Ok, the powerball lottery of tomorrow, Jan 13th, 2016. Some of the math has been already done here and there. So I was thinking another approach: Lets assume that a couple of Forbes-listed guys likes to play lotto in this favorable situation.
Think Gates and Buffet, or any of your preferred top ten or whatever corporations they own, or the corporations themselves (This is also considered by Andy Kiersz).
According numbers in «is-the-powerball-lottery-profitable«,
the total ticket cost is, at 2$ each number, $584M -$93.4 (prices) – $175M (self-jackpot) = 280M. And the jackpot price is 1.3 billion. Share between both players, it is still 650M
IRS tax is not problem for them, the corporations have an army of lawyers taking care of it, it is irrelevant for them, and same with other expenses as buying some lottery franchise, workers, management etc.
So should they play? Well, obviously yes if they are the only two players. They are still going to get $370M worst case. This is not very interesting.
What about two weeks ago, when the prize was only $528M? Sharing the price they get $264M, less than the cost of the, er, «ticket». So they are into a sort of
snowdrift or chicken game, (or hawks, doves or whatever bird you like):
G Passes G buys ticket B Passes 0, 0 0, 248M B Buys 248M, 0 -16M, -16M
(the numbers come from the differences 528-280=248 and 264-280=-16)
With pure strategies there are two situations with Nash equilibrium: either (G Buys, B Passes) or the reciprocal. In such situations, nobody has
incentive to change his move. But this fact is not help to decide what to play, specially
in a situation without memory of the previous games.
There is also a mixed equilibrium. Consider the
mixed strategies where G and B decision of buying the game has a
probability p, q, respectively. So G’s expectation of benefit is
p(1-q) 248M – p q 16M
and if B sets his mixed strategy to q= 248/264 = 0.9394 then G’s expectation is
zero independently of q. And same exchanging the roles.
Warning: I have an issue here: if I want to look for evolutionary stable strategies, I’d think I should set p=q, differentiate the above expectation, and look for extremes, and then I get p=q=124/264=0.4697. Obviously this value of (p,q) is not a Nash equilibrium because one player can move p fully up to 1.0 and get a benefit from it. But it is better: in the long run it gets about 58M of benefit for each player in average, instead of zero. Cooperation emerging from symmetry?
So, should G and B just throw a random number and then decide to play only if this
number is less than 0.9429? At least it looks rational.
Does it get more interesting with more players? Not a lot more, because the realistic number of tickets sold is estimated to be equivalent to three or four billonaire players. Well, lets see For a 1.3B jackpot to be unprofitable, we
need at least five, so that 1300/5 – 280 = -20 < 0. Lets say we have
other three players from the top 20 (hey, looking at the list there are some
math-minded guys in fact: Brin, Page, Bezos… rational enough to proceed
via game theory :-). Regrettably the calculation of the mixed strategy
becomes more complicated: we need to consider five separate probabilities
and the fact that we have situation with zero to five people sharing
the jackpot.
I think we could do a guess by assuming that the equilibrium solution sets everybody
to an expectation of zero losses (and zero benefit). This really works in the
previous case, setting p=q and asking p ( 248M – p *260M) = 0, but I am not sure about generalizing. Well, if we do, we ask
0= p^5 (1300/5 – 280)
+ 4 p^4(1-p) (1300/4 – 280)
+ 6 p^3(1-p)^2 (1300/3 – 280)
+ 4 p^2(1-p)^3 (1300/2 – 280)
+ p (1-p)^4 (1300 – 280)
to wolfram alpha, we see that there is only a root between zero and one:
p = 0.92857…
So the mixed strategy could be to play only if rnd() < 0.92857. Of course if the other four players adhere to this strategy, it is true that the extant player gets zero losses (and zero benefit) for any mixed strategy he chooses.
Again, consider two weeks ago, when we only had $528M in the jackpot. Then
0= p^5 (528/5 – 280)
+ 4 p^4(1-p) (528/4 – 280)
+ 6 p^3(1-p)^2 (528/3 – 280)
+ 4 p^2(1-p)^3 (528/2 – 280)
+ p (1-p)^4 (528 – 280)
gives p=0.32384…
I am not sure what those solutions mean. It is clear that if the players were able to
coordinate they could choose to form a cartel, only one of them playing each
week and thus extracting the same benefit but only at the cost of one «full ticket». If
they are able to evolve to such understanding, this is a topic for
Evolutionary Game Theory. But then, as I said in the red warning above, we perhaps want to define stability from the derivative of the expectation function across the diagonal.
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