Chance of 100th coin flip being heads Assuming I have flipped 99 coins all tails, what is the chance of the 100th coin being heads?
Yes I know it's still 50%, but is there any other formula that could help explain the risk/probability of a heads/win if I only start playing after X amount of losses? I would like to know how much better 'chances' I have of a heads if I start playing after a certain amount of losses.
 A: In the ideal world that mathematicians normally inhabit, there are perfectly fair coins.  With these, the previous $99$ tails would be totally irrelevant and the $100$th throw would be just as likely to be heads as the first.
In the real world, if I saw $99$ heads in a row then I would seriously doubt the fairness.  Either the coin was biased or the person throwing it was able to manipulate it.  You could use methods from statistics to judge whether the coin might really be fair.  I have not done the calculations but it will be very unlikely.
A: Let $X$ be the event that we throw $n$ tails in a run, and $Y$ the event we throw heads,
Using Bayes theorem we can write
$$P(Y|X) = \frac{P(X|Y)P(Y)}{P(X)} = \frac{\large(\frac12\large)^n\frac12}{\large(\frac12\large)^n} = \frac12$$
So it doesn't matter how many times we lose, the chance of a head remains the same.
A: Let $p$ be the probability that your coin has two tails. (Basically the probability that you've been tricked.) Now, the probability that your last coin will be tails is this:
$$p+(1-p)2^{-100}$$
out of the probability to get the $99$ tails in the first place:
$$p+(1-p)2^{-99}$$
so the conditional probability that you will get tails is:
$$\frac{p+(1-p)2^{-100}}{p+(1-p)2^{-99}}=1-\frac{1-p}{2^{100}p+2(1-p)}\ge 1-\frac{1}{2^{100}p}$$
which is very close to $1$, unless $p$ is of the order of magnitude of $2^{-100}$ or smaller! In other words, this confirms our intuition that, if there is any (reasonable) chance you've been tricked, you can well expect to have tails again.
(Of course, when $p=0$, the formula above gives $\frac{p+(1-p)2^{-100}}{p+(1-p)2^{-99}}=\frac{1}{2}$, as expected.)
You can expand this reasoning to the case where the coin might not really have two tails but be biased, and if there is any chance that the coin is heavily biased, this chance will again blow up by the fact that there were $99$ tails in a row. I would not provide a complete computation here.
