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?

  • 2
    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]?2012-06-20

1 Answers 1

5

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

  • 0
    @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)$2012-06-20