9 ms·

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)

by tromp 4y ago

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 (3.5).

thaumasiotes 4y ago
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. edit: here are results from sets of 1000 single-elimination 256-man tournaments where everyone brings a random die (in this case, draw/draw is equivalent to win/loss): [140, 424, 436] [144, 394, 462] [160, 417, 423] [123, 419, 458] [174, 411, 415] [150, 397, 453] The round of 8 is much more lopsided, generally featuring >3.4 Cs and <2.9 Bs. Bs get a boost in rate of becoming overall winner because there are so many Cs at high levels. Cs have an even higher rate of being overall winner, even though they have worse performance starting from the round of 8, because their overall record is so much better. As die. Obviously, in reality, people wouldn't bring random dice to this event. More people would bring C dice, because C is stronger than the other two.
- deleted 4y ago
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- tromp 4y ago
  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].

by tromp 4y ago

thaumasiotes 4y ago

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. edit: here are results from sets of 1000 single-elimination 256-man tournaments where everyone brings a random die (in this case, draw/draw is equivalent to win/loss): [140, 424, 436] [144, 394, 462] [160, 417, 423] [123, 419, 458] [174, 411, 415] [150, 397, 453] The round of 8 is much more lopsided, generally featuring >3.4 Cs and <2.9 Bs. Bs get a boost in rate of becoming overall winner because there are so many Cs at high levels. Cs have an even higher rate of being overall winner, even though they have worse performance starting from the round of 8, because their overall record is so much better. As die. Obviously, in reality, people wouldn't bring random dice to this event. More people would bring C dice, because C is stronger than the other two.

deleted 4y ago
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tromp 4y ago
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].