Ten Coin Flips, Four Heads

Date: 05/08/2001 at 18:21:06
From: Timi
Subject: If you flip a coin 10 times, what is the probability of 
getting at least 4 heads?

Hi!

My professor and I disagree about solving this problem:

If you flip a coin ten times, what is the probability of getting at 
least four heads?

The way I see it is that there are  2^10  possible outcomes, since 
each separate toss has 2 outcomes: head or tail, and there are 
10 tosses. So

10C4  (or C10,4?)  / 2^10 =  20.5%

The way my professor deals with this problem is:

"Add the ninth row and then the tenth row, by adding the two 
numbers above in Pascal's triangle :

  1  9  36  84  126  126  84  36  9  1
1 10 45  120 210 252  210  120 45 10  1

and we can write our equation:

1[(.5)^10] [(.5)^0] + 10[(.5)^9] [(.5)^1] + 45 [(.5)^8] 
[(.5)^2]+120[(.5)^7] [(.5)^3] +210[(.5)^6] [(.5)^4]+252[(.5)^5] 
    [(.5)^5]+210[(.5)^4] [(.5)^6]+ etc.  

but I can stop here because it says AT LEAST 4 and that means 
10,9,8..4 so I can stop.

Now a shortcut. Because our probabilities are .5 heads and .5 tails, 
each term is [(.5)^.5] or [1/2^10] or 1/(2^10) or 1/1024  so I can 
just add the coefficients I need and divide by 1024.

Adding 1 10 45  120 210 252  210  = 848  which we divide by 
1024  = 82.8% "

It seems a little bit complex to me, and our answers are different, 
too. Could you please let me know what you think about this problem?

Thank you!
Timi

Date: 05/10/2001 at 14:43:57
From: Doctor Twe
Subject: Re: If you flip a coin 10 times, what is the probability of 
getting at least 4 heads?

Hi Timi - thanks for writing to Dr. Math.

Your professor is right. Here is one way to look at the problem:

Since the total of the probabilities of all outcomes must be 1, the 
probability of getting at least four heads is:

     p(>=4H) = 1 - [ p(0H) + p(1H) + p(2H) + p(3H) ]

It's easier to add up these four probabilities and subtract from 1 
than to add up:

     p(>=4H) = p(4H) + p(5H) + p(6H) + ... + p(10H)

So what is the probability of getting no heads, p(0H) ? There's only 
one way to do that:

     T T T T T T T T T T

so p(0H) = 1/1024. (Your assessment that there are 1024 possible 
combinations is correct.)

What about the probability of getting exactly 1 head, p(1H)? There are 
ten ways to do that:

     H T T T T T T T T T
     T H T T T T T T T T
     T T H T T T T T T T
              :
     T T T T T T T T T H

so p(1H) = 10/1024.

What about the probability of getting exactly two heads, p(2H)? Here 
it starts to get tricky:

     H H T T T T T T T T
     H T H T T T T T T T
     H T T H T T T T T T   That's 9 ways that start with H
              :
     H T T T T T T T T H

     T H H T T T T T T T
     T H T H T T T T T T
     T H T T H T T T T T   That's 8 more ways that start with TH
              :
     T H T T T T T T T H

     T T H H T T T T T T
     T T H T H T T T T T
     T T H T T H T T T T   That's 7 more ways that start with TTH
              :
     T T H T T T T T T H

     T T T H H T T T T T
     T T T H T H T T T T
     T T T H T T H T T T   That's 6 more ways that start with TTTH
              :
     T T T H T T T T T H

     T T T T H H T T T T
     T T T T H T H T T T
     T T T T H T T H T T   That's 5 more ways that start with TTTTH
     T T T T H T T T H T
     T T T T H T T T T H

     T T T T T H H T T T
     T T T T T H T H T T
     T T T T T H T T H T   That's 4 more ways that start with TTTTTH
     T T T T T H T T T H

     T T T T T T H H T T
     T T T T T T H T H T   That's 3 more ways that start with TTTTTTH
     T T T T T T H T T H

     T T T T T T T H H T
     T T T T T T T H T H   That's 2 more ways that start with TTTTTTTH

     T T T T T T T T H H   That's 1 more way starting with TTTTTTTTH

So there are:

      9
     SUM i = 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45
     i=1

ways of getting exactly two heads, and so p(2H) = 45/1024.

We can do the same for exactly 3 heads, getting:

     H H H T T T T T T T ... H H T T T T T T T H   8 combinations
     H T H H T T T T T T ... H T H T T T T T T H   7 combinations
     H T T H H T T T T T ... H T T H T T T T T H   6 combinations
              :                                    :
     H T T T T T T T H H                           1 combination

     T H H H T T T T T T ... T H H T T T T T T H   7 combinations
     T H T H H T T T T T ... T H T H T T T T T H   6 combinations
     T H T T H H H T T T ... T H T T H T T T T H   5 combinations
              :                                    :
     T H T T T T T T H H                           1 combination

     T T H H H T T T T T ... T T H H T T T T T H   6 combinations
     T T H T H H T T T T ... T T H T H T T T T H   5 combinations
     T T H T T H H H T T ... T T H T T H T T T H   4 combinations
              :                                    :
     T T H T T T T T H H                           1 combination

and so on all the way to:

     T T T T T T T H H H                           1 combination

Thus we have:

      8       7       6       5       4       3       2       1
     SUM a + SUM b + SUM c + SUM d + SUM e + SUM f + SUM g + SUM h
     a=1     b=1     c=1     d=1     e=1     f=1     g=1     h=1

        8     i
     = SUM [ SUM j ] 
       i=1   j=1

     = 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36

     = 120

ways of getting exactly three heads, so p(3H) = 120/1024, and our 
probability of getting at least four heads is:

     p(>=4H) = 1 - [1/1024 + 10/1024 + 45/1024 + 120/1024]

             = 1 - 176/1024

             = 848/1024 = 53/64 ~= .828  (or 82.8%)

Notice that the values in the numerators, i.e. 1, 10, 45 and 120, are 
the first four values in the 10th row of Pascal's triangle; a fact 
that your professor used to save the time needed to count the 
combinations. For more information on this, check out the following 
entries in our Ask Dr. Math archives:

   An Explanation of Pascal's Triangle
   http://mathforum.org/dr.math/problems/pascal.html   

   Probability and Pascal's Triangle
   http://mathforum.org/dr.math/problems/ramirez6.4.98.html   

   Toss a Coin Six Times
   http://mathforum.org/dr.math/problems/beldon2.7.98.html   

   Binomial Expansions and Pascal's Triangle
   http://mathforum.org/dr.math/problems/fama.7.10.96.html   

I hope this helps! If you have any more questions or comments, write 
back again.

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