Mlodinow. The Drunkard's Walk. An example from the book. This excerpt is from Leonard Mlodinow's book The Drunkard's Walk:

And
  although Fortune is fair in potentialities, she is not fair in outcomes.
  That means that if each of 10 Hollywood executives tosses 10 coins,
  although each has an equal chance of being the winner or the loser,
  in the end there will be winners and losers. In this example, the
  chances are 2 out of 3 that at least 1 of the executives will score 8 or
  more heads or tails.

How can it be proved with math?
 A: Call an executive balanced if she got between $3$ and $7$ heads and call $b$ the chances that an executive is balanced. One asks for the chances $p$ that at least one executive is not balanced, that is, $p=1-q$ where $q$  are the chances that every executive is balanced, hence $q=b^{10}$. 
There are $1+10+45+45+10+1=112$ ways to be unbalanced on a total of $2^{10}=1024$ ways hence $b=1-112/1024=57/64$ and $p=1-(57/64)^{10}=0.6860$.
A: First let us look at the probability of one executive getting $8$ or more heads/tails when tossing a coin $10$ times. $$ P(\text{# of heads}\geq 8 \text{ or # of tails} \geq 8) = 2 \times \dfrac1{2^{10}} \left(\dbinom{10}8 + \dbinom{10}9 +\dbinom{10}{10} \right) = \dfrac7{64}$$
Hence, the probability that one executives will score $7$ or less heads/tails$ = 1 - \dfrac7{64} = \dfrac{57}{64}$.
Hence, the probability that at-least one of the executives will score $8$ or more heads/tails is nothing but $\left(1 - \text{ the probability that all the executives will score $7$ or less heads or tails} \right)$.
The probability that all executives will score $7$ or less heads or tails is $ \left(\dfrac{57}{64} \right)^{10}$. Hence, the desired probability is $$1 - \left(\dfrac{57}{64} \right)^{10} = 0.685986140417819036628477302741657695150934159755706787109375 \approx \dfrac23$$
A: First find the probability $p$ that any person tossing 10 coins will score between $3$ and $7$  heads (the number of heads follows a Binomial distribution).
Since they toss the coins independently, the probability that all of them score between $3$ and $7$ is $p^{10}$.
The event that at least one scores $8$ or more heads, or $2$ or fewer heads (so $8$ or more tails) happens then with probability $1 - p^{10}$.
A: Assume that each coin toss is independent, and that the coin is fair. 
Tossing 10 coins, the total number of heads $X$ follows a binomial distribution $\mathrm{Bin}\left(10,\frac{1}{2}\right)$. The probability that one executive scores 8 or more heads or tails equals:
$$
    \mathbb{P}( X \leqslant 2 \lor X \geqslant 8) = 1 - \mathbb{P}( 3 \leqslant X \leqslant 7) = 1 - \sum_{k=3}^7 \frac{1}{2^{10}} \binom{10}{k} = \frac{7}{64} = 1- p
$$
The probability that none of the executives obtains score 8 or more of head or tails equals:
$$
      1 - p^{10} = 1 - \left(\frac{57}{64}\right)^{10} \approx 0.685986
$$
