2
$\begingroup$

I am told that a fair coin is flipped $2n$ times and I have to find the probability that it comes up heads more often that it comes up tails.

Please, how do I find the required probability?

$\endgroup$
  • 2
    $\begingroup$ What is the probability of the number of heads equaling the number of tails? Then, wouldn't the probability of [more heads than tails] be the same as the probability of [more tails than heads]? $\endgroup$ – The Chaz 2.0 Jun 20 '12 at 18:32
5
$\begingroup$

Note that we have $$P(\text{# Heads} > \text{#Tails}) + P(\text{# Heads} = \text{#Tails}) + P(\text{# Heads} < \text{#Tails}) = 1$$ Assuming you are tossing a fair coin, by symmetry, we also have that $$P(\text{# Heads} > \text{#Tails})= P(\text{# Heads} < \text{#Tails})$$

If we want to get $k$ heads in $2n$ tosses, where the probability of getting a head is $p$ then the probability is $$\dbinom{2n}k p^k (1-p)^{2n-k}$$ In our case, if we want the number of heads to be the same as number of tails then $k = n$ and if we are tossing a fair coin then $p=1/2$. Hence, we get $$P(\text{# Heads} = \text{#Tails}) = \dbinom{2n}n \left(\dfrac12 \right)^n \left(\dfrac12 \right)^n = \dfrac1{2^{2n}} \dbinom{2n}n$$

Hence, we get that $$P(\text{# Heads} > \text{#Tails})= P(\text{# Heads} < \text{#Tails}) = \dfrac{1-\dfrac{\dbinom{2n}{n}}{2^{2n}}}2 = \dfrac12 - \dfrac{\dbinom{2n}n}{2^{2n+1}}$$

$\endgroup$
  • $\begingroup$ Thanks for your answer. could you please explain how you got $\binom {2n}{n}$ $\endgroup$ – Jay Jun 20 '12 at 18:40
  • $\begingroup$ @Jay Probability of getting $k$ heads when we toss a coin $2n$ times is $\dfrac1{2^{2n}} \dbinom{2n}k$. Are you familiar with the binomial distribution? en.wikipedia.org/wiki/Binomial_distribution $\endgroup$ – user17762 Jun 20 '12 at 18:43
  • $\begingroup$ Yes, I'm familiar with the binomial distribution. $\endgroup$ – Jay Jun 20 '12 at 18:47
  • $\begingroup$ If I understand correctly, the number of ways of getting the same number of heads as tails is really $\binom{2n}{n}\binom{2n-n}{n}$. Am I right? $\endgroup$ – Jay Jun 20 '12 at 18:49
  • $\begingroup$ @Jay Then if you want to get $k$ heads in $2n$ tosses, where the probability of getting a head is $p$ then the probability is $$\dbinom{2n}k p^k (1-p)^{2n-k}$$ In our case, if we want the number of heads to be the same as number of tails then $k = n$ and if we are tossing a fair coin then $p=1/2$. Hence, we get $$\dbinom{2n}n \left(\dfrac12 \right)^n \left(\dfrac12 \right)^n = \dbinom{2n}n \left(\dfrac1{2^{2n}} \right)$$ $\endgroup$ – user17762 Jun 20 '12 at 18:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.