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Nigerian Scammers Claim Another Victim
A Florida newspaper ran a story yesterday about a local retiree who fell hard for a 419 scam. The story goes into depth on the methods used to play on the target's beliefs and gain his confidence - in this case, the target (who lost $320,000) is still having a hard time accepting that they were thieves. Truly remarkable.
The Register used to do a great 419 T-shirt, but you can still get an "All my money went to Nigeria and all I got was this lousy t-shirt".
If the jackpot is large enough, each ticket bought that week will have an average return higher than the cost of the ticket. However, that fact alone does not make it a good bet, because the variance overwhelms the positive expectation.
Uhm...you aren't familiar with the formula
?You should have got this in the first week of your first statistics course. :-)
Or maybe you aren't familiar with how lottery prizes work? If no one wins, the prize gets added to the next lottery.
The probability of a given ticket winning is the same each week, but the prize goes up. However, more people tend to play as the prize goes up, so the chances of sharing the prize go up, but the net effect is the "prize if you win" term goes up each week.
If a lottery goes a few weeks without a winner, this can push the expected payoff higher than the price of a ticket.
The grass isn't necessarily greener on the other side of the fence
http://www.419eater.com/index.htm
Ok, I'm bored, so here's how I figure out when to play the lottery. First the main principle is that of the "progressive jackpot". That means that people play some game and if nobody wins, a pot grows and grows. Normally when playing poker or something with a progressive jackpot, you can't just sit down halfway through and start playing because you didn't put any money into the game. The lottery OTOH lets you start playing after many people have put their money in and lost. I think this is the same idea behind those teams that beat video poker by waiting for a huge jackpot, then playing all of the machines for days until they win the jackpot.
.6*.6 of the money, which is 36 percent, or roughly 1/3 of the total (About N dollars). Then you have the problem of if there's a 1/N chance of winning and 2N tickets are bought for that drawing, you will average 2 winners, and it could be more, but it could be less, so your real expected payoff is more like P(1 winner)*.36(Jackpot = J) + P(2 winners)*(.36(J)/2) + P(3 winners)*(.36(J)/3)...and so forth. It's a binomial distribution with p = 1/N, q = ((N-1)/N), so your P(X = 0) is ((N-1)/N)^2N, P(X=1) = C(2N,1)(1/N)((N-1)/N)^(2N-1), P(X=2) = C(2N,2)(1/N)^2((N-1)/N)^(2N-2).... All of the ((N-1)/N) terms are roughly the same and we can call them K, and we can simplify the combinations (By assuming that C(2N,P) is roughly (2N)^P/P!) for large N and small P to get
) +( 1/10)*(J/6)...)/7
.4.
If the lottery gives you a 1/N chance of winning the big prize per dollar ticket, and the jackpot is about 3N, then the tickets start getting worth it. Start with the 3N. First, they take away about 40 percent of the money if you pick the "lump sum" (you should consider this important since you pay for the tickets now and don't get to pay for your tickets over 25 years...), then you have taxes which will be about 40 percent of what's left, so you're looking at
P(X=P) = C(2N,P)(1/N)^P((N-1)/N)^P which is approx
(2^P)(N^P)/P!(1/N)^P(K) = (2^P)/P!(K), and the K is essentially constant over all P, so we can ignore it, so the P(X=P) is proportional to (2^P)/P!.
I will ignore the 0 winners case, since then you get a chance to play again next week, But the constants for the other numbers are : C1 = 2, C2 = 4/2 = 2, C3 = 8/6 = 4/3, C4 = 16/24 = 2/3, C5 = 32/120 ~= 1/4, C6 = 64/720 ~= 1/12, and it keeps going down.
Add those numbers up and make the last one a 1/10 or so to take care of the other numbers, and you see that the total is about 6.5 or 7. You have essentially a 2/7 chance of being the sole winner, a 2/7 chance of being a half winner, and so on, so your real expected value will look more like
(2*J+2*(J/2)+(4/3)*(J/3)+(2/3)*(J/4)+(1/4)*(J/5
which is approx ((2+1+4/9+1/6+1/20...)/7)*J ~= (1/2)*J, so you're looking at about half the jackpot being yours (ignoring the 0 winner case which lowers it even more to about
So, on top of the taxes and persent value which eat away about 2/3 of the value of the jackpot, the other winners make your jackpot about half or less of its value beyond that, so we're looking at about a 15-18 percent return on the actual "dollar value" of the jackpot. I tend to play when the jackpot is 3N where the chance of winning is 1/N, since I like poker and this situation only comes up every few years, but to take everything into account, you should wait until the jackpot is about 6N,. The only problem is that I was assuming 2N tickets bought for the current drawing, and if those numbers go way up, then the expected size of the jackpot keeps going down due to more players. So, I guess it will never be perfect, but it's nice to have better odds if you're going to play and the little prizes increase the expected value, as well, so it might be worth playing once in a while. And no I never won except for the little stuff.
Best. Comment. Ever. Enjoy!