Getting More HeadsDate: 05/29/2000 at 23:15:56 From: Ernesto Ruiz Subject: Probability Bob flips 850 fair coins. Alice flips 851 fair coins. What is the probability that Alice gets strictly more heads than Bob? Briefly but clearly explain your answer. Date: 05/30/2000 at 06:40:32 From: Doctor Floor Subject: Re: Probability Hi, Ernesto, Thanks for writing. The probability that Bob gets n heads and Alice gets more heads is found by: C(850,n)*(1/2)^850 * SUM{m>n} C(851,m)*(1/2)^851 = (1/2)^1701 * C(850,n) * SUM{m>n} C(851,n) So the requested probability is (1/2)^1701 * SUM{n} [ C(850,n) * SUM{m>n} C(851,m) ] ......[1] SUM{n} is short for SUM{n = 0 to 850} and SUM{m>n} is likewise short for SUM{m = n+1 to 851}. Now note that SUM{n} [ C(850,n) * SUM{m>n} C(851,m) ] = SUM{n} [ C(850,n) * (2^851 - SUM{m<=n} C(851,m)] = 2^850*2^851 - SUM{n} [ C(850,n) * SUM{m<=n} C(851,m) ] Observe that by the symmetry of Pascal's triangle we see SUM{n} [ C(850,n) * SUM{m>n} C(851,m) ] = SUM{n} [ C(850,n) * SUM{m<=n} C(851,m) ] So we have 2*SUM{n} [ C(850,n) * SUM{m>n} C(851,m) ] = 2^1701 SUM{n} [ C(850,n) * SUM{m>n} C(851,m) ] = 2^1700 Substituting this into [1] we find the requested probability is equal to 1/2. If you have more questions, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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