How many hands to reduce the probability of no royal flush in n hands to less than 1/e? I get close but can't figure out how they get the answer to this question -- Introduction to Probability, Grinstead, Snell, Chapter 5.1, exercise 17:

The probability of a royal flush in a poker hand is p = 1/649,740. How large
  must n be to render the probability of having no royal flush in n hands smaller
  than 1/e?

My answer is: since this is for one success, use a geometric distribution instead of a negative binomial.


*

*success p(flush) = 1/649740 = .0000016

*failure p(no flush) = 1 - 1/649740 = .9999984

*1/e = 1 / 2.71828  = .36788


For geometric distribution, I set them equal to find the minimum n:


*

*q^n = 1/e

*(.9999984)^n = .36788

*log _.9999984 (.36788) = n   (base is .9999984, the failure probability)

*log .36788 / log .9999984 = n   (found this technique for using base 10 for both)

*n = 624998.55


Therefore, my answer is 624999 (just over the equal n by rounding up). However, the answer in the text book says 649741. Using my above process, I get that too -- if I change 1/e from .36788 to .3536. Is my 1/e incorrect? Or have I used the wrong distribution and formula? Thanks in advance for any help.
 A: This is just an answer due to rounding.  You correctly solve the equation
$$
q^n = 1/e
$$
for $n$ to get:
$$\begin{align}
n = \frac{\log(1/e)}{\log(q)} = \frac{-1}{\log(q)} \approx 649739.5715,
\end{align}$$
according to my calculator.  Notice that there is no need to explicitly solve for $e$ since $\log(1/e) = -1$.
A: It's a rounding error.  You can get closer results as follows:
$$
\begin{align}
   q^n     &= e^{-1}
\\ n \ln q &= -1
\\ n       &= 1 / (-\ln q)
\end{align}
$$
Note that $-\ln(1 - x)$ is approximately $x$ for small x.  Just plugging in - 1/649740 gives a result of 649740.  The actual series is $x + x^2 / 2 + x^3/3 ...$, which means that the $-\ln q > 1/649740$, but only very slightly, which means that $-1/\ln q < 649740$, so 649740 is sufficient, rather than the 649741 that the book claims.
A: When $p$ is small, the probability of at least one success in $n$ trials is approximately $1-1/\mathrm e$ when $n = 1/p$.  Your $p = 1/649740$ should be plenty small enough for this approximation to be very good.
To be exact, the probability of no success in $n$ independent trials, with probability $p$ of success in each, is $(1-p)^n = \exp(n \log(1-p))$.  When $n$ is large and/or $p$ is small, this is quite well approximated by $\exp(-np)$.  Equivalently, writing $k = np$ for the expected number of successes in the $n$ trials, the probability of no success is approximately $\exp(-k)$ regardless of $n$ (as long as $n$ isn't very small).
