How many coin flips

How many coin flips?

Let's say I can flip a fair coin as many times as I want, and for simplicity, I will call heads each time. I am looking for a correct result at least 10 times. (in other words, 10 or more results of heads) I would like to know the minimum number of flips it would take to expect to get at least 10 correct coin flips more than half the time. In other words, p(#Heads => 10) => .5, and consequently, p(#Tails < n-10) <= .5, where n = number of flips. If this is found easily enough, feel free to tack on a number of flips to make it even more likely, say .75 or .95. Thanks in advance!

Other Answers:

Nice question but I don't think there's an easy way to solve it algebraically - except by slogging it out.

If you wanted the probability of all heads or no heads or at least one head it would be easy.

But your question is much harder. (I think) Someone else might know better.

For yours:
Let X be the number of heads in n throws of the coin.
X follows a binomial distribution with n trials, p=1/2

You want
P(X>=10) > 1/2
ie P(x<10) < 1/2

P(X<10) is the sum of the probabilities that X=0, X=1,X=2,...X=9

nC0(1/2)^n +nC1(1/2)^n + nC2(1/2)^n +... +nC9(1/2)^n <1/2

ie nC0 +nC1 +nC2 +... +nC9 < 1/2 ^(1-n)

ie nC0 +nC1 +nC2 +... +nC9 < 2^(n-1)

I think the only way to do this is by trial and error. Probably setting up some lists in Excel or a graphing calculator.

Good luck.

I've just set up some lists and found different values for n.
Here goes.
For P(x>=10)>1/2 n is 20

For P(X>=10) > 3/4
You would solve
P(X<10) <1/4
nC0 +nC1 +nC2 +... +nC9 < 2^(n-2)
in which case n=23

and P(X>=10) > 0.95

You would solve
P(X<10) <0.05
nC0 +nC1 +nC2 +... +nC9 < 2^(n-2)/5
in which case n=28

Again, good question. I hope my reasoning is correct.