|
Edited on Mon Sep-18-06 08:10 PM by ryanus
Suppose we compare this to roulette. Suppose that instead of 36 black or red numbers with one or two green zeros, we have 100 numbers. 99 numbers are black, representing non-9/11 DVD's. 1 number is red, representing the a single 9/11 DVD. The chance of any particular number coming up is 1/100 (given the assumption that a number will be chosen and that it is random. Hold on to this).
Although the chance of any particular number coming up is the same chance as an other number, the chance of a particular type of number coming up is not equal. The probability of a non-9/11 DVD coming up is 99 out of 100. The chance of the 9/11 DVD coming up is 1/100. So where do you place your bet?
Let's say you bet on some non-9/11 DVD, but the 9/11 DVD comes up. That doesn't imply any causation right? Of course it doesn't.
But let's pretend that there are 1 million DVD's. Forget the first roulette spin. It never happened. 1 number represents a 9/11 DVD. You bet on a non-9/11 DVD and you have a 999,999 / 1,000,000 that the number that comes up will not be the 9/11 DVD. But let's just say the 9/11 DVD comes up. Just coincidence right? Possibly. The more non-DVD's you buy, the less likely it will be that the 9/11 DVD comes up.
At what number do you start to question the randomness of the game?
Again forget that we did any previous spins. 100 numbers, 80 non-9/11 DVD's, 20 9/11 DVD's. You bet on non-9/11 DVD's and you have a 8/10 of winning, 2/10 chance of losing. 9/11 DVD comes up. You lose, just coincidence. You bet again the same way, you lose the same way. Ok so at the outset, your chances of losing twice were 1/5 times 1/5 which is 1/25. Unlikely, but possible. You bet the same way and again you lose. 1/5 x 1/5 x 1/5 = 1/125. Hmm, pretty unlikely, but you have bad luck I guess. One more bet, you lose, the odds of this happening were 1/625, or 0.16%. At what point do you question the randomness of the game? After only three times, the chances of the 9/11 DVD numbers coming up was 0.8%!
Remember the assumption that any number would come up at all. I am not guaranteed that any of my DVD packages will be opened at all. So let's say I have a 1 in 20 chance that I will have a DVD package opened. If I bought 99 non-9/11 DVD's and one 9/11 DVD, the chances that any particular DVD package will be opened is 1/20 times 1/100 = 1/2000. Pretty low odds for any paricular DVD. The chances that a non-9/11 DVD is opened is 1/20 * 99/100 = 99/2000, but the chances that a 9/11 DVD is opened is still 1/2000, or 0.5%. At what point do you question the randomness?
Now what if we did this game twice, the chances at the outset that only the 9/11 DVD's would be opened would be 1/2000 times 1/2000 which is 1/4000000 or 0.000025%. So if two times I ordered 100 DVD's and I had a 1 in 20 chance that one of them would be opened, and each time only 1 of the DVD's was a 9/11 DVD, the chances that only those two 9/11 DVD's is 1 in 4,000,000. Four million. Say another person has the same result. At what point do you question the randomness?
So the questions I should have gotten where: how many DVD's have your ordered? How many of those DVD's were 9/11 DVD's? How many DVD's in total were opened? How many of the opened DVD's were 9/11 DVD's?
If only the chances of 9/11 official story could be asked the same way...
|