You have rediscovered the Poisson approximation to the Binomial 
distribution, or, alternatively, an approximation of compounded 
interest by continuously compounded interest.

If p = probability of death, N = number of trials, then the binomial 
probability of survival is the term

S = (1 - p)^N

For the Poisson distribution, the probability of no adverse events is

S' = exp(- N p)

The approximation is

S ~ S' or (1-p)^N ~ exp(- N p)

which becomes better as p becomes smaller. Adequate results are 
obtained with p = 0.05 and good results with p = 0.01. (The error, 
which can be found by computing the Maclaurin series expansion of the 
difference, has a leading term of O(p^2)).

Because exp( - N p) = 1/2^(N p / ln 2), your discovered factor of 
"70" would be better "69.314718" = 100 ln 2.

At 07:28 AM 7/16/2009, Russell McMahon wrote:
>It occurred to me today that a rearrangement of a classic numerical "trick"
>usually used for mental calculation of interest rate returns could be put to
>more honourable , or fun, purposes.
>
>Simplistically, if an object is subject to a series of N trials with the
>probability of destruction at each trial being x  (0 < x < 1) then the
>probability of surviving the N trials is
>
>               Survival = (1-X/100)^N
>
>This is easily calculated with a calculator but not so easy to work in your
>head for larger values of N.
>
>However, a surprisingly good result for a good range of values of N and x is
>given by
>
>
>             Survival = 1 / 2 ^ (x.N / k )
>
>k varies for best accuracy depending on the values of x and N.
>A good figure for typical x and N values is k=70, so
>
>              Survival = 1 / 2 ^ (x.N / 70)
>
>
>As all competent embedded development engineers can calculate powers of 2
>mentally, even for fractional indices, and as most other people liable to be
>interested can multiply by 2 repeatedly, this becomes almost useful.
>
>eg
>
>G for George is a Lancaster bomber, now resident in the Canberra war
>memorial museum in Australia.
>G for George flew 90 combat missions over Europe. (Of the survivors only S
>for Sally at 137 odd flew more).
>
>If the average mission loss rate was 5% (about right on average, highly
>variable in practice) what are the chances of G for George having survived?
>
>90 missions x 5 percent / 70 ~= 6.4
>2^6.4 ~= 80 (actually 84.4)
>1/80 = 1.25%
>
>Standard calculation gives (1-5/100)^90 == 1%.
>
>Not too bad for a mental calculation.
>Using the correct 84.4 gives 1.2%
>
>Use with care and some [pretesting if of value to you. Errors can blow way
>out for certain areas of input space. Adjusting k for your area of interest
>helps.
>
>The "method" uses the classic "rule of 72" - here modified to 'rule of 70'
>:-).
>
>If Scott Dattalo sees this he will no doubt provide a far superior method in
>moments.
>
>Working out how and why the rule of 72 works should give people many an hour
>of mindless diversion,.
>
>
>
>
>               Russell
>--
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================================================================
Robert A. LaBudde, PhD, PAS, Dpl. ACAFS  e-mail: ral@lcfltd.com
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