Results 1 to 1 of 1

Thread: Dice odds. Risk style.

  1. #1
    Nov 2007

    Dice odds. Risk style.

    Ok, so you might know the game of Risk. If you don't here is a summary of the relevant rules.
    The attacking player may, if attacking with more than 3 armies, throw 3 dice to descide the results of a battle. The defender may, if the defending territory holds 2 armies, defend with 2 dice. The higher 2 of the Attacker's dice are matched against the 2 of the defender.
    The result of the rolls descide the amount of downsizing in the military force of the battleing territories.

    For example, these rolls, will result as follows.
    6, 6, 6 vs 6, 6 will result in a win of 2 armies as the defence has the favor in any tie.
    If the highest of the attacker is 3 and the highest of the defender 3 the attacker will lose at least 1 army.

    So only if the highest of the attacker and it's 2nd are both higher than the matching defender's dice. (5 3 1 vs 4, 2) (the 4 doesnt beat the 3 because it's matched vs the other highest, the 5)

    So there you have it, the relevant rules of Risk. Now for the actual problem.

    I want to calculate the odds of the result of a single 3 vs 2 battle of the dice in Risk. 5 dice, it's n/7776 chance for the attacker making the defender lose 2 armies. p/7776 for a tie, (each makes the other lose 1 army, for example 6 4 1 vs 6 3) and an q/7776 chance the defender makes the attacker lose 2 armies.

    I already found a promising way of getting my answer.
    I wanted to use the following, I actually NEED to use the following, because if I can find out the correct use of the figures I figured out I can use it in a decision algorhythm in case the defender gets to choose wether to use 1 or 2 dice.

    So here it is,
    I found out that the odds for the higest of a roll by the defender being 6 is 11/36
    5: 9/36
    4: 7/36
    3: 5/36
    2: 3/36
    1: 1/36

    And exactly the opposite for the odds of the lower of a defenders roll being:
    6: 1/36
    5: 3/36
    4: 5/36
    3: 7/36
    2: 9/36
    1: 11/36

    The Highest for the attacker has it's distribution as follows.
    6: 91/216
    5: 61/216
    4: 37/216
    3: 19/216
    2: 7/216
    1: 1/216

    The second highest for the attacker has the following odds.
    6: 16/216
    5: 40/216
    4: 52/216
    3: 52/216
    2: 40/216
    1: 16/216

    Now does anyone know how I can easily pair these odds to get to n/7776: p/7666 and q/7666????
    Last edited by J0057; Nov 30th 2007 at 01:01 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Odds with 2 normal dice and 1 unusual?
    Posted in the Statistics Forum
    Replies: 3
    Last Post: Apr 24th 2011, 01:56 PM
  2. bookies odds vs real odds
    Posted in the Statistics Forum
    Replies: 10
    Last Post: Feb 18th 2011, 04:49 PM
  3. AP Style Question, Help?
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Oct 19th 2009, 05:21 PM
  4. Help, Dice odds with Risk.
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: Dec 4th 2007, 10:43 PM

Search Tags

/mathhelpforum @mathhelpforum