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### 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
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 -

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
```
Associated Topics:
College Probability
High School Probability

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