Top Document: Axis & Allies FAQ v1.4 Previous Document: 15. What are the effects of using the 2nd Edition optional rules? Next Document: 17. Are there any game conventions that include Axis & Allies games? See reader questions & answers on this topic!  Help others by sharing your knowledge This task can be quite daunting. For the most part, the difficulty depends on the number of units involved. To correctly determine the odds of a particular result, you multiply the chance of hitting for each unit involved. For example, take two battleships (keeping the odds balanced for now). The attacking battleship has a 4/6 (everything will be in sixths here) chance of hitting. No matter that roll, the defending battleship also has a 4/6 chance of hitting. With two results per die (hit or miss, though the chances are weighted), there are a total of four possible outcomes. They are: Attacker Defender Probability Result 1 Hit Hit ==> (4/6) * (4/6) = 16/36 Mutual Annihilation 2 Hit Miss ==> (4/6) * (2/6) = 8/36 Attacker Wins 3 Miss Hit ==> (2/6) * (4/6) = 8/36 Defender Wins 4 Miss Miss ==> (2/6) * (2/6) = 4/36 Draw; continue Note that the sum of the probabilities is 1. This makes for a simple means of verifying your work. Adding more units makes this calculation much more complicated. To calculate such odds, you must take into consideration all rolls. This task can be simplified by creating a binary tree. To generate a binary tree, draw a branch for each possible outcome (hit or miss) for each unit involved for each round of combat. Keep them all in a "path" within the tree (e.g., you should be able to go from the left (top) of the tree all the way to the right (bottom) and have a representation of every die rolled during a specific battle). So, for example, if there were two attacking units and one defending, you would show both attackers one after the other, then the defender (by now, there are eight outcomes), then continue based on the outcomes of those rolls. Be sure to label each branch (Mutual Annihilation, Attacker Wins, etc.). I always work sideways and make the upward branch the hit branch and label it by placing the number of ways (out of six!) that a hit could be rolled. Likewise, the downward branch is for misses and is labelled with the number of ways (out of six!) to miss. For the battleship example above: Result Probability Defender 4 / Mutual Annihilation (16/36) / Attacker < / \ 4 / 2 \ Attacker Wins (8/36) / Start < \ 2 \ 4 / Defender Wins (8/36) \ / < \ 2 \ Draw (4/36) To get the probabilities, divide each number by 6 and multiply all numbers on a path from left (top) to right (bottom). Write this result at the left (bottom) of the tree next to the label of the outcome at the end of that path. When you have all those numbers, add all fractions of similar labels (like Mutual Annihilation or Attacker Wins), no matter where in the tree it is. This sum is the probability of that outcome. The sum of all outcomes is 1 (or a mistake has been made). But, what to do at "Draw"? By the rules, you continue the battle (as we are concerned about the odds of possible outcomes we will not consider withdraw options). If units have been lost, they will no longer appear in the tree. Each "Draw" could actually be considered the "Start" of a new battle with only the units that survived to get to that "Draw". In this example, the entire tree could be copied and placed where the "Draw" is located, though this would become a recursive loop (which would never end), making calculations quite difficult. There is a simplification! Recall that to find the odds for an outcome when the tree is complete, you will add the fractions of all occurances of that outcome. Under the "Draw" node, you will find the same ratios of results as in the parent branch immediately above it, thus as you add the fractions that occur under it, you do so in the same proportions as the top tree (so, as the number of recursive branches approaches infinity, you reach a limit for the other outcomes and the "Draw" probability becomes zero). With that, you can ignore the "Draw" branch with one provision: Instead of counting the other outcomes out of 36, count them out of 36 minus the 4 occurances of draw. This is of course 32. This simplification will work at each "Draw", but take care to normalize each "Draw" node separately. This is done so that the sum of probabilities is still 1 (called normalization). So, the odds of a mutual annihilation with two warring battleships is 0.50, and of either combatant winning is 0.25. While this is how you can calculate the exact odds of a result, it is obviously no easy task when there are more than a few units! While all you have to do is separate between the three results above, you can distinguish more. For example, Attaker wins but loses two units, Defender wins but loses three units, or Mutual Annihilation. With more categories you can get a more detailed breakdown of the outcomes. This obviously takes more time, but is quite feasible. For a more indepth discussion, or if more explanations of the odds mechanism is needed, please email Dewey Barich (barich@tam2000.tamu.edu). User Contributions:Comment about this article, ask questions, or add new information about this topic:Top Document: Axis & Allies FAQ v1.4 Previous Document: 15. What are the effects of using the 2nd Edition optional rules? Next Document: 17. Are there any game conventions that include Axis & Allies games? Single Page [ Usenet FAQs  Web FAQs  Documents  RFC Index ] Send corrections/additions to the FAQ Maintainer: pgoudswa@jumppoint.com
Last Update March 27 2014 @ 02:11 PM
