simple math problem

Ideas for expansions and improvements to Defcon

Moderator: Defcon moderators

User avatar
Posts: 1288
Joined: Wed Sep 19, 2007 3:02 pm
Location: Germany

simple math problem

Postby Nightwatch » Thu Sep 23, 2010 5:38 pm

iam looking for optimzed naval combat strategies atm
unfortunately i realized that my ten years of math in school ages ago are not enough and iam rusty as hell

I need to calculate the probability of a event (not) to happen. I think its acutally very easy but i never learned it and so i have no real idea how to do it.
Could someone point me in the right direction please?

For event A to happen, 1 out of z attemps needs to be successfull.
Statistical speaking, 1 in 3 attemps is successfull.
How likely is it that event A will not happen, if you try X z times?

if i draw up every possible case on paper and add everything up, i get a probability of event A occurring of
0,33 for z = 1 (obviously since one in three attempts will result in a kill )
0,56 for z = 2
0,7 for z = 3
0,81 for z = 4

Or in Defconian:
If a bomber is being shot at (with a fighter) four times, the bomber will be killed in 81 of 100 attemps.
The probability of surviving is below 20%.
Is that right?
And if yes, how does it look if you shoot at a bomber 12 times?
or to be precise, how does it look like if 12 fighters shoot at the same bombers once simultaneously?

How can i calculate that one?

*cries for help*
User avatar
Posts: 2750
Joined: Thu Jun 28, 2007 5:52 pm
Location: Russia, St. Petersburg

Postby rus|Mike » Thu Sep 23, 2010 5:49 pm

The probability of an even happening in one attempt is x (in the format 1=surely will happen, 0=surely will not happen).

Then the probability of it NOT HAPPENING is 1-x.

The probability of it NOT HAPPENING in y attempts is: (1-x)^y.

Example: I'm throwing a coin 5 times. Let's calculate the probability of side "A" NOT happening 5 times in a row:


Probability= (1-0,5)^5 = 1 / (2^5) = 1/32.

Same example with a fighter-bomber case: x=0,(6) (survives 2 out of 3 attempts), y=4 (4 attempts)
then the probablility of it surviving 4 shots is (2/3)^4=16/81=0,1975

For 12 attempts, it looks bad for a bomber :P

User avatar
Posts: 1288
Joined: Wed Sep 19, 2007 3:02 pm
Location: Germany

Postby Nightwatch » Thu Sep 23, 2010 6:19 pm

ahhhh, so easy, just like i thought :?

thanks so much!

Return to “Think Tank”

Who is online

Users browsing this forum: No registered users and 1 guest