Why is that? The probability of a random event happening is independent of previous events. It is still a 50% chance of flipping heads no matter the result of the previous 25 flips.
Another math opportunity.
Emperical Probability Formula:
P(E) = # of times event E occurs / total # of occurences
Say someone gets injured pushing a car to the line once every 1000 runs
P(E) = 1 / 1000 = .001 or .1%
and
P(Enot) = 1-.001 = .999 or 99.9% (probably of not getting injured on a run)
Each run as you mentioned is independent. Putting the probabilities of two independent events together we get the following:
P(E AND F) = Probability of E and F happening = P(E) * P(F)
Lets say event F = event E
then P(E AND E) = Probability of E happening twice = P(E)*P(E) or P(E)^2
Or generically P(E) happening N times (keeping in mind each E is independent) = P(E)^N where N equals number of times E happens
For our example P(Enot) = .999 and P(Enot) happening N times is .999^N
(a) N=1 P(Enot)^N = .999^1 = .999 = 99.9%
(b) N=10 P(Enot)^N = .999^10 = .990 = 99.0%
(c) N=100 P(Enot)^N = .999^100 = .905 = 90.5%
(d) N=1000 P(Enot)^N = .999^1000 = .367 = 36.7%
(a) Probability of not getting injured on each run is 99.9%
(b) Probability of not getting injured in any of 10 runs is 99.0%
(c) Probability of not getting injured in any of 100 runs is 90.5%
(d) Probability of not getting injured in any of 1000 runs is 36.7%
This just means that the more times you push a car to the line, even if the probability is remote, your pressing your luck further as time goes on. (Note that it is not a sure thing that you will get injured in 1 of 1000 runs even if the odds are 1 in 1000, its only 63% that you would)
I think my math is correct - I didn't fully understand my statistics class...
One bit of common sence, the probability of getting injured while pushing the car to the line if you do not push the car to the line is 0 no matter how many runs you do.
