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