Optimal strategy for a two player game.
$begingroup$
Player one has a black ace (worth 1 point) and a red four (worth 4 points)
Player two has a black two (worth 2 points) and a red three (worth 3 points)
Both players select a card and show each other. If the colors match, player two gives player one the sum of the cards in dollars. Otherwise, player one gives player two the sum of the cards in dollars.
Is there an optimal strategy to this game? It seems to me like the expected value of either player is zero dollars.
game-theory
edited Dec 7 '18 at 2:28
David M.
1,709418
asked Dec 7 '18 at 0:00
SneakosSneakos
1
$endgroup$
add a comment |
$begingroup$
Player one has a black ace (worth 1 point) and a red four (worth 4 points)
Player two has a black two (worth 2 points) and a red three (worth 3 points)
Both players select a card and show each other. If the colors match, player two gives player one the sum of the cards in dollars. Otherwise, player one gives player two the sum of the cards in dollars.
Is there an optimal strategy to this game? It seems to me like the expected value of either player is zero dollars.
game-theory
edited Dec 7 '18 at 2:28
David M.
1,709418
asked Dec 7 '18 at 0:00
SneakosSneakos
1
$endgroup$
add a comment |
$begingroup$
Player one has a black ace (worth 1 point) and a red four (worth 4 points)
Player two has a black two (worth 2 points) and a red three (worth 3 points)
Both players select a card and show each other. If the colors match, player two gives player one the sum of the cards in dollars. Otherwise, player one gives player two the sum of the cards in dollars.
Is there an optimal strategy to this game? It seems to me like the expected value of either player is zero dollars.
game-theory
edited Dec 7 '18 at 2:28
David M.
1,709418
asked Dec 7 '18 at 0:00
SneakosSneakos
1
$endgroup$
Player one has a black ace (worth 1 point) and a red four (worth 4 points)
Player two has a black two (worth 2 points) and a red three (worth 3 points)
Both players select a card and show each other. If the colors match, player two gives player one the sum of the cards in dollars. Otherwise, player one gives player two the sum of the cards in dollars.
Is there an optimal strategy to this game? It seems to me like the expected value of either player is zero dollars.
game-theory
game-theory
edited Dec 7 '18 at 2:28
David M.
1,709418
asked Dec 7 '18 at 0:00
SneakosSneakos
1
edited Dec 7 '18 at 2:28
David M.
1,709418
asked Dec 7 '18 at 0:00
SneakosSneakos
1
edited Dec 7 '18 at 2:28
David M.
1,709418
edited Dec 7 '18 at 2:28
David M.
1,709418
edited Dec 7 '18 at 2:28
David M.
1,709418
1,709418
asked Dec 7 '18 at 0:00
SneakosSneakos
1
asked Dec 7 '18 at 0:00
SneakosSneakos
1
asked Dec 7 '18 at 0:00
SneakosSneakos
1
1
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In fact, Player 2 can guarantee an average payout of $$0.15$
Let's make a table of the payoffs for Player 1. Since this is a zero-sum game, we can treat Player 2 as trying to minimize Player 1's payoff.
begin{matrix}
& B2 & R3\
B1 & $3 & -$4 \
R4 & -$6 & $7
end{matrix}
This obviously has no pure strategy Nash equilibria, so a mixed strategy will be optimal.
Now suppose Player 1 plays his red card with probability $p$ and Player 2 plays her red card with probability q. Then the expected payoff is
$$
E(P) = $3(1-p)(1-q) -$4(1-p)q - $6(1-q)p + $7pq = $3 - $9p - $7q + $20pq = $20left(p-0.35right)left(q-0.45right)-$0.15
$$
Player 2 thus has a winning strategy, as if she plays his red card 45% of the time and his black card 55% of the time, Player 1 will on average give her $$0.15$ every game no matter how he chooses his card. For Player 1, the winning move is not to play.
answered Dec 7 '18 at 3:02
eyeballfrogeyeballfrog
6,208629
$endgroup$
add a comment |
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$begingroup$
In fact, Player 2 can guarantee an average payout of $$0.15$
Let's make a table of the payoffs for Player 1. Since this is a zero-sum game, we can treat Player 2 as trying to minimize Player 1's payoff.
begin{matrix}
& B2 & R3\
B1 & $3 & -$4 \
R4 & -$6 & $7
end{matrix}
This obviously has no pure strategy Nash equilibria, so a mixed strategy will be optimal.
Now suppose Player 1 plays his red card with probability $p$ and Player 2 plays her red card with probability q. Then the expected payoff is
$$
E(P) = $3(1-p)(1-q) -$4(1-p)q - $6(1-q)p + $7pq = $3 - $9p - $7q + $20pq = $20left(p-0.35right)left(q-0.45right)-$0.15
$$
Player 2 thus has a winning strategy, as if she plays his red card 45% of the time and his black card 55% of the time, Player 1 will on average give her $$0.15$ every game no matter how he chooses his card. For Player 1, the winning move is not to play.
answered Dec 7 '18 at 3:02
eyeballfrogeyeballfrog
6,208629
$endgroup$
add a comment |
$begingroup$
In fact, Player 2 can guarantee an average payout of $$0.15$
Let's make a table of the payoffs for Player 1. Since this is a zero-sum game, we can treat Player 2 as trying to minimize Player 1's payoff.
begin{matrix}
& B2 & R3\
B1 & $3 & -$4 \
R4 & -$6 & $7
end{matrix}
This obviously has no pure strategy Nash equilibria, so a mixed strategy will be optimal.
Now suppose Player 1 plays his red card with probability $p$ and Player 2 plays her red card with probability q. Then the expected payoff is
$$
E(P) = $3(1-p)(1-q) -$4(1-p)q - $6(1-q)p + $7pq = $3 - $9p - $7q + $20pq = $20left(p-0.35right)left(q-0.45right)-$0.15
$$
Player 2 thus has a winning strategy, as if she plays his red card 45% of the time and his black card 55% of the time, Player 1 will on average give her $$0.15$ every game no matter how he chooses his card. For Player 1, the winning move is not to play.
answered Dec 7 '18 at 3:02
eyeballfrogeyeballfrog
6,208629
$endgroup$
add a comment |
$begingroup$
In fact, Player 2 can guarantee an average payout of $$0.15$
Let's make a table of the payoffs for Player 1. Since this is a zero-sum game, we can treat Player 2 as trying to minimize Player 1's payoff.
begin{matrix}
& B2 & R3\
B1 & $3 & -$4 \
R4 & -$6 & $7
end{matrix}
This obviously has no pure strategy Nash equilibria, so a mixed strategy will be optimal.
Now suppose Player 1 plays his red card with probability $p$ and Player 2 plays her red card with probability q. Then the expected payoff is
$$
E(P) = $3(1-p)(1-q) -$4(1-p)q - $6(1-q)p + $7pq = $3 - $9p - $7q + $20pq = $20left(p-0.35right)left(q-0.45right)-$0.15
$$
Player 2 thus has a winning strategy, as if she plays his red card 45% of the time and his black card 55% of the time, Player 1 will on average give her $$0.15$ every game no matter how he chooses his card. For Player 1, the winning move is not to play.
answered Dec 7 '18 at 3:02
eyeballfrogeyeballfrog
6,208629
$endgroup$
In fact, Player 2 can guarantee an average payout of $$0.15$
Let's make a table of the payoffs for Player 1. Since this is a zero-sum game, we can treat Player 2 as trying to minimize Player 1's payoff.
begin{matrix}
& B2 & R3\
B1 & $3 & -$4 \
R4 & -$6 & $7
end{matrix}
This obviously has no pure strategy Nash equilibria, so a mixed strategy will be optimal.
Now suppose Player 1 plays his red card with probability $p$ and Player 2 plays her red card with probability q. Then the expected payoff is
$$
E(P) = $3(1-p)(1-q) -$4(1-p)q - $6(1-q)p + $7pq = $3 - $9p - $7q + $20pq = $20left(p-0.35right)left(q-0.45right)-$0.15
$$
Player 2 thus has a winning strategy, as if she plays his red card 45% of the time and his black card 55% of the time, Player 1 will on average give her $$0.15$ every game no matter how he chooses his card. For Player 1, the winning move is not to play.
answered Dec 7 '18 at 3:02
eyeballfrogeyeballfrog
6,208629
answered Dec 7 '18 at 3:02
eyeballfrogeyeballfrog
6,208629
answered Dec 7 '18 at 3:02
eyeballfrogeyeballfrog
6,208629
answered Dec 7 '18 at 3:02
eyeballfrogeyeballfrog
6,208629
6,208629
add a comment |
add a comment |
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