Fortune teller

[Kaljamaha]

secretperson said:
Interesting! Thanks for the responses!

It's been wayyyy too long since high school statistics... How would this look statistically?

It's binomial distribution. Look up the correct row from Pascal's triangle. So, for example, if you have three fortune tellers (5) in a province, the chance that at least one of them prevents an event is:

0.05^3 + 3 * 0.05^2 * 0.95 + 3 * 0.05 * 0.95^2 = 14.2625%

Or, if you prefer the easy way, its:

1 - (1 - 0.05)^3


K.

[llamabeast]

its better than adding.
2 5% chances > 1 10% chance, as you have the small possibility of cancelling two events.

This isn't really correct. The chance of blocking at least one event is less than 10% for two 5% fortune tellers. Of course there is a chance of blocking two events, but most often that will be irrelevant as there won't be two events trying to happen there anyway.

[Agrajag]

If an event attempts to get past fortune tellers, then 20 fortune tellers with 5% will have ~64% of stopping that event.
The formula goes: 1-(1-P)^n, P is the probability of stopping an event (0.05 in this case), and n is the number of fortune tellers (20 in this case).

Explanation: The chance for a fortune teller to prevent the event is P, so the chance of the event getting past that fortune teller is (1-P), so the chance for getting past all of the fortune tellers is (1-P)^n, so if we want to see what the chance of the event not getting past all of the fortune tellers we get 1-(1-P)^n.

(Feel free to correct me if I'm mistaken )

[Sombre]

You're mistaken.

The goat is actually behind door number 4, the door you came in through.