# Equilibrium in a Lottery

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?

• It seems that the prize will be given even if only one ticket is sold. – user Jan 23 at 9:19
• Yes, this is correct. And? – 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: $$u_i(s) = \frac{1}{N_B} * 999 + \frac{N_B - 1}{N_B} * (-1) = \frac{1000}{N_B} - 1$$

• 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? – Orpheus Jan 23 at 10:06
• You're right, I made an error while computing it; I took N as 1000 in my calculations earlier. Edited now. – Kaind Jan 23 at 10:13
• 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? – Orpheus Jan 23 at 10:27
• 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.) – Kaind Jan 23 at 10:40
• Right, thanks a lot for your answer. – Orpheus Jan 23 at 10:41