Rock, Paper, Scissors..
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  Rock, Paper, Scissors..
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Author Topic: Rock, Paper, Scissors..  (Read 309 times)
tik 🪀✨
ComradeCarter
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« on: April 07, 2014, 04:54:46 AM »
« edited: April 07, 2014, 05:02:45 AM by Tik: Suppository of All Wisdom »

Via Here

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muon2
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« Reply #1 on: April 07, 2014, 11:26:46 AM »
« Edited: April 07, 2014, 12:07:44 PM by muon2 »

As the link notes, the natural reaction to always play paper can be met by your opponent playing scissors the half of the time that rock is not required. That means you each win half the time and your net gain is zero.

Since your opponent is using rock more often than a third of the time it doesn't make sense to pick scissors. However, if you pick some combination of rock and paper, when you pick rock you aren't hurt when your opponent picks rock and you crush if scissors is used against you. Your opponent can select any strategy so let's consider all the options. I assume that you will pick paper a fraction p times and rock the remaining fraction 1-p times. Your opponent is stuck with rock 1/2 of the time and can pick paper x of the time and scissors (1/2 - x) of the time, since it makes no sense to load up even more on rock and give you a greater edge.

Opponent picks rock and you pick rock: you gain zero
Opponent picks rock and you pick paper: you gain at a fraction equal to (1/2)*p
Opponent picks paper and you pick rock: you lose at a fraction equal to x*(1-p)
Opponent picks paper and you pick paper: you gain zero
Opponent picks scissors and you pick rock: you gain (1/2 - x)*(1-p)
Opponent picks scissors and you pick paper: you lose (1/2 - x)*p

You fractional winnings are W = (1/2)*p - x(1-p) + (1/2 -x)*(1-p) - (1/2 - x)*p
And multiplying this out W = p/2 - x + px +1/2 - x - p/2 + px - p/2 + px = 1/2 - p/2 - 2x + 3px
Notice what happens when it is written W = 1/2 - p/2 + x*(3p - 2), if p = 2/3 then the quantity in parenthesis is zero and W = 1/2 - (2/3)/2 = 1/6 regardless of the choice of x.

In other words, if you choose paper 2/3 of the time and rock 1/3 of the time you will see an average fractional gain of 1/6 of the wager with every play.

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True Federalist (진정한 연방 주의자)
Ernest
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« Reply #2 on: April 07, 2014, 06:59:04 PM »

Actually, you've missed an important point, muon.  As you've shown, as the player you can adopt a strategy that guarantees you to win more often than you lose under the stated conditions. (Indeed, under the rules it is impossible with a fair coin for you to lose more often than you win, provided you play long enough.)  Since the opponent will adapt to pick the optimal strategy, his optimal strategy is to not play the game and thus no money will change hands.
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