First of all, congratulations to the winners. Lets try to collect here the final data of the jackpot run: Reminder, the jackpot prize comes from an extraction of 5 numbers from 69 plus another of 1 number between 26. This was chosen purposefully last year, to produce this kind of big prizes once in the yer. (69 5) times 26 is 292 201338, ticket price is $2 and you can expect a return from minor prizes setting the price to a lower $1.58. The number of players was unknown, but from the increase of the jackpot it can be napkin-guessed to be around 635M, (no idea why the estimation of wired is a lot lower, say 170M, and on the other hand (1500-947.9)/0.34/2 is even greater, 800M. Perhaps it should be estimated from the cash-payout, which is a 62% of the jackpot?). There has been, or it is claimed so, three winners that will share the big prize. The number of players increases faster than linear with the size of the jackpot, fitting well with quadratic (!) or exponential function.

Now, back to the game-theoretical considerations of yesterday.

We assume a god-given jackpot, a “resource” to be harvested, having always the same value, and players having no memory of previous winners, so that they only can choose between buying or not the offered tickets. They can do it by having a internal preference q1,q2,… between 0 and 1 and drawing a random number to decide.

One point of the lottery is that our result if we do not play is simply zero, not losses -except moral ones if some friend wins, of course-. In the general formula for the expected win, with p our own mixed strategy,

p f(q1,q2,q3,…) + (1-p) g( q1,q2,q3,…)

the function g is zero. This means that whatever mixed strategy the other players choose, the total expected win will always be proportional to our own probability p of playing. This also means that if f(…) were non zero, we should always counter by moving to a pure strategy, either p=1 or p=0. So this kind equilibrium happens only when p is cancelled out because f(…) itself is equal to g(…) ; in this case g() is zero, and then the expected benefit in the long run for all the players will be zero. Generically the benefit will be exactly the same as the pure strategy of not playing at all.

Another argument for not to play is to minimize your expected loss if you are completely unsure of the other partners strategy: if g(…) is zero but f(q1,q2,q3…) can be less than zero, then p=0 is your escape option.

Of course all of this is a tragedy because a god-given jackpot could be harvested. Ideal goal is to coordinate all players to buy only exactly the 292 201338 combinations, and share the prize. The question is, can it be done, or approximated, without a fully forceful central coordination? I leave it open.

What about populating the lotto world with hawks and doves? This is, the players can be either fanatic, always play for this amount of jackpot, or pessimistic, never playing. A initial population of doves could gradually mutate to hawks in order to extract some benefit for the resource, but at some point the probability of fighting hawk vs hawk should stop the mutation and keep a fixed level of players against no players. Is this point also given by the same equilibrium that above? If so, is the total resource extraction again zero? I have run some simulations and it seems that the proportion of hawks evolves to be slightly smaller that the expected point of zero benefit, but still near enough to attribute the discrepancy to finite size effects, and finite number of trials. Of course, that implies another philosophical question, given the fact that real games have a finite number of trials, should we recalculate for only a few trials?

There is, to me, some lack of intuition here. One could think that the idea in a lottery is to keep adding players until the probability of benefit becomes zero again, and that in hawks and doves we keep turning doves into hawks also in the same way, until the benefit extracted from the game becomes zero. This seems consistent with setting to zero the above equation, for the case g(q) =0, f(q)= q (P/2-T) + (1-q) (P-T), is

p (P-T) + p q (-P/2)

and then f(…) =0, exhausting but if there is some small prize or penalisation K when we do not play, the intuition fails, as then we need the partial respect to p of

p f(q1,q2,q3,…) + (1-p) K

and we get f(…) = K and the total expected benefit equal to K, not zero. So in principle if K is still positive there is room to turn doves to hawks even beyond the equilibrium point. The general solution q (P/2-T) + (1-q) (P-T) = K is q= 2 (1-T/P) – 2K/P