JudyH said:
Rudy(CA) said:
I'll show you mine if you show me yours.
I don't have one. 8)
Rudy(CA) said:
No what I am saying is that random chance would produce on average a probability of success of 6.25% over many repeated trials of 4 attempts each.
How many repeated LRL trials would it take to reach the 50% mark, Dr. Rudy?
How many repeated coin toss trials would it require to reach that same mark?
I don't understand your question. The 50% mark of what?
Just to be clear, in the example I gave I was just trying to show how the compound probabilities would
build up by showing a string of 4 tests. Later, when explaining to Hung, I said that the result is really
a binomial distribution. So, if we were doing the test for real, I wouldn't recommend doing sequences
of 4 tests, instead it would be smarter to do "n" tests, where "n" is picked to drive the probability of success
"P" due to random chance down to an acceptable level.
Doing it this way, would then allow the simple calculation of the mean and standard deviation of the binomial
distribution. The mean is μ= n * P
and the standard deviation is σ = sqrt[ n * P * ( 1 - P ) ]
Since in our test, there is a 50 50 of success for each trial so the standard deviation
becomes σ = sqrt[0.25n] = 0.5*sqrt(n)
If we set n to 16 (the number of tests), then the
Average expected result is μ=16*0.5 = 8 successful detections
with a standard deviation of σ = 2
And the probability of detecting the gold all 16 times without fail is 0.5^16 or 0.0015258789063%
Assuming the binomial distribution with 16 trials is close enough to the normal (gaussian) distribution,
then 99.6% of all attempts at taking the test sequence should fall within ± 3σ of the mean, so 99.6%
of the tests should fall in the interval of μ ± 3σ
In other words, 99.6% of the test sequence results will yield from 2 to 14 successes in 16 trials.
Conversely, only 0.4% of the test sequence takers will have results that lie on the tail of the distribution.
0.2% in the 0 & 1 success interval and 0.2% in the 15 & 16 success interval.
So, the odds of successfully detecting the gold target 16 times in a row is 0.0015258789063%
and the chance of it happening randomly is less than 0.2%