Probability of at Least 45 Heads in 100 Tosses of Fair Coin

Date: 05/15/2004 at 08:14:21
From: Joe
Subject: A different type of coin toss probability question

What is the probability of getting AT LEAST 45 HEADS out of 100 
tosses of a fair coin?

I assumed that the answer would be P(45) + P(46) + .... + P(100).  
When I calculated that, it came out to .86437.  However, an "expert"
suggested the answer was P(45) + P(46) + .... + P(49) + P(50) + 0.5,
where the 0.5 represents SUM P(51): P(100).  Doing it that way, I came
up with .90417.

My concern is that if I were to "modify" the problem to "at least 41 
heads", when I add the P(44) + P(43)+ P(42)+ P(41), I have already 
exceeded 100% probability with the .90417 answer.  It leads me to 
believe that my original answer of .86437 might be correct.  Please 
help by confirming if either is correct, or provide the correct answer 
if neither is correct.  Thanks.

Date: 05/15/2004 at 08:38:05
From: Doctor Douglas
Subject: Re: A different type of coin toss probability question

Hi Joe -

Thanks for writing to the Math Forum.  Your answer is correct.

There is a subtle detail that slightly complicates the shorthand 
method.  Since there are an EVEN number of coins, the number of 
possible outcomes is odd (e.g. 4 coins:  0,1,2,3,4 heads), and so the 
middle probability must be bisected at its halfway point:

  0.5 = Pr(50)/2 + [Pr(51) + ... + Pr(100)]

This explains the difference in your results:

  Pr(50)/2 = C(100,50)/2^101 = 100!/(50! x 50! x 2^101) = 0.0397946,

equal to the discrepancy between the two numbers you calculated above.

- Doctor Douglas, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 05/16/2004 at 06:44:26
From: Joe
Subject: Thank you (A different type of coin toss probability question)

Dr. Douglas,

Thanks for your prompt response and your explanation as to why the
shortcut method does not work.  It's great to have real experts when
one needs help.

Sincerely, Joe