Probability of n heads in 2n tosses, given at least one head in n tosses Suppose we toss a fair coin $2n$ times. Conditioned on the fact that the coin came up heads at least once in the first $n$ tosses, what is the probability that we get precisely a total of $n$ heads over all $2n$ heads?
 A: By the $P(B \mid A) = P(A \cap B) / P(A)$ definition of conditional probability, you need to compute

*

*the probability of getting $n$ heads over all tosses, with at least one of the heads occurring in the first $n$ tosses, and

*the probability of getting at least one head in the first $n$ tosses.

For the second term, it is easier to compute the probability of the complement.
For the first term, it might be helpful to consider the probability of getting $n$ heads over all tosses, with no heads occurring in the first $n$ tosses.
A: The magic words are: Bayes' rule.
$$P(A|B) = \frac{P(B|A) P(B)}{P(A)}.$$
In this case: $A$ is the event of getting exactly $n$ heads, so $P(A) = \frac{\binom{2n}{n}}{2^{2n}}.$ $P(B)$ is the probability of getting at least one head in the first $n$ tosses: $1-\frac1{2^n}.$
$1-P(B|A) = \frac1{2^{2n}}$ (since this is the event of getting all tails in the first $n$ tosses and all heads in the last $n$ tosses).
A: 

*Suppose you perform $2n$ coin flips. There are $2^{2n}$ possible outcomes, because each of the $2n$ flips can be either heads or tails.


*Suppose you perform $2n$ coin flips and got exactly $n$ heads. Combinatorically, this involves picking $n$ out of $2n$ coins to come up heads; there are ${2n \choose n}$ such outcomes (out of $2^{2n}$ possible outcomes).


*Suppose you perform $2n$ coin flips and get $n$ tails followed by $n$ heads: TTT...TTT HHH...HHH.  This is the one and only outcome where you have exactly $n$ heads but none of them occur within the first $n$ tosses.


*Hence by (1) and (2), there are exactly $\left[{2n \choose n} - 1\right]$ outcomes that have exactly $n$ heads and also at least one head in the first $n$ flips.
Because the outcomes are all equally likely by assumption, the probability of this event is $ \left[{2n \choose n} - 1\right] / 2^{2n}$. This is the probability of getting exactly $n$ heads, at least one of which occurs within the first $n$ tosses.


*Let $X$ denote the event "You flip a coin $2n$ times". Let $A$ denote the event "...and get exactly $n$ heads". Let $B$ denote the event "...and there are no heads in the first $n$ tosses." Then:

*

*The quantity you want to compute is "The probability of getting exactly $n$ heads, conditioned on getting at least one head in the first $n$ tosses." That is, $P(A|\overline B) = \frac{P(A\cap \overline B)}{P(\overline B)}$.

*By (3), the probability $P(A\cap \overline B)$ is $\left[{2n \choose n} - 1\right] / 2^{2n}$.

*The probability of $B$ is the probability of getting no heads in the first $n$ tosses. How many outcomes are like this? : they have $n$ tails for the first $n$ tosses, and either heads or tails for the  next $n$ tosses; hence $2^n$ possibilities overall.  Because each outcome is equally likely, their combined probability is $P(B) = 2^n / 2^{2n} = \frac{1}{2^n}$. Therefore, the quantity we really want, $P(\overline B)$, is equal to $1-2^{-n}$.

*Hence, completing the division, the probability is $$P(A\cap\overline B)/P(\overline B) = \frac{\left[{2n \choose n} -1\right]}{2^{2n}}\left[\frac{1}{1-2^{-n}}\right].$$
