Strategy for the game of Buttonmen. Maybe other buttonmen-related topics.
Monday, December 7, 2020
not deep, but I was mildly surprised when I did the computation
I have four possible attacks. One of them is obviously suboptimal, but the others all give roughly similar chances of winning. Or do they?
Which one is best?
Going on gut instinct, I'd choose to try to keep my two dice and capture the opponent's two dice, so that would be (12) captures p(20) and then hope that I can capture the (X=4) also.
On second thought, I guess the option of using the (12) to capture the (X=4) first leaves a pathway to winning no matter whether the (12) rolls high or low, so that's probably preferable.
you have two ways to win: capture both dice without losing anything, or capture just the (X=4) and lose both dice.
option 1: (10) takes p(20): 70% chance of re-rolling high enough to win option 2: (12) takes p(20): clearly better than option 1, 75% chance of re-rolling high enough to win. option 3: (10) takes (X=4) 60% chance of rolling high enough to then take the poison and win PLUS 40% * 65% = 26% chance of rolling low enough to be captured, but then the p(20) rolls high enough to be forced to capture the (12), too. Overall winning percentage: 86%! option 4: (12) takes (X=4) 66.7% [i.e. 2/3] chance of rolling high enough to then take the poison and win PLUS 33.3% * 55% = 18.3% [i.e. 11/60] chance of rolling low enough to be captured, but then the p(20) rolls high enough to be forced to capture the (10), too. Overall winning percentage: 85%, not QUITE as good as option 3.
So it's slightly better to capture the (X=4) first and the surprise is that it's actually best to do so with the (10), though the difference is slight. [if I tweaked the values on each die in the problem, we could make the effect more dramatic]
4 comments:
That is surprising. Quite a big difference in win percentage too.
Going on gut instinct, I'd choose to try to keep my two dice and capture the opponent's two dice, so that would be (12) captures p(20) and then hope that I can capture the (X=4) also.
On second thought, I guess the option of using the (12) to capture the (X=4) first leaves a pathway to winning no matter whether the (12) rolls high or low, so that's probably preferable.
you have two ways to win: capture both dice without losing anything, or capture just the (X=4) and lose both dice.
option 1: (10) takes p(20): 70% chance of re-rolling high enough to win
option 2: (12) takes p(20): clearly better than option 1, 75% chance of re-rolling high enough to win.
option 3: (10) takes (X=4) 60% chance of rolling high enough to then take the poison and win PLUS 40% * 65% = 26% chance of rolling low enough to be captured, but then the p(20) rolls high enough to be forced to capture the (12), too. Overall winning percentage: 86%!
option 4: (12) takes (X=4) 66.7% [i.e. 2/3] chance of rolling high enough to then take the poison and win PLUS 33.3% * 55% = 18.3% [i.e. 11/60] chance of rolling low enough to be captured, but then the p(20) rolls high enough to be forced to capture the (10), too. Overall winning percentage: 85%, not QUITE as good as option 3.
So it's slightly better to capture the (X=4) first and the surprise is that it's actually best to do so with the (10), though the difference is slight. [if I tweaked the values on each die in the problem, we could make the effect more dramatic]
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