Kdtyjb

The probability is $frac{1}{2}$ because the last flip determines it.

If there are an even number of heads then there must be an odd number of tails. But heads and tails are symmetrical, so the probability must be $1/2$.

Your approach is good also, you probably made a mistake in calculations. The number of favorable outcomes is
$$binom{9}{0}+binom{9}{2}+binom{9}{4}+binom{9}{6}+binom{9}{8}=1+36+126+84+9=256$$
The number of all possible outcomes is $512$ thus the probability to get even number of heads is $0.5$.

All the possible sequences of tosses (which are all equally likely) may be divided into the two categories "even number of heads" and "odd number of heads". You can then pair up each sequence in one category with the "flipped" version (flip every coin in the sequence around) in the other category. This shows that there are equally many sequences in each of the two categories. So the probability of landing in a specific one of them must be $frac12$.

There are two cases here:

• There's an even number of heads: 0, 2, 4, 6 or 8 heads


• There's an odd number of heads: 1, 3, 5, 7 or 9 heads

But notice that having an odd number of heads means having an even number of tails (e.g. 5 heads means 4 tails), so the second case is the same as:

• There's an even number of tails: 0, 2, 4, 6 or 8 tails

Since heads and tails are equally probable, we can, by symmetry, see that these two cases have the same probability. Therefore each must have probability $1/2$.

The easiest way to see this : Consider the number of heads we have in the first $8$ coins.

• If this number is even, we need a tail, we have probability $frac{1}{2}$


• If this number is odd, we need a head, we have probability $frac{1}{2}$

Hence no matter what the $8$ coins delivered, we have probability $frac{1}{2}$ , which is the answer.

The probability generating function of a Binomiall random variable $Xsim text{Bin}(n, 1/2)$ with probability of success $1/2$ is given by
$$
g_{X}(t)=Et^X=sum_{k=0}^nP(X=k)t^k=sum_{k=0}^nbinom{n}{k}frac{t^k}{2^n}=frac{1}{2^n}(1+t)^n
$$
In particular the probability that $X$ is even is given by
$$
sum_{0leq kleq n, k,{text{even}}}P(X=k)=frac{g(1)+g(-1)}{2}=frac{1+0}{2}=frac{1}{2}.
$$

If $h$ denotes getting a heads and $p$ denotes getting tails, let's write $frac{1}{2}h+frac{1}{2}p$ for the notion of a fair coin: half the time it turns up heads, and half the time tails.

If we expand out the following product while keeping track of multiplication order,
$$left(frac{1}{2}h+frac{1}{2}pright)left(frac{1}{2}h+frac{1}{2}pright)=frac{1}{4}hh+frac{1}{4}hp+frac{1}{4}ph+frac{1}{4}pp,$$
we see that the sequences $hh$, $hp$, $ph$, and $pp$ are equally likely. Forgetting about multiplication order, which we want to do because we only care about how many times heads or tails showed up, corresponds to just treating this like a polynomial:
$$=frac{1}{4}h^2+frac{1}{2}hp+frac{1}{4}p^2.$$
We can keep multiplying copies of a fair coin together to find the probability that a certain number of heads or tails happened, where the coefficient in front of $h^kp^ell$ is the probability of $k$ heads and $ell$ tails.

Nine coins is the expansion
$$left(frac{1}{2}h+frac{1}{2}pright)^9=sum_{k=0}^9binom{9}{k}left(frac{1}{2}hright)^kleft(frac{1}{2}pright)^{9-k}=sum_{k=0}^9frac{1}{2^9}binom{9}{k}h^kp^{9-k}.$$
So far, all this has done is explain why you were adding up $2^{-9}binom{9}{k}$ for $k=0,2,4,dots,8$. Here, now, is a nice trick. If we formally set $h=1$ and $p=1$, then we get
$$1=sum_{k=0}^9frac{1}{2^9}binom{9}{k},$$
and if we formally set $h=-1$ and $p=1$, then we get
$$0=sum_{k=0}^9frac{1}{2^9}binom{9}{k}(-1)^k.$$
The average of these two equations is
$$frac{1}{2}=sum_{k=0,ktext{ even}}^9frac{1}{2^9}binom{9}{k},$$
since $frac{1}{2}(1+(-1)^k)$ is $1$ or $0$ depending on whether $k$ is even or odd. Thus the probability of an even number of heads is $frac{1}{2}$.

Notice that this did not use the fact nine coins were tossed at any point! (Other than the fact that at least one coin was tossed. In the case of tossing no coins, an even number of heads happens with probability $1$. What part of my argument goes wrong for the case of zero coins?)

There's a way to do it with barely any maths:

It's clear that if there's an odd number of heads, there's an even number of tails and vice versa, so P(even number of heads) + P(even number of tails) = 1.

Formally rename "heads" to "tails". The problem remains unchanged.

So P(even number of heads) = P(even number of tails) = 1/2.