If the sampling is large enought, 1/6 of the answers are forced "yes", 1/6 are forces "no", and only 4/6 of the answers are reals.
So to get the actual frecuency of "yes", you should remove the 1/6 of the forced "yes", but the total size of the sampling is only 4/6 of the original, so the formula is
(If Measured_YES+Measured_NO=1=100%, then Actual_YES+Actual_NO=1=100%)
More technical details:
You have to be carefull, because the number of forced "yes" and "no" will not be exactly 1/6 of the total sampling. For example, supouse that you ask 60 persons if they are "aliens" using this method.
If you get 11 "yes" answers, it doen'n mean that (11/60-1/60) * 6/4 = 2.5% of the people are "aliens".
If you get 9 "yes" answers, it doen'n mean that (9/60-1/60) * 6/4 = -2.5% of the people are "aliens"!
And each of this sampling results happens ~13% of the times that you make the pool.
If the sampling is large enought, 1/6 of the answers are forced "yes", 1/6 are forces "no", and only 4/6 of the answers are reals.
So to get the actual frecuency of "yes", you should remove the 1/6 of the forced "yes", but the total size of the sampling is only 4/6 of the original, so the formula is
(If Measured_YES+Measured_NO=1=100%, then Actual_YES+Actual_NO=1=100%)More technical details:
You have to be carefull, because the number of forced "yes" and "no" will not be exactly 1/6 of the total sampling. For example, supouse that you ask 60 persons if they are "aliens" using this method.
If you get 11 "yes" answers, it doen'n mean that (11/60-1/60) * 6/4 = 2.5% of the people are "aliens".
If you get 9 "yes" answers, it doen'n mean that (9/60-1/60) * 6/4 = -2.5% of the people are "aliens"!
And each of this sampling results happens ~13% of the times that you make the pool.