Hi everyone. My question is what is the probability of getting 10
heads from a fair coin flip 20 times?

Please show as much work as the amount I have paid will allow.

Magnoliatwin ?

Getting exactly 10 heads is relatively straightforward.  (Calculating
AT LEAST 10 heads is harder.)

Let?s start with how many ways you can toss the coin and get results. 
Each coin flip has 2 outcomes: H or T, so it?s
2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2

That?s 2 to the 20th power, often written as 2^10 = 1,048,576


But how many ways are there to get 10H?s?  In statistics this is a
combination.  Actually, because you can choose only H or T each time,
it?s a ?combination without repetition? --

Wikipedia
?Permutations and Combinations? (Aug. 30, 2005)
http://en.wikipedia.org/wiki/Combinations_and_permutations

Your combinations are often written 20C10 and it?s defined as:

nCr = n!/r! (n-r!)

or 20C10 = 20! / 10! 10! = 670,442,527,800 / 3,628,800 = 184,756 combinations


To figure your probability, it?s 184,756 combinations of 10 Heads ?
out of 1,048,576 = 17.62% or about 1 in 6.

Google search strategy:
Permutation combination ?coin toss?

The Wikipedia article cited above is very good at describing the
probabilities, permutations and computations and is one of the
clearest on the web in its use of symbols.

Best regards,

Omnivorous-GA

Thank you very much for this straightforward and speedy reply.

Hello,

This is a simple case of binomial probabilty distribution.

This link will give you the details and the formula.
http://science.kennesaw.edu/~jdemaio/1107/binomial_probability_distributio.htm

In your case, 

n- number of trials is 20
x- number of successes is 10
p- probability of success is 0.5 as you say it is a fair trial.

With these the binomial formula yields an answer 0.176.
So your probability is 17.6%.

Regards,
Krishnan

Logically, if the odds of exactly 10 heads is 17.62%, then the 
odds of '10 or more' heads is 58.81%

My reasoning is thus: 

There are 10 possible outcomes with less than 10 (0 - 9) and 
10 possible outcomes with more than 10 (11 - 20)

The odds (whatever they may be) for no heads or 20 heads are the 
same, as are the odds of 9 heads or 9 tails. So the 'greater than' 
and 'less than' odds are equal. All outcomes (100%) - exactly 10 
heads (17.62%) leaves 82.38% other outcomes, if 1/2 (41.19%) are 
'greater than' then 17.62 + 41.19 or 58.81% must be equal or greater.

I've always been confused by this question.
You take your coin and flip it, having a 1/2 chance of it landing on heads.
You flip it again, having a 1/2 chance of it landing on heads.
So on and so forth until your 100th flip.
Now assume you have managed to get 99 heads up to this point. What is
the probability to get another head in the 100th toss?
Theoretically, because each toss is independent, it should be again
50%. But probability theory also tells us that in the long run, the
tendency should be half heads and half tails. So, we got 99 heads
already, but it doesn't increase the chance of getting a tail next
toss. Then, how could the probability structure of half-half be
demonstrated anyway?