Even Number of HeadsDate: 02/12/2002 at 09:11:18 From: Courtney Subject: Probability of getting an even number of heads when tossing a biased coin In my book, it says that the probability of getting an even number of heads when a coin is tossed N times, when the probablity of getting a head is p and tail is q = 1-p, is .5(1+(q-p)^N). Can you explain how they got this? Thanks, Courtney Date: 02/12/2002 at 19:38:20 From: Doctor Anthony Subject: Re: Probability of getting an even number of heads when tossing a biased coin Let u(r) be the probability that there is an even number of heads after r throws. So 1-u(r) is the probability of an odd number of heads after r throws. The probability of an even number of heads after r throws is q times the probability of an even number of heads after r-1 throws plus p times the probability of an odd number of heads after r-1 throws. So u(r) = q.u(r-1) + p.[1-u(r-1)] u(r) = u(r-1).(q-p) + p = [u(r-2).(q-p) + p](q-p) + p = u(r-2).(q-p)^2 + p(q-p) + p = [u(r-3).(q-p) + p](q-p)^2 + p(q-p) + p = u(r-3).(q-p)^3 + p(q-p)^2 + p(q-p) + p and so on down to u(0).(q-p)^r + p[(q-p)^(r-1) + (q-p)^(r-2) + ...+ 1] 1 - (q-p)^r u(0).(q-p)^r + p[-------------] 1 - (q-p) 1 - (q-p)^r u(0).(q-p)^r + p[-------------] p+q - q + p u(0).(q-p)^r + (1/2)[1 - (q-p)^r] and u(0) = 1 so this reduces to (q-p)^r + 1/2 - (1/2).(q-p)^r = 1/2 + (1/2)(q-p)^r = (1/2)[1 + (q-p)^r] as required - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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