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?

  • $\begingroup$ 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

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: $u_i(s) = \frac{1}{N_B} * 999 + \frac{N_B - 1}{N_B} * (-1) = \frac{1000}{N_B} - 1$

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    $\begingroup$ 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

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