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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.

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    Nice design, I guess, but too clever. On physical calculators, $e^x$ is permanently written under the key.2011-07-28

3 Answers 3

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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 $\log(1/e) = -1$.

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    You're quite welcome. That just comes from $\log_b(b) = 1$. Indeed, when a mathematician writes $\log$ (s)he means the natural log $\ln = \log_e$. Since all logs are the same (up to constants), we just stick with the notation $\log$.2011-07-27
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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.

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    thanks -- I get the same answer with your solution if I hit "ln" on my calculator, not "log". I was using "log" before, and did not know e inverse = -1.2011-07-27
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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).