Fortune teller
 

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

Re: Fortune teller
 

yes.

Re: Fortune teller
 

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

Re: Fortune teller
 

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

Re: Fortune teller
 

Interesting! Thanks for the responses!

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

Re: Fortune teller
 

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.

Re: Fortune teller
 

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.

Re: Fortune teller
 

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.

Re: Fortune teller
 

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 http://forum.shrapnelgames.com/image...ies/tongue.gif)

Re: Fortune teller
 

You're mistaken.

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