Mixing
Equilibrium in a Lottery
$begingroup$
We have the following game with 1700 participants: Each participant can buy a lottery ticket for 1$ (when a participant buys a ticket, he does not know how many other participants are buying tickets) . Then a winner is selected uniformly at random from those who bought tickets, and he gets 1000$ (if no tickets were bought, no one gets the prize). What are the pure strategy equilibria in this game?
I am not sure how to interpret the fact that when a participant buys a ticket, he does not know how many other participants are buying tickets. I think that without that detail, an equilibrium point would be one where exactly 1000 people buy a lottery ticket. Am I correct? and how do you solve the question as given?
game-theory
asked Jan 23 at 8:52
OrpheusOrpheus
234111
$endgroup$
add a comment |
$begingroup$
We have the following game with 1700 participants: Each participant can buy a lottery ticket for 1$ (when a participant buys a ticket, he does not know how many other participants are buying tickets) . Then a winner is selected uniformly at random from those who bought tickets, and he gets 1000$ (if no tickets were bought, no one gets the prize). What are the pure strategy equilibria in this game?
I am not sure how to interpret the fact that when a participant buys a ticket, he does not know how many other participants are buying tickets. I think that without that detail, an equilibrium point would be one where exactly 1000 people buy a lottery ticket. Am I correct? and how do you solve the question as given?
game-theory
asked Jan 23 at 8:52
OrpheusOrpheus
234111
$endgroup$
$begingroup$
It seems that the prize will be given even if only one ticket is sold.
$endgroup$
– user
Jan 23 at 9:19
$begingroup$
Yes, this is correct. And?
$endgroup$
– Orpheus
Jan 23 at 9:28
add a comment |
$begingroup$
We have the following game with 1700 participants: Each participant can buy a lottery ticket for 1$ (when a participant buys a ticket, he does not know how many other participants are buying tickets) . Then a winner is selected uniformly at random from those who bought tickets, and he gets 1000$ (if no tickets were bought, no one gets the prize). What are the pure strategy equilibria in this game?
I am not sure how to interpret the fact that when a participant buys a ticket, he does not know how many other participants are buying tickets. I think that without that detail, an equilibrium point would be one where exactly 1000 people buy a lottery ticket. Am I correct? and how do you solve the question as given?
game-theory
asked Jan 23 at 8:52
OrpheusOrpheus
234111
$endgroup$
We have the following game with 1700 participants: Each participant can buy a lottery ticket for 1$ (when a participant buys a ticket, he does not know how many other participants are buying tickets) . Then a winner is selected uniformly at random from those who bought tickets, and he gets 1000$ (if no tickets were bought, no one gets the prize). What are the pure strategy equilibria in this game?
I am not sure how to interpret the fact that when a participant buys a ticket, he does not know how many other participants are buying tickets. I think that without that detail, an equilibrium point would be one where exactly 1000 people buy a lottery ticket. Am I correct? and how do you solve the question as given?
game-theory
game-theory
asked Jan 23 at 8:52
OrpheusOrpheus
234111
asked Jan 23 at 8:52
OrpheusOrpheus
234111
asked Jan 23 at 8:52
OrpheusOrpheus
234111
asked Jan 23 at 8:52
OrpheusOrpheus
234111
asked Jan 23 at 8:52
OrpheusOrpheus
234111
234111
$begingroup$
It seems that the prize will be given even if only one ticket is sold.
$endgroup$
– user
Jan 23 at 9:19
$begingroup$
Yes, this is correct. And?
$endgroup$
– Orpheus
Jan 23 at 9:28
add a comment |
$begingroup$
It seems that the prize will be given even if only one ticket is sold.
$endgroup$
– user
Jan 23 at 9:19
$begingroup$
Yes, this is correct. And?
$endgroup$
– Orpheus
Jan 23 at 9:28
$begingroup$
It seems that the prize will be given even if only one ticket is sold.
$endgroup$
– user
Jan 23 at 9:19
$begingroup$
It seems that the prize will be given even if only one ticket is sold.
$endgroup$
– user
Jan 23 at 9:19
$begingroup$
Yes, this is correct. And?
$endgroup$
– Orpheus
Jan 23 at 9:28
$begingroup$
Yes, this is correct. And?
$endgroup$
– Orpheus
Jan 23 at 9:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Model this as the following strategic form game $(N , S_i , u_i )$
$N = 1700$
$S_i = { B, DB } forall i$, where B represents player $i$ buying a ticket and DB represents player i not buying a ticket.
$u_i(s) = begin{cases} 0 &text{ if } s_i = DB \ (frac{1}{N_B}*1000) - 1 &text{ if } s_i = B end{cases}$, where $N_B$ represents the number of players buying the ticket i.e. $N_B = { i in N | s_i = B } $.
There are $binom{1701}{1000} $ total pure Nash equilibria strategies here:
$binom{1700}{1000} $ strategies corresponding to - when exactly 1000 players buy the ticket and $binom{1700}{999} $ strategies corresponding to - when exactly 999 players buy the ticket.
Calculation for utility value when player $i$ buys ticket:
Now since the lottery ticket winner is picked randomly, player $i$ gains money $ 1000-1 = 999$ $ with probability $frac{1}{N_B}$ and gains money $ -1 $ $ (i.e. loses a dollar) with probability $frac{N_B - 1}{N_B}$. Hence his utility in this regard is the weighted mean of the money he gains with the weights being the probabilities, hence:
$u_i(s) = frac{1}{N_B} * 999 + frac{N_B - 1}{N_B} * (-1) = frac{1000}{N_B} - 1$
answered Jan 23 at 9:35
KaindKaind
749414
$endgroup$
1
$begingroup$
Sorry, I am a bit confused by your answer. Why does the strategy B weakly dominates the strategy DB? Doesn't it depend on how large $N_B$ is?
$endgroup$
– Orpheus
Jan 23 at 10:06
$begingroup$
You're right, I made an error while computing it; I took N as 1000 in my calculations earlier. Edited now.
$endgroup$
– Kaind
Jan 23 at 10:13
$begingroup$
This makes more sense, however I am still confused. Doesn't your answer imply that there is no difference between the situation where every participant knows how many others bought a ticket before deciding whether to buy a ticket himself and the situation where he doesn't know that?
$endgroup$
– Orpheus
Jan 23 at 10:27
$begingroup$
There is a big difference; All players decide to buy a ticket independant of the other's opinion i.e. without knowing the other's strategy. Note that these strategy profiles are just Pure nash equilibria, it doesn't mean that a player knows before hand which strategy of his to choose, so that he reaches a PNE, all it means is that from any non-PNE, players can find an improvement path to a PNE (as you can check the game is weakly acyclic.)
$endgroup$
– Kaind
Jan 23 at 10:40
$begingroup$
Right, thanks a lot for your answer.
$endgroup$
– Orpheus
Jan 23 at 10:41
add a comment |
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1 Answer
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$begingroup$
Model this as the following strategic form game $(N , S_i , u_i )$
$N = 1700$
$S_i = { B, DB } forall i$, where B represents player $i$ buying a ticket and DB represents player i not buying a ticket.
$u_i(s) = begin{cases} 0 &text{ if } s_i = DB \ (frac{1}{N_B}*1000) - 1 &text{ if } s_i = B end{cases}$, where $N_B$ represents the number of players buying the ticket i.e. $N_B = { i in N | s_i = B } $.
There are $binom{1701}{1000} $ total pure Nash equilibria strategies here:
$binom{1700}{1000} $ strategies corresponding to - when exactly 1000 players buy the ticket and $binom{1700}{999} $ strategies corresponding to - when exactly 999 players buy the ticket.
Calculation for utility value when player $i$ buys ticket:
Now since the lottery ticket winner is picked randomly, player $i$ gains money $ 1000-1 = 999$ $ with probability $frac{1}{N_B}$ and gains money $ -1 $ $ (i.e. loses a dollar) with probability $frac{N_B - 1}{N_B}$. Hence his utility in this regard is the weighted mean of the money he gains with the weights being the probabilities, hence:
$u_i(s) = frac{1}{N_B} * 999 + frac{N_B - 1}{N_B} * (-1) = frac{1000}{N_B} - 1$
answered Jan 23 at 9:35
KaindKaind
749414
$endgroup$
1
$begingroup$
Sorry, I am a bit confused by your answer. Why does the strategy B weakly dominates the strategy DB? Doesn't it depend on how large $N_B$ is?
$endgroup$
– Orpheus
Jan 23 at 10:06
$begingroup$
You're right, I made an error while computing it; I took N as 1000 in my calculations earlier. Edited now.
$endgroup$
– Kaind
Jan 23 at 10:13
$begingroup$
This makes more sense, however I am still confused. Doesn't your answer imply that there is no difference between the situation where every participant knows how many others bought a ticket before deciding whether to buy a ticket himself and the situation where he doesn't know that?
$endgroup$
– Orpheus
Jan 23 at 10:27
$begingroup$
There is a big difference; All players decide to buy a ticket independant of the other's opinion i.e. without knowing the other's strategy. Note that these strategy profiles are just Pure nash equilibria, it doesn't mean that a player knows before hand which strategy of his to choose, so that he reaches a PNE, all it means is that from any non-PNE, players can find an improvement path to a PNE (as you can check the game is weakly acyclic.)
$endgroup$
– Kaind
Jan 23 at 10:40
$begingroup$
Right, thanks a lot for your answer.
$endgroup$
– Orpheus
Jan 23 at 10:41
add a comment |
$begingroup$
Model this as the following strategic form game $(N , S_i , u_i )$
$N = 1700$
$S_i = { B, DB } forall i$, where B represents player $i$ buying a ticket and DB represents player i not buying a ticket.
$u_i(s) = begin{cases} 0 &text{ if } s_i = DB \ (frac{1}{N_B}*1000) - 1 &text{ if } s_i = B end{cases}$, where $N_B$ represents the number of players buying the ticket i.e. $N_B = { i in N | s_i = B } $.
There are $binom{1701}{1000} $ total pure Nash equilibria strategies here:
$binom{1700}{1000} $ strategies corresponding to - when exactly 1000 players buy the ticket and $binom{1700}{999} $ strategies corresponding to - when exactly 999 players buy the ticket.
Calculation for utility value when player $i$ buys ticket:
Now since the lottery ticket winner is picked randomly, player $i$ gains money $ 1000-1 = 999$ $ with probability $frac{1}{N_B}$ and gains money $ -1 $ $ (i.e. loses a dollar) with probability $frac{N_B - 1}{N_B}$. Hence his utility in this regard is the weighted mean of the money he gains with the weights being the probabilities, hence:
$u_i(s) = frac{1}{N_B} * 999 + frac{N_B - 1}{N_B} * (-1) = frac{1000}{N_B} - 1$
answered Jan 23 at 9:35
KaindKaind
749414
$endgroup$
1
$begingroup$
Sorry, I am a bit confused by your answer. Why does the strategy B weakly dominates the strategy DB? Doesn't it depend on how large $N_B$ is?
$endgroup$
– Orpheus
Jan 23 at 10:06
$begingroup$
You're right, I made an error while computing it; I took N as 1000 in my calculations earlier. Edited now.
$endgroup$
– Kaind
Jan 23 at 10:13
$begingroup$
This makes more sense, however I am still confused. Doesn't your answer imply that there is no difference between the situation where every participant knows how many others bought a ticket before deciding whether to buy a ticket himself and the situation where he doesn't know that?
$endgroup$
– Orpheus
Jan 23 at 10:27
$begingroup$
There is a big difference; All players decide to buy a ticket independant of the other's opinion i.e. without knowing the other's strategy. Note that these strategy profiles are just Pure nash equilibria, it doesn't mean that a player knows before hand which strategy of his to choose, so that he reaches a PNE, all it means is that from any non-PNE, players can find an improvement path to a PNE (as you can check the game is weakly acyclic.)
$endgroup$
– Kaind
Jan 23 at 10:40
$begingroup$
Right, thanks a lot for your answer.
$endgroup$
– Orpheus
Jan 23 at 10:41
add a comment |
$begingroup$
Model this as the following strategic form game $(N , S_i , u_i )$
$N = 1700$
$S_i = { B, DB } forall i$, where B represents player $i$ buying a ticket and DB represents player i not buying a ticket.
$u_i(s) = begin{cases} 0 &text{ if } s_i = DB \ (frac{1}{N_B}*1000) - 1 &text{ if } s_i = B end{cases}$, where $N_B$ represents the number of players buying the ticket i.e. $N_B = { i in N | s_i = B } $.
There are $binom{1701}{1000} $ total pure Nash equilibria strategies here:
$binom{1700}{1000} $ strategies corresponding to - when exactly 1000 players buy the ticket and $binom{1700}{999} $ strategies corresponding to - when exactly 999 players buy the ticket.
Calculation for utility value when player $i$ buys ticket:
Now since the lottery ticket winner is picked randomly, player $i$ gains money $ 1000-1 = 999$ $ with probability $frac{1}{N_B}$ and gains money $ -1 $ $ (i.e. loses a dollar) with probability $frac{N_B - 1}{N_B}$. Hence his utility in this regard is the weighted mean of the money he gains with the weights being the probabilities, hence:
$u_i(s) = frac{1}{N_B} * 999 + frac{N_B - 1}{N_B} * (-1) = frac{1000}{N_B} - 1$
answered Jan 23 at 9:35
KaindKaind
749414
$endgroup$
Model this as the following strategic form game $(N , S_i , u_i )$
$N = 1700$
$S_i = { B, DB } forall i$, where B represents player $i$ buying a ticket and DB represents player i not buying a ticket.
$u_i(s) = begin{cases} 0 &text{ if } s_i = DB \ (frac{1}{N_B}*1000) - 1 &text{ if } s_i = B end{cases}$, where $N_B$ represents the number of players buying the ticket i.e. $N_B = { i in N | s_i = B } $.
There are $binom{1701}{1000} $ total pure Nash equilibria strategies here:
$binom{1700}{1000} $ strategies corresponding to - when exactly 1000 players buy the ticket and $binom{1700}{999} $ strategies corresponding to - when exactly 999 players buy the ticket.
Calculation for utility value when player $i$ buys ticket:
Now since the lottery ticket winner is picked randomly, player $i$ gains money $ 1000-1 = 999$ $ with probability $frac{1}{N_B}$ and gains money $ -1 $ $ (i.e. loses a dollar) with probability $frac{N_B - 1}{N_B}$. Hence his utility in this regard is the weighted mean of the money he gains with the weights being the probabilities, hence:
$u_i(s) = frac{1}{N_B} * 999 + frac{N_B - 1}{N_B} * (-1) = frac{1000}{N_B} - 1$
answered Jan 23 at 9:35
KaindKaind
749414
edited Jan 23 at 10:18
answered Jan 23 at 9:35
KaindKaind
749414
answered Jan 23 at 9:35
KaindKaind
749414
answered Jan 23 at 9:35
KaindKaind
749414
749414
1
$begingroup$
Sorry, I am a bit confused by your answer. Why does the strategy B weakly dominates the strategy DB? Doesn't it depend on how large $N_B$ is?
$endgroup$
– Orpheus
Jan 23 at 10:06
$begingroup$
You're right, I made an error while computing it; I took N as 1000 in my calculations earlier. Edited now.
$endgroup$
– Kaind
Jan 23 at 10:13
$begingroup$
This makes more sense, however I am still confused. Doesn't your answer imply that there is no difference between the situation where every participant knows how many others bought a ticket before deciding whether to buy a ticket himself and the situation where he doesn't know that?
$endgroup$
– Orpheus
Jan 23 at 10:27
$begingroup$
There is a big difference; All players decide to buy a ticket independant of the other's opinion i.e. without knowing the other's strategy. Note that these strategy profiles are just Pure nash equilibria, it doesn't mean that a player knows before hand which strategy of his to choose, so that he reaches a PNE, all it means is that from any non-PNE, players can find an improvement path to a PNE (as you can check the game is weakly acyclic.)
$endgroup$
– Kaind
Jan 23 at 10:40
$begingroup$
Right, thanks a lot for your answer.
$endgroup$
– Orpheus
Jan 23 at 10:41
add a comment |
1
$begingroup$
Sorry, I am a bit confused by your answer. Why does the strategy B weakly dominates the strategy DB? Doesn't it depend on how large $N_B$ is?
$endgroup$
– Orpheus
Jan 23 at 10:06
$begingroup$
You're right, I made an error while computing it; I took N as 1000 in my calculations earlier. Edited now.
$endgroup$
– Kaind
Jan 23 at 10:13
$begingroup$
This makes more sense, however I am still confused. Doesn't your answer imply that there is no difference between the situation where every participant knows how many others bought a ticket before deciding whether to buy a ticket himself and the situation where he doesn't know that?
$endgroup$
– Orpheus
Jan 23 at 10:27
$begingroup$
There is a big difference; All players decide to buy a ticket independant of the other's opinion i.e. without knowing the other's strategy. Note that these strategy profiles are just Pure nash equilibria, it doesn't mean that a player knows before hand which strategy of his to choose, so that he reaches a PNE, all it means is that from any non-PNE, players can find an improvement path to a PNE (as you can check the game is weakly acyclic.)
$endgroup$
– Kaind
Jan 23 at 10:40
$begingroup$
Right, thanks a lot for your answer.
$endgroup$
– Orpheus
Jan 23 at 10:41
1
1
$begingroup$
Sorry, I am a bit confused by your answer. Why does the strategy B weakly dominates the strategy DB? Doesn't it depend on how large $N_B$ is?
$endgroup$
– Orpheus
Jan 23 at 10:06
$begingroup$
Sorry, I am a bit confused by your answer. Why does the strategy B weakly dominates the strategy DB? Doesn't it depend on how large $N_B$ is?
$endgroup$
– Orpheus
Jan 23 at 10:06
$begingroup$
You're right, I made an error while computing it; I took N as 1000 in my calculations earlier. Edited now.
$endgroup$
– Kaind
Jan 23 at 10:13
$begingroup$
You're right, I made an error while computing it; I took N as 1000 in my calculations earlier. Edited now.
$endgroup$
– Kaind
Jan 23 at 10:13
$begingroup$
This makes more sense, however I am still confused. Doesn't your answer imply that there is no difference between the situation where every participant knows how many others bought a ticket before deciding whether to buy a ticket himself and the situation where he doesn't know that?
$endgroup$
– Orpheus
Jan 23 at 10:27
$begingroup$
This makes more sense, however I am still confused. Doesn't your answer imply that there is no difference between the situation where every participant knows how many others bought a ticket before deciding whether to buy a ticket himself and the situation where he doesn't know that?
$endgroup$
– Orpheus
Jan 23 at 10:27
$begingroup$
There is a big difference; All players decide to buy a ticket independant of the other's opinion i.e. without knowing the other's strategy. Note that these strategy profiles are just Pure nash equilibria, it doesn't mean that a player knows before hand which strategy of his to choose, so that he reaches a PNE, all it means is that from any non-PNE, players can find an improvement path to a PNE (as you can check the game is weakly acyclic.)
$endgroup$
– Kaind
Jan 23 at 10:40
$begingroup$
There is a big difference; All players decide to buy a ticket independant of the other's opinion i.e. without knowing the other's strategy. Note that these strategy profiles are just Pure nash equilibria, it doesn't mean that a player knows before hand which strategy of his to choose, so that he reaches a PNE, all it means is that from any non-PNE, players can find an improvement path to a PNE (as you can check the game is weakly acyclic.)
$endgroup$
– Kaind
Jan 23 at 10:40
$begingroup$
Right, thanks a lot for your answer.
$endgroup$
– Orpheus
Jan 23 at 10:41
$begingroup$
Right, thanks a lot for your answer.
$endgroup$
– Orpheus
Jan 23 at 10:41
add a comment |
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$begingroup$
It seems that the prize will be given even if only one ticket is sold.
$endgroup$
– user
Jan 23 at 9:19
$begingroup$
Yes, this is correct. And?
$endgroup$
– Orpheus
Jan 23 at 9:28