In my quest to find the answer to this question, I have decided to start smaller. The rules of Button Men are here, and the characters are listed below.

  • Avis: d4, d4, d10, d12, dX (where X is d4, d6, d8, d10, d12, or d20)

  • Iago: d20, d20, d20, dX (where X is d4, d6, d8, d10, d12, or d20)

Note: Technically speaking according to the Button Men rules, the dX swing die can be any number within the range of 4 to 20 inclusive. For the purposes of this question, we will assume that the swing die will be one of the polyhedral dice listed.

If you want to take a stab at this question, be my guest. I will be attempting to brute force an answer over the next couple of days. If you want to post whether you believe brute forcing an answer is possible, and the total number of permutations that need to be examined, that would be nice (I will attempt to make a guess myself).


1 Answer 1


According to Swing Dice theory, Avis has the advantage.

If both players choose the same size X, Iago needs to keep 20 points to win (the 2/3 difference between his 60+X points and Avis' 30+X). So all things being equal, Iago needs to win and hold on to one of his 20-siders.

The bad news for Iago is that even in the best case (Avis takes a d20, Iago takes a d4), he still needs to hold 9.3 points (which means he still needs that d20, since the d4 won't cut it here). The point difference between the two buttons is wide enough that Iago must always win - there's no case where Avis can end up with a higher point total.

The news gets worse - if Iago ends up with a larger X than Avis, he now needs to hold two dice. And it doesn't help that Avis has five dice vs. Iago's four - given that a normal game of BM involves taking dice each round, he needs Avis to pass at least once.

To sum up - Iago's best action is to take a d4 (regardless of what Avis has), win initiative, force Avis to pass at least once each battle, take all of Avis' dice, and hold one of the d20s.

(Avis, on the other hand, can take a d20 and still force Iago to win and hold a d10 or better to win. There's no bad matchup for him.)

Source: a quick Excel spreadsheet to calculate win conditions for each combination of Xs. :)

  • 1
    You might want to post the tables if they aren't too large, or at least link to the swing dice theory page (and possibly quote the formula).
    – user1873
    Commented Jul 24, 2013 at 17:04
  • Added the link, but the spreadsheet was very quick-and-dirty (basically iterated through the various formulas). If I get a moment I'll do up a Pretty GDocs version and link. Commented Jul 24, 2013 at 17:15
  • This is a good start towards answering the question, but it isn't a definitive answer. This answers the question about what swing die is optimal for Iago (d4), which allows Iago to only have to save 1 die and gives a slightly less than 25% chance of going first. You still need to determine how likely it is for Iago to have his d20s below Avis's dice (or cumulative sum). I think Avis has an advantage, but I have a suspicion it isn't more than a 60-40% split.
    – user1873
    Commented Jul 26, 2013 at 5:22
  • @user1873 The problem is that you rapidly fall down the rabbit-hole of probability - Iago(4) has 32000 different possible rolls, which combined with Avis(4)'s 7680 different openings means that there are 245,760,000 different ways the opening roll could play out. And then you start tacking on the branches as you re-roll dice. And that's for just one of the 289 combinations of each player's X die... Commented Jul 26, 2013 at 15:28
  • it actually isn't 245+ million if you do some pruning. First, if you recognize that one d20 is indistinguishable from another d20 (or a d4), you can prune it down to about 3,536,820 game state nodes (already doubled to account for turn order). Additional pruning would bring this down considerably. I have been trying to think of a way to solve this problem with a computer, and I think working backwards (from possible end states) is the way to go. There is only 31 end-game nodes, and using pruning techniques and working backwards can probably reduce that 3 million game nodes further
    – user1873
    Commented Jul 27, 2013 at 4:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .