Conquer Club

Umm I'm doing a math project on RISK of course, just measuring probability and such. THough I stink at math, and Voila! that's why I'm asking for help right now

How would you go about calculating the odds of an attack? For example, if you wanted to find the probability of the attacker winning, say 3 armies vs 2 defenders, what mathematical process would you use?

I've found web site with calculators and stuff, but I need to show actual work so if anyone could bother explaining how this guy got this...

Attacker: three dice; Defender: two dice:

Attacker wins both: 2890 out of 7776 (37.17 %)
Defender wins both: 2275 out of 7776 (29.26 %)
Both win one: 2611 out of 7776 (33.58 %)

http://www.plainsboro.com/~lemke/risk/#dice

Any help would be greatly appreciated...thanks in advance!

just a couple of things you should make sure to differentiate..

Are you attacking with 1 - 2 - 3 armies? This will change the number of dices you can throw.

Is the defender using 1 or 2 armies (1 dice or 2)

Unless you only take the case where attacker has 3 dices and defender 2.

Once this is done... you could draw a tree (huge...) of all the possible outcomes.

Dice one: 1/6 chance to roll a 1... 1/6 to rolla 6
Then, from each number, possible, dice 2 rolls 1 to 6, with 1/6 possibiity.


You do this until dice 5. This will give 6 x 6 x 6 x 6 x 6 outcome = in other words... 7,776 possibilities to measure. At the end of each branch (scenario, you assign a value (2 wins att, 2 win def, 1-1) in the end you sum the results for each type and divide by the total possibilities

Doing the case with 2 dices attackers and 1 dice defender would give ony 216 results to calculate. And 1 dice attacker and 1 dice defender gives only 36 possible results.

Anyhow... I know there are formulas to help, but due to the size of possibilities, calculations still takes lots of time

Me, being just OK at Math, shall valaintly attempt to help! Hurrah!

Okay...let's see...the basics. Attacker rolls 2 dice, Defender rolls 2 dice, Highest two are compared. There are 1296 possible combinations, I believe, and out of them, the attacker-favourable would be...something. Sorry, it's midnight, and I'm tired...I'll figure it out in the morning. >.<

EDIT: Whoops, didn't see you there, Maverikc. I do believe that he's thinking of how ConquerClub plays it, though...it's set to the most dice possible. If he isn't...well...that makes it a whole lot more complicated. Ouch, my brain.

Oh yeah.. bye the way... if you want to talk about random number generator... Many studies, at masters and PhD level in advanced mathematics are devoted to provng right or wrong effective randomness of random number generators, even among the most popular such as the one in Excel, or in calculators.

ugh this is what i have so far...

okay i started off small...the probability of 1 defender surviving a 2 army attack

Prob of Defender* Prob of Attacker 1 losing* Prob of Attacker 2 losing
Odds of rolling a 6--- 1/6 * 6/6 * 6/6 = 36/216
Odds of rolling a 5--- 1/6 * 5/6 * 5/6 = 25/216
Odds of rolling a 4--- 1/6 * 4/6 * 4/6 = 16/216
Odds of rolling a 3--- 1/6 * 3/6 * 3/6 = + 9/216
Odds of rolling a 2--- 1/6 * 2/6 * 2/6 = 4/216
Odds of rolling a 1--- 1/6 * 1/6 * 1/6 = 1/216

91/216


which is the correct answer.

Now to find the probablity of the attacker winning in that situation, simply subtract 91 from 216 to get a probablity of 125/216, which is also correct.

RESURRECTED

This was the oldest thread I could find lol. Yay math!

-Sully

Please don't bump threads without adding anything new to the conversation.

Locked, Ljex