The probability that all coins are in the same phase 
(Feller Volume 1, P.316) Let $r \ge 2$ be an arbitrary integer and consider a sequence of simultaneous independent tosses of $r$ coins. Let $\zeta$ stand for the recurrent event that all $r$ coins are in the same phase (that is, the accumulated numbers of heads are the same for all $r$ coins). The probability that this occurs at the $n$th trial is 
  $$u_n = 2^{-rn}\left( {n\choose 0 }^r + {n\choose 1 }^r + \cdot\cdot\cdot + {n\choose n }^r\right).$$
  On the right we recognize the terms of the binomial distribution with $p= 1/2$, .... 

I know that when $r=1$, $u_n = {n \choose n/2} 2^{-n}$ for $n$ even (when $n$ is odd, $u_n =0$). But, I don't understand how does the author derive the probability in the display above. Could you elaborate this? 
Thanks in advance. 
 A: The first term is if all the flips have $0$ heads; the second term is if all the flips have just $1$ head, ....
A: $u_n$ is the probability that at $n$th trial all the coins were tossed same number of times.

Deriving $u_n$: At the $n$th trial if all are in same phase, then they all may have $1$ or $2$ or $3$ or ... or $n$ heads. For some coin probability of $k$ heads is $p_k=\binom{n}{k} 2^{-n}$. So, probability that all the $r$ coins will have $k$ heads is $(p_k)^r ={\binom{n}{k}}^r2^{-nr}$
So, the total probability to be in same phase is the sum of $p_k$ for all $k=0,1,2,.....n$.
That means  $u_n=2^{-rn}\left({\binom{n}{0}}^r+{\binom{n}{1}}^r+.....+{\binom{n}{n}}^r \right)$

But this contains the case when all the coins were flipped same number of times at $n-1$, $n-2$, ....., $1$ th trial. More specifically $u_n$ is the probability that the coins were in same phase before $n+1$ th trial.
So it is required to subtract the probability that those were in same phase before $n$th trial to get the probability that those are at the same phase at exactly $n$th trial.
If $u_{n-1}$ is the probability that those were in same phase before $n$th trial , then the probability that after this event , they will also be at same phase after one more flip, is $2×(\frac{1}{2})^r{u_{n-1}}$. Because at $n$th uniphase trial all the coins may become head or tail.
So, $P_n=u_n- 2^{-rn}(2u_{n-1})$.
