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Show HN: Probabilistic Tic-Tac-Toe

csun.io

21–30 of 113 posts

Re: Show HN: Probabilistic Tic-Tac-Toe

#22

Nice! And irritating! I would make it a lot faster though. It takes so much time waiting for the animations to finish.

Agreed. Taking the time to roll the dice is important the first few times, to fully cement the idea of the game. After that it gets annoying.

To be specific, you could probably even leave the roll time as-is, to give you that suspense, but the time it takes to move the die to the center, flash it, and flash the result, is too long and gets irritating.

Re: Show HN: Probabilistic Tic-Tac-Toe

#23

Nice! And irritating! I would make it a lot faster though. It takes so much time waiting for the animations to finish.

An extra click that would stop the animation immediately might be helpful.

Or to turn this into a different game: the d20 stops fast by default, but extra click to cheat and keep it rolling if you feel that it's about to stop on an unfavorable face.

Re: Show HN: Probabilistic Tic-Tac-Toe

#25

Nice! UI suggestion: show the probabilities for a move as a point in a triangle, with your outcome labels on the vertices. (Or maybe as red/green/neutral colors in the triangle's interior.) This representation is called the "probability simplex". It would look less busy, quicker to scan, I think.

Or a pie chart.

Re: Show HN: Probabilistic Tic-Tac-Toe

#27
Harder for humans, but easy to make a really strong AI for this. Even overcounting because of illegal board states (multiple winners) and not even bothering to eliminate symmetries, there are at most 2 * 3^9 = 39366 board states.

There are cycles in the board state graph, although they are of a very specific form (the only kind of cycle that exists is for board B with O and X alternating turns). So it is probably possible to make a completely deterministic and optimal algorithm for this probabilistic game, but it does sound complicated. You can't naively apply expectiminimax.

However after marking the winning board states as 0 or 1 respectively if O or X wins I would expect value iteration to very quickly converge to an optimal strategy here.

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