On the correct way to estimate a proportion given that x events were observed in n trials.
We intend to discuss the method to estimate the proportion through p= (r+1)/(n+1) as shown in the paper: A Note on the Calculation of Empirical P Values from Monte Carlo ... B. V. North, D. Curtis and P. C. Shame Am. J. Hum. Genet 71-439-441, 2002
The reason is that we expect that the next occurrence is a success, a forecast that is always hazardous mainly when r is very large compared with x. In terms of expectation, E(r), one have _p´= (r+E(r))/(n+1) = (r+r/n)/(n+1) = r* (1+1/n)/(n+1) = r/n
Therefore, following this criterion, p=r/n needs not at all to be *corrected* and is independent of how many trials are performed. The current formula is neatly an estimate by excess of the future proportion. . .
An humorist note I do not know at what extent the AA. have enjoyed to prove numerically trough simulations (!) that (see *) ____________ r/n <= (r+1)/(n+1) A High School pupil did do : Since (n+1)/n <= (r+1)/r or more evident (1+1/n) <= (1+1/r), 1/n<=1/r, n>=r and the *problem* is solved . . . * Apparently with *inverted* results : 12´103 against 10´106.
More profitable they were involved in a golf game.