Optimal RPS vs. RP play

Logic Level 4

Alice and Bob are playing 900 games of Rock-Paper-Scissors, but Alice is not allowed to choose scissors in any of the games. If both players choose their strategies optimally (i.e. Nash equilibrium is reached), what is the expected number of games Bob will win?

"Optimally" means that both players want to maximize the difference between the number of games they win and the number of games the opponent wins.

The answer is 400.

2 solutions

Anthony Ritz
Mar 25, 2016

	$R$	$P$	$S$
$R$	$(0,0)$	$(1,-1)$	$(-1,1)$
$P$	$(-1,1)$	$(0,0)$	$(1,-1)$

Bob's choices are along the top, and his payouts are the first number in each ordered pair. Alice's choices are down the side, and her payouts are the second number in each ordered pair.

No pure strategy is optimal for either player, since that player would never win given an opponent with foreknowledge of such a strategy. So mixed strategies are optimal.

Bob should never play Rock; Bob can never win playing Rock, and Rock is strictly worse for Bob than Paper. Assume that Bob plays Paper with probability $x$ , and Scissors with probability $1-x$ . Assume that Alice plays Rock with probability $y$ and Paper with probability $1-y$ .

	$R (0)$	$P (x)$	$S (1-x)$
$R (y)$	$(0,0)$	$(1,-1)$	$(-1,1)$
$P (1-y)$	$(-1,1)$	$(0,0)$	$(1,-1)$

Then Bob's payout (and the negative of Alice's payout, since it's a zero-sum game) is:

$xy-(1-x)y-(0)(1-y)+(1-x)(1-y) =$

$xy-y+xy+1-x-y+xy =$

$3xy-x-2y+1$ .

There must be a Nash equilibrium in which neither player can improve even by knowing the other player's strategy. This means that both $3xy-2y=0$ (Alice cannot improve by changing $y$ ) and $3xy-x=0$ (Bob cannot improve by changing $x$ ). Simplifying both gives $x=\frac{2}{3}$ and $y=\frac{1}{3}$ .

	$R (0)$	$P (\frac{2}{3})$	$S (\frac{1}{3})$
$R (\frac{1}{3})$	$(0,0)$	$(1,-1)$	$(-1,1)$
$P (\frac{2}{3})$	$(-1,1)$	$(0,0)$	$(1,-1)$

Bob wins $\frac{2}{3}*\frac{1}{3}+\frac{1}{3}*\frac{2}{3}=\frac{2}{9}+\frac{2}{9}=\frac{4}{9}$ of the games. Since $900$ games are played, Bob wins $\boxed{400}$ .

Great solution, Anthony -- well explained. Thanks for sharing!

Eli Ross Staff - 5 years, 2 months ago

I sorta followed until you said 3xy - 2y = 0

Andrew Darley - 1 year, 4 months ago

why didn't you just put the values in his expected payout, It turns out to be 1/3 so he stands to win 1/3 of the games or 300 games?

Ajinkya Shivashankar - 4 years, 7 months ago

Bob wins 400, Alice 100 and 400 ties; hence expected payout of Bob equals +400-100+0=+300 i.e. +1/3 of 900 games

Willy Bracke - 2 years, 6 months ago

Riker Wachtler
Mar 22, 2016

I can't LaTeX, sorry for bad formatting, if applicable.

Scissors = S Rock = R Paper = P

We know that Bob can choose from RPS. However, R is a very bad idea. If he chooses rock he either loses or ties, neither of which he wants.

So Bob choose from PS, Alice from RP.

If Bob constantly chooses P, then he either wins or ties. That is good, but Alice can just constantly choose paper also and then nobody wins.

If Bob choose S, then he either wins or loses. This is good, but there is a chance of losing.

So the solution is a compromise of S and P, with P $2/3$ of the time, and S $1/3$ of the time.

Alice's optimal solution to combat this is similar, with P $2/3$ of the time and R $1/3$ .

So the 9 permutations of this are $(PP, PP, PR, PP, PP, PR, SP, SP, SR)$ , with Bob's move first and Alice's after.

We can easily see that $4$ of these are wins for Bob, $4$ are ties, and $1$ is a loss, so $4/9$ of the games are wins for Bob.

$4/9*900 = 400$ , and that is our answer.

Where do you get 2/3rds from?

Alex Li - 5 years, 2 months ago

Intuition. If there is a better way, I am ignorant. Sorry.

It happens to coincide with the number of different moves they can play though, I don't know if that means anything.

Riker Wachtler - 5 years, 2 months ago

I think 2/3 is because paper also gets the wieghtage of scissor as it cannot loose but can draw!

Prince Loomba - 5 years, 1 month ago

Optimal RPS vs. RP play

The answer is 400.

2 solutions

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