Fortune teller

[secretperson]

Simple question really- Do the effects of multiple fortune tellers stack?

[chrispedersen]

yes.

[Wick]

It depends what you mean by stack. The chances aren't added but each one gets a chance of stopping each bad event.

[Edi]

Yes, in the sense that each fortune teller checks separately for prevention. No, as in their abilities are not additive.

[secretperson]

Interesting! Thanks for the responses!

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

[chrispedersen]

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

Well, I meant "is better" by the > sign.

I grant that the math has been elegantly presented and is right. However, I wonder if there is a way to test this.
I raise the question for the following reasons:

If you mod 5 10% chances of a dominions magic random - they get summed into 1 50% chance. (Try it). I wonder if that same approach hasn't been done to fortune tellers.

The second reason I wonder is because around 20 fortune tellers, I don't seem to ever get bad random events.
If I get industrious.. I'll run a test on that.

[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.