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Here’s what I’m thinking: The neural network doesn’t have to be correct about which one you pick, it has to be correct about which you don’t pick. Only one option they pick is a loss, so if the other party can be somewhat certain you won’t pick a specific option, it can at least tie. So if a randomizer has pretty even distribution, I think it can win more than half the time, because it can gather roughly how likely the same choice is to be played in a row.

I’m basically suggesting a predictable distribution can be exploited in RPS.

Honestly, the bot could always be winning against the RNG by dumb luck. More experimentation would be needed to be sure. I am just making guesses.




> I think it can win more than half the time, because it can gather roughly how likely the same choice is to be played in a row.

With a random choice, the chance of playing the same choice in a row is 1/3. This does not give you any advantage over having no information (where each choice has a 1/3 chance.)

I think the misunderstanding is in > a randomizer has pretty even distribution

Having an even distribution over a long time does not make any specific choice less random. https://en.wikipedia.org/wiki/Gambler%27s_fallacy


Here's a thought experiment that might help: imagine what you say is truly the case - that would mean you could "charge up" a dice by rolling it until you got a long run of a given number - lets pick something arbitrary, say you roll until you get 5 twos in a row. According to what you've said the chance of the next number rolled being a two is now lower than it was when you started "charging up" your dice.

How is this possible? Nothing is physically changing about the dice between rolls.


> So if a randomizer has pretty even distribution, I think it can win more than half the time, because it can gather roughly how likely the same choice is to be played in a row.

No.




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