
Edited on Wed Apr0704 01:52 AM by TruthIsAll
... virtually 100%, depending on the number of polls. Lets calculate the probability assuming a total universe of anywhere from 1 to 10 polls.
Lets calculate the probability that Kerry would win the election, based on final polling results: 53% Kerry, 47% Bush.
This is a useful calculation because we want to determine the probability that Bush stole the election, if he had 47% final polling numbers, so that each of n polls had him losing to Kerry by 6%.
We assume the following: 1 the winner is the one who receives the most popular votes (i.e. over 50%). There is no electoral college. 2) there are no 3rd party candidates. 3) the polls are independent of each other. 4) Kerry is ahead by 5347 in each of n polls, where n = 1,2,5,7,10. 5) The polls are given the night before the election. 6) the Margin of Error (MOE) for each poll is = +/3%. 7) the polls are all fair and unbiased.
This means that for any single poll, there is a 95% probability that Kerry would receive 5056% and Bush 4450%.
In addition, There is a .05 probability the result will fall outside the MOE, evenly split with a .025 probability that Kerry's election percentage will be over 56% and a corresponding .025 probability Bush's percentage will be over 50%.
Thus the total probability that Kerry would receive over 50%, given that he is ahead by 5347 in a given poll is: p(Kerry)= 0.975
The probability Bush would receive over 50%, given that he has 47% in a given poll is: p(Bush)= 0.025
Here are the probabilities that Bush would win fairly (no fraud) where n = total number of final polls:
n Probability (that Bush won fairly) 1 0.025 or 1 out of 40 (unlikely) 2 0.000625 or 1 out of 1600 (extemely unlikely) 5 9.76563E09 or 1 out of 102,400,000 (1 out of 102 million) 7 6.10352E12 (forget it) 10 9.53674E17 (really forget it)
We will know the fix is in if the 10 final poll numbers have the race at 5149%, for we all know that Bush's numbers are heading to the basement. Kerry should receive a minimum of 53%.
