Getting More Heads

Date: 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/