0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.