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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?

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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]? –  The Chaz 2.0 Jun 20 '12 at 18:32
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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 $$P(\text{# Heads} > \text{#Tails})= P(\text{# Heads} < \text{#Tails}) = \dfrac{1-\dfrac{\dbinom{2n}{n}}{2^{2n}}}2 = \dfrac12 - \dfrac{\dbinom{2n}n}{2^{2n+1}}$$

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Thanks for your answer. could you please explain how you got $\binom {2n}{n}$ –  Jay Jun 20 '12 at 18:40
    
@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 –  user17762 Jun 20 '12 at 18:43
    
Yes, I'm familiar with the binomial distribution. –  Jay Jun 20 '12 at 18:47
    
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? –  Jay Jun 20 '12 at 18:49
    
@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 $$\dbinom{2n}n \left(\dfrac12 \right)^n \left(\dfrac12 \right)^n = \dbinom{2n}n \left(\dfrac1{2^{2n}} \right)$$ –  user17762 Jun 20 '12 at 18:49
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