion to the distance from the starting point. So we can expect that the probabilities of reaching 0 before we reach 7 are in the ratio 4:3. So the probability of reaching 0 before we reach 7 is 4/7. We can prove this by a series of linear equations. Suppose that x1 is the probability of reaching 0 first if we are at position 1, x2 is the probability of reaching 0 if we are at position 2, and so on. We need to find the value of x3. We have x1 = 1/2 + (1/2)x2 That is, if we are at position 1, there is 1/2 probability of 'winning' and 1/2 probability of moving to position 2 where the probability of winning is now x2. So we get the following sequence of six equations: x1 = 1/2 + (1/2)x2 ......(1) x2 = (1/2)x1 + (1/2)x3 ......(2) x3 = (1/2)x2 + (1/2)x4 ......(3) x4 = (1/2)x3 + (1/2)x5 ......(4) x5 = (1/2)x4 + (1/2)x6 ......(5) x6 = (1/2)x5 + 0 ......(6) The last entry is 0 because if you move to position 7 you can't win. From these 6 equations with 6 unknowns we can find any of the probabilities we require. Working up the list of equations, and putting (6) into (5) we get x5 = (1/2)x4 + (1/2)(1/2)x5 (3/4)x5 = (1/2)x4 x5 = (2/3)x4 Then from (4) x4 = (1/2)x3 + (1/3)x4 (2/3)x4 = (1/2)x3 x4 = (3/4)x3 Then from (3) x3 = (1/2)x2 + (3/8)x3 (5/8)x3 = (1/2)x2 x3 = (4/5)x2 Then from (2) x2 = (1/2)x1 + (2/5)x2 (3/5)x2 = (1/2)x1 x2 = (5/6)x1 Then from (1) x1 = 1/2 + (5/12)x1 (7/12)x1 = (1/2) x1 = 6/7 Then x2 = (5/6)(6/7) = 5/7 and x3 = (4/5)(5/7) = 4/7 So we get the result x3 = 4/7 as we predicted. This method could of course be used if there were a difference in the probabilities of moving left or right where the simple ideas of method (1) would not work. If you are familiar with difference equations you can use the following method to solve this. We can make up a linear recurrence relation to find the required probability. Let u(n) be the probability that we end up at position 0 before position 7 when we are at position 'n'. The recurrence relation is u(n) = (1/2)u(n-1) + (1/2)u(n+1) or u(n+1) - 2.u(n) + u(n-1) = 0 This is a homogeneous difference equation whose solution will be given by solutions of the quadratic x^2 - 2x + 1 = 0 (x-1)^2 = 0 or x = 1 (twice) So the expression for u(n) will be u(n) = (A + Bn)1^n = A + B.n We have u(0) = 1 and u(7) = 0 so we get A + 0.B = 1 -> A = 1 A + 7.B = 0 -> B = -1/7 Therefore u(n) = 1 - n/7 We require u(3) which we can see is 1 - 3/7 = 4/7 as required. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/