4 heads in 8 tosses If someone asked me the odds of getting 4 heads in 8 flips of a fair coin.
I would initially think to do something like this:
$\dfrac{2^8 - \left( \binom{8}{0} + \binom{8}{1} + \binom{8}{2} + \binom{8}{3} \right)}{2^8} = \dfrac{163}{256} \simeq 63.67\%$
In that there's $2^8$ total different outcomes of 8 flips, and of those (in the numerator) I subtract out the outcomes that have 0 heads, 1 heads, 2 heads, or 3 heads.  (Or equivalently, I could have subtracted out those with 8, 7, 6, or 5 tails).
However, I think this is more like asking "getting at least 4 heads in 8 fair flips"?  I assume if you were tossing these as soon as you hit 4 heads you'd be in a "win" state, so it wouldn't matter if the rest of the flips would result in greater than 4 heads for that particular outcome?
However, suppose that as soon as 4 heads were obtained, the flips would stop, so it would not be possible to get greater than 4 heads in any outcome.
So wouldn't that mean the total outcomes would be less?
Either 0 heads happen, 1 heads, 2 heads, 3 heads, or 4 heads:
$\binom{8}{0} + \binom{8}{1} + \binom{8}{2} + \binom{8}{3} + \binom{8}{4} = 163 $
Then, I only care about those that have 4 heads:
$\dfrac{\binom{8}{4}}{163} = \dfrac{70}{163} \simeq 42.94\%$
So why are the probabilities different? Shouldn't this be the same sort of question?
 A: Imagine I watch you while you flip the coin. If you reach four heads before the eighth flip, I distract you by shouting, "Look! Your events are not equally likely!" While you are distracted, I steal your coin. Then I go where you can't see flip the coin enough times so that you and I have made eight flips altogether.
Now the coin has been flipped eight times, so we know how likely it has four heads altogether. And you have observed all the cases with heads. So yes, the probability you observe should be the same as the probability I observe.
Something to note is that in your second setup, you count the events HHHH and TTTTHHHH equally (i.e., each contributes $1$ to the numerator $70$). But the HHHH event is much more likely to occur than TTTTHHHH. Therefore your total probability does not compute correctly under the assumption that you have found equally likely outcomes.
In fact, any given unique result of a series of $n$ coin tosses has a probability 
$2^{-n}$ to occur. For example, HHTHH is one unique result that can occur from the first five coin tosses, and it has probability $2^{-5} = \frac1{32}$.
The probabilities actually add up like this, dividing the "successful" outcomes 
into five cases depending on how many times you toss the coin in order to achieve
four heads:
The 4th head occurs on the 4th toss: This can happen only one way: HHHH. The probability of that event is 
$$P_4 = \frac1{16}.$$
The 4th head occurs on the 5th toss: This means there are $3$ heads in the first $4$ tosses, and the last toss is a head. There are $\binom{4}{3} = 4$ unique ways this can happen (for example, HHHTH). Each of these unique outcomes has probability $\frac1{32}$, so the probability that the outcome will be in this set is
$$P_5 = \frac4{32} = \frac18.$$
The 4th head occurs on the 6th toss: That is, $3$ heads in the first $5$ tosses, then a head. There are $\binom{5}{3} = 10$ unique ways this can happen. Each of these unique outcomes has probability $\frac1{64}$, so the probability that the outcome will be in this set is
$$P_6 = \frac{10}{64} = \frac5{32}.$$
The 4th head occurs on the 7th toss: There are $\binom{6}{3} = 20$ unique ways this can happen. Each of these unique outcomes has probability $\frac1{128}$, so the probability that the outcome will be in this set is
$$P_7 = \frac{20}{128} = \frac5{32}.$$
The 4th head occurs on the 8th toss: There are $\binom{7}{3} = 35$ unique ways this can happen. Each of these unique outcomes has probability $\frac1{256}$, so the probability that the outcome will be in this set is
$$P_8 = \frac{35}{256}.$$
The total probability of tossing $4$ heads is therefore
$$P_4 + P_5 + P_6 + P_7 + P_8 = \frac{163}{256} \approx 0.6367,$$
just like the first result.
Alternatively, consider the probability of failing to get four heads. Each unique outcome in this set requires you to toss the coin $8$ times, so it occurs with probability $\frac1{256}$. The total probability of all these cases is
$$P_\mbox{fail} = 
\frac1{256} \left(\binom{8}{0} + \binom{8}{1} + \binom{8}{2} + \binom{8}{3} \right)
= \frac{93}{256}.$$
The probability that you do toss four heads is therefore
$$1 - P_\mbox{fail} = \frac{163}{256}.$$
A: You are misinterpreting what the second probability, $70/163\approx42.94\%$, means.  It's actually a conditional probability:  It's the probability that you got exactly four heads, given that you got at least four heads.  (As DavidK points out in comments, it's also the conditional probability of getting exactly four heads given that you got no more than four heads, which is closer to the OP's derivation.  Both interpretations, however, require that the coin be tossed all eight times, which is counter to what the OP wants to do.  DavidK's answer shows the proper way to analyze things if you insist on stopping as soon as the fourth head appears.)
Incidentally, a quicker way to get the number $163$ is to consider that you either get your four heads within the first seven flips, or else you get the fourth head on the eighth flip.  After the first seven flips, you either have more heads than tails or more tails than heads; by symmetry each probability is $1/2$.  To get the fourth head on the eighth flip, you need to have gotten the other three in the first seven, and then flip a head.  In total, the probability is
$${1\over2}+{1\over2}\cdot{{7\choose3}\over2^7}={1\over2}+{35\over256}={163\over256}$$
A: $\newcommand{\+}{^{\dagger}}
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$$
{8 \choose 4}\pars{\half}^{8 - 4}\pars{\half}^{4}
={8\cdot 7\cdot 6\cdot 5 \over 4\cdot 3\cdot 2}\,{1 \over 2^{8}} = \color{#88f}{\Large{35 \over 128}}
\approx 0.2734
$$
A: If you want exactly 4 Heads at the 8th toss, that means you have 3 heads somewhere in the initial 7 tosses.
So $\binom 73 p^3 (1-p)^4$, then there is the final eighth trial which must be head.
so we get $\binom 73 p^3 (1-p)^4 p$
$$=\binom 73 p^4(1-p)^4$$
$$=\binom 73 \left(\frac{1}{2}\right)^8$$
I would suggest looking up the negative binomial distribution.  There are many ways of expressing it. Above, I have shown the version where we specifiy the number of desired successes, the paramater of success, $p$. Which means, the random variable in question examines the number of trials to make tha thappen:
$$P(X=k)=\binom{k-1}{r-1}p^r(1-p)^{k-r}$$
Where of course $k\geq r$, $r$ is number of successes.
This is a more general case for the geometric distribution:
NegativeBinomial(r=1,p) ~ Geometric(p)
A: The Question is stated as "If someone asked me the odds of getting $4$ heads in $8$ flips of a fair coin".
The probability is easily shown to be $\frac{35}{128}$ as some of you have correctly answered, and the odds would be $35$ to $93$.
