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.