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I believe I've finally found a workable solution. It involves the Unknown Formula published by Murderous Maths. BOTH the 2x2 and 3x3 can be solved by a table of unique two-dice combinations for both the Attacker and Defender. The key is a set of "frequencies" of how often each unique pair occurs, and there are just 21 pairs with two-dice. The frequencies differ based upon the number of dice tossed by each player, but the 21 x 21 grid can be populated with four symbols, like LB,L1,L2,LN (both,1st,2nd,neither). The sum of the Attacker frequencies on each line for any particular symbol, times the Defender's frequency for that line, when added together gives the numerator for LB,L1,L2,LN. I showed the process for L2. Using other formulas that gives the overall probability for winning (or loosing) the 1st battle, or the 2nd battle, leads to a complete solution. No computer is required. I just used the Unknown Formula, and some logic to reduce duplicates involving the 3rd dice in the Attacker's 3-dice roll, to obtain the frequencies. Visit: web.stanford.edu/group/spires/risk.notes.html #here
I may have found a breakthrough, but I can't prove it. Visit: http://web.stanford.edu/group/spires/risk.notes.html#here where you'll find a summary of the problem, and a SPECIAL NOTE about a discovery that, if proven, leads to a solution.
I must say, this has turned into a spirited discussion, and I'm truly getting the "Math is Fun" feeling.
bobbym, 'I agree with a "series solution"' IF ... I added a condition.
I still think this is a good example of a Dependent rather than Independent Probability. Winning the 1st, or winning the 2nd battle are almost Independent events. They rely on the probabilities of the "highest" rolled for the 1st, and "next highest" for the 2nd. But BOTH wins is asking for the probability of the 2nd win given the 1st win occurred. That's a Dependent Probability. Right? (or wrong)
bobbym, I agree with a "series solution" if you mean something like the "(A and B) or (C and D) ..." chains I have shown on my webpage for related solutions to this problem with Risk, like the probability of "losing the 1st battle".
ShivamS, you overrate me. I'm a retired Stanford Staff Emeritus, and was never a professor. I worked at Stanford for nearly 41 years, as a Systems Programmer. My specialty was databases, and I'm a co-designer of SPIRES, the "Stanford Public Information REtrieval System". It still runs today on a variety of platforms.
bobbym, I am NOT surprised. I've worked on this off-and-on for 30 years, with NO success. I'm curious about what you found..."posed on other forums" or "monographs". My computer program "rolls" all combinations of 3-dice for the attacker, like an odometer; sorts each result into a separate descending list, and then does all combinations of 2-dice for the defender, and sorts each result into another list. It compares 1st-to-1st in each list (1st battle), then 2nd-to-2nd (2nd battle), and then takes 0 and adds 1 for the 1st attacker win, and 2 for the 2nd attacker win. The result is 0,1,2,3, an index into a table of four counters that record: attacker wins another at this index. After doing all combinations for the defender, I then go back and get another attacker list, and run the defender combinations against that, etc. When I'm all done (all attacker combinations with all defender combinations), I have a table of 4-values for attacker wins: none, 1st only, 2nd only, or both. For the 3-on-2 scenario, that table contains: 2890, 1834, 777, 2275 which matches the Attacker wins columns in my document.
bobbym, the # of armies only determines the maximum number of dice each player can throw. Assume that has been determined yielding 1-1, 1-2, 2-1, 2-2, or 3-2 dice for A-D. So, yes, I'm looking for "Attacker wins both" probabilities of 295/1296 with (2-2), and 2890/7776 with (3-2).
bobbym, ask ma any questions you like. I recommend regular email, rather than post/reply on this forum. My forum profile should show you my new email address which won't get censored.
I've registered again, with the same 8-character username, but a different email-address. I got a new Password via that email, and now I can login. I wish I could edit my original post, but I'm new at this site and couldn't see any way to Edit. But at least now I have a Reply button. If you can edit the post, I'd like the URL changed to a new location: http://web.stanford.edu/group/spires/risk.notes.html
That should be accessible to anyone, anytime. It is a detailed description of what I'm trying to determine, the probability "formula" for "The attacker wins BOTH battles in a 3-on-2 attack in Risk". By formula, I mean the (AB) + (CD) ... type of expression where (AB) represents (A and B), and "+" represents "OR". A and B are concurrent events, like A = (attacker rolls 6), and B = (defender rolls less than 6). (CD) would substitute 5 for 6, etc. I describe the probability formulas for "Attacker wins the 1st dice compare", and "Attacker wins the 2nd dice compare", but can't get the probability "Attacker wins BOTH 1st AND 2nd dice compares".
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