Past coin tosses affect the latest one if you know about them? Suppose Mark and Paul are sitting on a table, and Mark starts tossing an unbiased and fair coin. He tosses it for 99 times, and he gets 99 consecutive tails. At this point Mark asks Paul: "let's bet $100 on the next toss, do you want to pick tail or head?"
Question: In order to maximize his expected return, should Paul pick tail, head, or it doesn't matter?

 A: The restriction is that the coin is fair. There does not seem to be a restriction that the flipping is fair.
With practice I used to be able to get the same result from coin flips - mostly heads or mostly tails. One needs to start with the coin in the smae head up or tail up position, but it is not difficult to make the flipping unfair. 
A: If, as you said, it's an unbiased and fair coin, then it has an equal chance of coming up heads as tails on the 100th flip, by definition, because that's what an unbiased fair coin is: it is a coin that has an equal chance to come up heads or tails on any flip. 
On the other hand, if you see a coin come up tails 99 times in a row you had best revisit your assumption that it is an unbiased fair coin. It would be foolish to bet on it coming up heads after that.
Addendum: Here's the reference you asked for: Gambler's fallacy
A: To elaborate on Mark's answer. Let's say you don't necessarily believe that the coin is unbiased, but instead believe it has probability $p$ of coming up heads.
You degree of belief in the value of $p$ can be specified by providing the parameters $a$ and $b$ of a beta distribution. Intuitively, they correspond to the number of times you've seen heads and tails come up already. A typical 'default belief' is given by $a=b=1$, which essentially expresses maximum ignorance - you believe that the true value of $p$ is distributed uniformly between 0 and 1.
The benefit of doing this is that there is a simple method of updating your belief about the distribution of $p$ whenever you see a new coin toss - you simply increment $a$ by 1 whenever you see a head, and increment $b$ by one whenever you see a tail.
The mean of the distribution is simply $a/(a+b)$ and its variance is $ab/[(a+b)^2(a+b+1)]$. If you see 99 tails in a row, then your new values of $a$ and $b$ are $a=1$ and $b=100$, giving an expected value for $p$ of
$$E(p) \approx 0.01$$
$$\mathrm{StDev}(p) \approx 0.01 $$
So with a very high degree of certainty, you believe that the true value of $p$ is 0.01, and certainly isn't much more than 0.04. You would therefore be very naive to bet on heads coming up on the next toss (unless you were given very favorable odds!)
A: Due to the law of large numbers, the proportion of heads to tails seen with an unbiased coin will converge to $\frac12$ as the number of flips tends towards infinity. If you were to make the bet before 99 coins have been tossed, then it would be apparent that the probability of seeing all tails is $(\frac12)^{99}$. If you have already achieved this then the probability of seeing a subsequent tail is $\frac12^{100}-\frac12^{99}$, it is not about the coin having a 'memory' but rather the probability of observing 99 events involving the coin with identical values. It would be the same if you decided to flip the coin 100,000 times and chose to observe it 0.1% of the time. Even if the average proportion of tails to heads of the 100,000 were 0.5, the probability of observing 99 consecutive tails would still be $(\frac12)^{100}-(\frac12)^{99}$. The same would also be true if you selected a new coin every time.
