Mathematicians Roll Dice and Get Rock-Paper-Scissors

Re:Mathematicians Roll Dice and Get Rock-Paper-Scissors by tromp on Saturday January 21, 2023 @03:25AM (#34464851)

All your program does is pair A vs B 2/9 of the time, B vs C 2/9 of the time, and A vs C 2/9 of the time (pairing a die against itself the remaining 3/9 times).

So it approximates wins = 10000 * 2/9 * [58%+31%, 42%+58%, 42%+69%] = 20000/9 * [89%, 100%, 111%].

(Note these percentages are rounded, e.g 21/36 to 58%).

This is just averaging the win and loss percentages of each coin against the 2 others, not "a conventional metric" like expected value of a roll which are all the same as a regular die.

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Re:Mathematicians Roll Dice and Get Rock-Paper-Scissors by thaumasiotes on Saturday January 21, 2023 @05:11AM (#34465327)

That's what the game is. Rolling a particular value doesn't get you anything; the expected value of a roll is not informative. If I gave you a die with all 6s, a die with all 3s, and a die with all 1s, and told you the rules of the game were that I give you $600 minus $100 for each point you roll, which die would be the strongest? Which die would have the highest expected value?

Note that your math is incorrect; B gets more wins from being paired against itself than either A or C do.

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Re:Mathematicians Roll Dice and Get Rock-Paper-Scissors by tromp on Saturday January 21, 2023 @07:23AM (#34465937)

Right; I forgot about wins from a die paired with itself. The proportion of wins is [5+21+11, 15+9+21, 25+15+5] = [37, 45, 45].

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