Does the expected number of rounds when tossing N biased coins until each lands heads at least once scale with logarithm of N? This game has been considered before here and here and the expected number I am interested in has been found, but the expression is recursive.
Let there be $N$ biased coins with a probability $p$ to come up heads and probability $q=1-p$ to come up tails. Initially all coins are tails. Each round every coin still tails is flipped. The game is over when all coins are heads.
The expression in the linked answers, which I am sure is correct, is
$$ E_N = \dfrac{1+\sum\limits_{k=1}^{N-1}\binom{N}{k}p^{N-k}q^{k}E_k}{1 - q^N} $$
with $ E_1 = \frac{1}{p} $. From this form it is hard to see how it scales with $N$.
Can it be proven that $E_N$ grows with $\log(N)$? If not logarithmic, what is the scaling?
I suspect the scaling is logarithmic because each round we can expect the number of coins still tails to shrink by a factor of $q$, that's where the recursion comes from. Notice that I am not asking for a closed form solution (although that would be very nice), I'm only interested in the scaling.
 A: You can interpret the waiting time as the maximum of n independent geometric waiting times, i.e.
$$Y=\max(X_1,\ldots,X_n)$$ where $X_i\sim Geo(p).$
Then, the limit laws for extreme values will show, that $Y$ converges to a Gumbel distribution, when scaled with log(n).
I asked a similar question here, but with three dice waiting for the last die to show six for the first time. The formula there holds for the coin example, too. It is explicit without recursion.
A: Here is an approximation that gives a fairly easy way to see the relation to log growth:
Let $y_0=N$ be the initial number of coins showing tails. Let $y_n$ be the number of coins still showing tails after $n$ rounds of flipping.
We expect $y_n\approx N\cdot q^n$. With this approximation we will never get $y_n$ to be $0$, but we will get arbitrarily close. It seems reasonable to say that in this model, once $y_n$ falls below $0.5$ (less than half a tail) that we are done. [Alternatively we could get $y_n$ to be $1$ (or just below) and then we are left with a geometric random variable for getting the last coin to be heads.]
Accepting this model, we can say that the number of rounds of flipping ($n$) until we expect all heads should be approximated by the solution of
$$0.5=N\cdot q^n$$
That is: $n=\frac{\ln\frac{.5}{N}}{\ln q}=\frac{-\ln(2N)}{\ln q}$.
[Or, using the alternative model suggested above, one gets that the expected number of flipping rounds is $n=\frac1p+\frac{-\ln N}{\ln q}$.]
Remark:
I find it interesting to note that this closely follows the exponential decay model for radioactive material, as often seen in first-year calculus. In fact the radioactive-decay problem is more like this dice problem than it is like the continuous deterministic approximation given in calculus.
