October 26, 2005
NY lottery expectations
The jackpot in the New York lottery has now reached $147 million. On top of
that, there are smaller sums ranging from $2 to $250,000 which can also be won,
at various different odds as delineated here.
A quick and dirty way of working out how much you can expect to win or lose
is to simply divide the winnings by the odds of winning them, and add them all
result? An expected profit of just under 2 cents on every dollar
"invested" in a lottery ticket.
Of course, that doesn't mean that a lottery ticket is a remotely sensible investment.
There are taxes on winnings, for one thing, and in fact if you want cash you
don't win $147 million at all, but something much smaller. Plus there's the
possibility you'll have to split the jackpot with someone else even if you do
win it. And I suspect there's a bit of double-counting in the calculation: if
you get four numbers right, for instance, you only get the $150 prize for that,
and not an extra $28 for the three different ways in which you got three numbers
right. But that might be reflected in the odds, I'm not sure.
Still, it's a rare enough occurrence that a lottery even nominally has a positive
expected payout. I reckon there's going to be a hell of a lot of tickets sold
in the next couple of days.
at 04:43 PM GMT
" I reckon there's going to be a hell of a lot of tickets sold in the next couple of days."
Which of course means that your expected earnigs go down, as the likelihood that you have to share the jackpot with 1 or more other people goes up proportionally.
If there are twice as many people playing this week as the result of this news, expected returns will halve. Which goes too show that in a prefect information is no deterrent for irrational people.
You should stand on the street and tell people you will offer them a better expected return for their dollar than the lottery this week... Offer them 60c on every dollar they give you, in cash right now. Anybody's who's bought a ticket should take you up on it.
Posted by: Stefan Geens on October 26, 2005 06:17 PM
Tuesday's drawing, which also had a nine-figure jackpot, had about 30 million tickets sold. (There were 766,544 winning tickets, and there's a one in 39.89 chance that any given ticket will win something.) Even if this drawing goes up to 40 million tickets, the chances of two tickets both managing the one-in-175 million probability of hitting the jackpot is pretty slim. Even if the number of players doubles, the expected return will not halve, since the chance of having to share your jackpot will remain very small.
By the way, tickets for this lottery aren't only sold in New York: they're also sold in California, Georgia, Illinois, Maryland, Massachusetts, Michigan, New Jersey, Ohio, Texas, Virginia and Washington.
Posted by: Felix on October 26, 2005 06:54 PM
"Even if this drawing goes up to 40 million tickets, the chances of two tickets both managing the one-in-175 million probability of hitting the jackpot is pretty slim. "
yes, but not very much less slim than the chance of exactly one ticket hitting the jackpot. If you double the number of tickets sold, you halve the expected value. basic maths.
Posted by: Stefan Geens on October 26, 2005 07:05 PM
This isn't a raffle, Stefan, where tickets are pulled out of a hat. In that case, if the number of tickets doubles, then the probability of winning halves. But in a lottery, it doesn't work like that.
Here's a thought experiment, simplifying to make it easier for you to understand. Same lottery, with chances of hitting the jackpot one in 175 million. Let's say there are no other prizes, and the jackpot is $147 million. And let's say one ticket is sold. The expected value of that ticket is about 84 cents. Now let's say a second ticket is sold. Are you seriously saying that because the number of tickets outstanding has doubled, the expected value of the first ticket has dropped to 42 cents? And that by the time the 128th ticket is sold, the expected value of each ticket is less than one cent?
Posted by: Felix on October 26, 2005 07:58 PM
I was wrong. But so is Felix. We need to simplify further
If there was only one $1 ticket sold, the actual prize if you hit the jackpot would only be $1, assuming no profit motive. If the chance of winning is one in 10, then your expected value is 10 cents.
Now assume that a second ticket is sold before the lottery is drawn. The prize money doubles to $2. The chance of each individual ticket winning is still 1/10. Is the expected value 20 cents?
No, because there is the possibility that both tickets win and share the prize, in addition to exactly one or none.
Chance of no ticket winning: 9/10 * 9/10 = 81% (coz these events are unrelated, like coming up with a random number). Payout = 0
Chance of 2 winning tickets: 1/10 * 1/10 = 1%. Payout = $1 to each ticket.
Chance your ticket winning but not the other one: 1/10 * 9/10 = 9%. Payout = $2 to the winning ticket. (This has an equal 9% chance of happening to the other ticket.)
Expected value for $1 "invested" is then just 0*0.81 + 1*0.01 + 2*0.09 = 19 cents.
That difference might appear negligible, and it is if the number of tickets sold is far below the odds of winning. But what if the number of tickets sold is the same as the odds of winning, i.e. 10 $1 tickets are sold for a $10 pot, chance of winning still 1/10.
Chance of no tickets winning: (9/10)^10 = 34.9% Payout = 0
Chance of 1 specific ticket winning but no other ones: 1/10 * (9/10)^9 = 3.87%. Payout = $10
Chance of 1 specific ticket winning and just one other: (1/10)^2 * (9/10)^8 * 9 = 3.87%. Payout $5 to each ticket
Chance of 1 specific ticket winning and 2 others: (1/10)^3 * (9/10)^7 * (9 * 8) / (2 * 1)= 1.72%. Payout = $3.33 each
1 Specific, 3 others: (1/10)^4 * (9/10)^6 * (9 * 8 * 7) / ( 3 * 2 * 1) = 0.45%. Payout $2.50 each
1 Specific, 4 others: (1/10)^5 * (9/10)^5 * (9! / 5!) / 4! = 0.074%. Payout $2.50 each
yadda yadda yadda
Chance of 1 specific, 7 others winning: (1/10)^8 * (9/10)^2 * (9! / 2!) / 7! = 0.000029%. Payout $ 1.25 each
Chance of 1 specific, 8 others winning: (1/10)^9 * 9/10 * (9! / 1!) / 8! = 0.00000081%. Payout $1.11 each
Chance of all 10 winning: (1/10)^10 * (9! / 0!) / 9! = 0.00000001%. Payout $1 each
So what can we tell from this?
Trivially, that your expected value is just a little over 65 cents. But more important is the fact that the contribution to expected value from the chance that you have to share your prize with precisely one other winning ticket is exactly half that from the chance that you win it all by yourself (assuming that the odds of winning are (1 / the number of tickets sold), not that it's a raffle). This is a fact, and it is independent from the odds (again, as long as it is equal to the number of tickets sold).
In a real life scenario, like the NY lottery where the number of tickets sold approaches the odds, the contribution to expected value is not negligible. Specifically, it is not true that "the chance of having to share your jackpot will remain very small." So Felix, you are wrong.
Posted by: Stefan Geens on October 26, 2005 10:53 PM
You guys are total nerds. And po' ones at that. Hurry up and buy those tickets.
Posted by: michelle on October 26, 2005 11:32 PM
Stefan, don't try to blind me with science. What's all this bullshit about if only one $1 ticket is sold then then jackpot is $1? Why would anybody buy a ticket in that case? I can tell you for a fact that offering 1 in 175 million odds of winning $147 million is a good business, no matter how many tickets are sold- whether you're selling 1 ticket or 100 million of them.
What's more, it seems in your case that the expected return if one ticket is sold is 10 cents, while the expected return if 10 tickets are sold is 65 cents. The return is going up, not down.
Posted by: Felix on October 27, 2005 04:42 AM
A 1 in 175 million odds of winning $147 million is or is not a good business depending on how you see it:
1) You buy one ticket (at $1). You stand a 174,999,999 chance in 175,000,000 of losing your dollar. You do however also have the 1 in 175 million chance of having all your hedonistic fantasies come true. While this is a tiny probability, it is at least non-zero. Personally, I can afford a dollar (or equivalent) every now and then to keep the dream of international playboyery alive.
2) You buy 87,500,000 tickets. You then have a 1 in 2 chance of making a $59,500,000 profit. Otherwise you've lost $87.5 million. Now, if I could afford to lose $87.5 million, I think I'd rather spend them on nice things than risk losing them for the sake of another $60 million.
Oh... is this a lottery where they draw random numbers and match them against tickets? (rather than drawing one-to-one matched tickets) Then never mind. I just wanted to make the point that expected value isn't that useful unless you buy lots of tickets and have lots of different prizes.
Posted by: Simon on October 27, 2005 11:01 AM
Felix, really, where are you going to get the money from to pay the winnings if not from ticket sales? You only have so much money to give away as you make through selling tickets.
And I've proven that the number of tickets you sell makes a big difference to expected value. In the case of 1 ticket, 1/10 chance of winning and $10 jackpot, the expected value is obviously $1, but in the case of 10 tickets 1/10 chance of winning and $10 jackpot, the expected value is 65 cents because of the likelihood you will have to share your winnings. It isn't half, but it's getting there.
Posted by: Stefan Geens on October 27, 2005 04:27 PM
I've found a proof to explain why your math is completely fucked up but I don't have the time to explain it.
Posted by: Sterling on October 27, 2005 04:40 PM
Stefan, you can get the money from anywhere. Your bank account, for one, if you're that rich. More likely, from an insurance company. Basically, you insure yourself against the 1-in-175 million chance that you'll have to pay out $147 million. Once you've done that, at a cost of say 90 cents per ticket, the other 10 cents are pure profit for yourself, no matter how many tickets you sell. There are lots of sources of money, not just ticket sales.
Posted by: Felix on October 27, 2005 04:45 PM
There were 766,544 winning tickets for Tuesday's drawing, and 889,071 winning tickets for Friday's drawing, which implies that the number of tickets sold increased by about 16%. Now that the jackpot has gone up to $165 million, is there going to be another increase? Do people care about the difference between $147 million and $165 million?
Posted by: Felix on October 29, 2005 03:24 PM
There are smarter ways of playing the lottery. I would like to point out a very cool program that I am a member of. Basically, you can get paid to play the lottery which is VERY cool. If you check out LottoMagik.com and click on ìIncomeî it explains everything. To be a captain sounds expensive ($50 a month) but with just two team captain referrals you are already getting paid to play the lottery every month. Plus, since you are on a team, you increase your odds of winning. Iím not saying you have to join, or that you even have to give it the time of day. Iím just letting you know that there are options out there for playing the lottery smartly.
Posted by: Michael on October 29, 2005 06:49 PM
Michael , really now. I just read the FAQ. It's a pyramid scheme coupled to a lottery pool where a percentage is skimmed off your winnings. So while you share the winnings of 8 times as many tickets, your only get 10% of each ticket, leaving you with an expected value that is 80% of what you'd get it you played by yourself. And then there is the pyramid scheme... ISn't it just simpler to sell your friends' emails to a spammer?
Playing the lottery is dumb. But then there are dumber ways of playing the lottery.
Posted by: Stefan Geens on October 29, 2005 07:07 PM
BTW I'm retracting my concession of wrongness earlier. I'm actually correct, with the proviso that the number of tickets sold approaches infinity.
For example, to get back to our case where there tickets are $1, the chance of winning is 1/10 and they payout is $10. If 100 tickets were sold, we can expect for every possible combination to be "hit" on average 10 times. The expected value here is hence intuitively near 10 cents. Sell 200 tickets, however, and the expected value will be nearer 5 cents, as each combination will on average be hit 20 times. Because of all the permutations that are possible regarding the number of winners, and the skewed probabilities for each of these happening (it's not a bell curve), expected value E as a function of total payout P, the number of tickets sold N and the number of possible combinations X (of which only one is the winner in this example) is E -> P / (N . X), as N -> infinity, or more precisely, as N / X -> infinity.
Another way of putting it: Expected value approaches total payout times the odds of winning divided by the number of tickets sold as the number of tickets sold times the odds approaches infinity.
Posted by: Stefan on October 29, 2005 07:56 PM
Stefan, who are you trying to kid here? In comment #5, the jackpot is the same as the number of tickets sold. In comment #15, the jackpot stays constant as the number of tickets sold goes up. Which is it going to be?
Posted by: Felix on October 29, 2005 10:53 PM
In comment 5, I was trying to model a situation similar to the one in real life, where the $ amount of tickets sold equals the prize money, but at the same time use a small number so I could make the calculations and derive a pattern and hence an equation. In #15, I generalize from that specific case.
Posted by: Stefan Geens on October 29, 2005 10:58 PM
Explain, please. In #5, each additional ticket sold adds to the jackpot. In #15, each additional ticket sold adds nothing to the jackpot. In what way is #15 a generalization of #5?
Posted by: Felix on October 29, 2005 11:06 PM
Stop it. These are models. We can make them do whatever we want. the size of the jackpot is irrelevant. I made it go up at first to make the model simpler.
Posted by: Stefan on October 29, 2005 11:33 PM
The size of the jackpot is irrelevant???? Of course it's bloody relevant. If the jackpot increases proportionally to the number of tickets sold, then your theory about expected value decreasing with the number of tickets sold goes straight out the window. Basically, you're talking about a situation in which the jackpot is fixed, and the number of tickets sold is significantly greater than the inverse of the probability of winning. Which is a state of affairs which exists in precisely no lottery anywhere in the world. Some model.
Posted by: Felix on October 29, 2005 11:41 PM
It's irrelevant in the sense that the expected value is directly proportional to the total payout in all calculations, which makes it the trivial part of calculating expected value. What's far more complicated is effect of changing the total number of possible outcomes X and the number of tickets sold N.
What we can tell, however, is that as N becomes bigger compared to X, the the relationship between expected value and N becomes increasinly inversely proportional. Which is another way of saying that if you double the number of tickets sold, you halve the expected value.
I went and confirmed this on Mathworld. What happens as N increases with X fixed is that we move from a Poisson distribution for the likely number of winning tickets to a standard binomial distribution.
Posted by: Stefan Geens on October 30, 2005 11:50 AM
Stefan, you don't even need to increase the number of variables. Just put jackpot=N, and see what happens. I'm telling you, your theory goes up in a puff of smoke.
Posted by: Felix on October 30, 2005 03:35 PM
Stefan, all members only receive 10% of the ticket winnings from any ticket that Lotto Magic purchases for them in the Florida Lottery. The other 70% is divided up between the 7 members that are on your team. 10% goes to your sponsor; 10% goes to their sponsor. This totals 100%. Lotto Magic does not receive any share of the ticket winnings.
Posted by: Michael on October 31, 2005 05:44 AM
A very Belgian thread, this.
Posted by: Claude de Bigny on October 31, 2005 07:54 AM
Michael, while Stefan has been completely wrong in nearly every comment he's left on this thread, he was completely right about you. Lotto Magic is a pyramid scheme, and frankly the only reason I didn't delete your comment as spam was that I have enough faith in MF's readers that I'm sure none of them would ever fall for it.
But just in case you care, Stefan never accused LM of receiving a share of the ticket winnings -- it doesn't need to, since it's making more than enough with its share of the monthly fees. That said, since I'm sure the owner/founders of LM are at the top of the pyramid, they get a significant chunk of the ticket winnings themselves in any case.
Posted by: Felix on October 31, 2005 02:11 PM
Felix, I just don't understand what is wrong with the LM program? What constitutes a pyramid scheme? And... if it is a pyramid scheme, how is it possible that it has been in business since 1996 while maintaining its membership in the chamber of commerce and BBB? I'm not trying to say you are wrong, or that I am right... but it is a cool program. The referral commissions are an option and not required. You can opt to just be a member of the lotto pool. That would be ok for a lot of people, but I like the idea of actually getting paid to play the lottery. Right now, that is the case... and I like it. If that is wrong, then I don't know what is right.
Posted by: Michael on October 31, 2005 11:08 PM
"Stefan, you don't even need to increase the number of variables. Just put jackpot=N, and see what happens. I'm telling you, your theory goes up in a puff of smoke."
Felix, that makes negative amount of sense. If only you could visualize it... It's a beautiful thing to see the distribution change as N goes from 1 to much larger than X (the number of possible outcomes). When N=X, Poisson's influence is still very much in evidence, but it is waning.
Posted by: Stefan Geens on October 31, 2005 11:35 PM
I defy you to find a lottery where N>>X.
But why do you insist on dodging the question? You are not only postulating a lottery where N>X, which is ridiculous, you are also postulating a lottery where the jackpot doesn't grow as the number of tickets sold rises. In other words, your model bears zero relation to reality. The really weird thing is that it was you who made the point about the correlation of ticket sales and jackpot in the first place. Only to subsequently ignore it.
Posted by: Felix on October 31, 2005 11:58 PM
I haven't the slightest idea the nature of your disagreement, but aren't scratch-off games and number games lottery?
I don't know if it will resolve your dispute, but what if you both applied your argument to a Pick Three game? 1-in-500 odds, $1 bet. Even though you don't split winnings, the pool is still self-funding, and you can easily calculate expected value, right?
Posted by: 99 on November 1, 2005 12:07 AM
Well, in a game like that, 99, Stefan's theory is trivially false. If you scratch of a winning ticket you win a certain amount of money, regardless of how many other tickets have been bought. Expected value and number of tickets sold are utterly unrelated.
Posted by: Felix on November 1, 2005 12:10 AM
I didn't really think I had anything to add to the conversation. But that's never stopped me before.
Posted by: 99 on November 1, 2005 01:08 AM
"Well, in a game like that, 99, Stefan's theory is trivially false."
What's with the emotive language? My "theory", which unlike evolution is actually a fact in this case as it is mathematics, has nothing to do with scratch and stiff cards.
Posted by: Stefan on November 2, 2005 08:02 AM
"Scratch and stiff" is good.
Still, the longer this thread goes on, the more tempted I am by Michael's excellent-sounding idea of "being paid to do the lottery". Where can I sign up, Michael?
Posted by: Claude de Bigny on November 2, 2005 10:23 AM
Being paid to do the lottery. Not a bad idea -- just ask who's doing the paying.
Posted by: Felix on November 2, 2005 02:08 PM
Felix, I am looking at all possible simple lotteries (where only one combination wins). And what we find is that expected value is below 1/X.N when N>X.
At around X=N, the distribution for likely number of winners is in transition from a Poisson distribution to a normal binomial distribution.
Posted by: Stefan on November 2, 2005 03:36 PM
Why are you so obsessed with hypothetical lotteries where N>X, Stefan, when real-world lotteries have N much lower than X? For intsance, the mega millions jackpot has rolled over again, and is now $192 million. (Bringing the calculation above up to an EV of $1.27 per $1 lottery ticket.) There were still less than 40 million tickets sold, with 175 million possible results.
It's also worth noting the number of tickets sold increases with the jackpot. So in the biggest jackpot ever, there were two winning tickets, splitting $363 million. But that's still better than winning a normal jackpot which hasn't been rolled over dozens of times.
Posted by: Felix on November 2, 2005 03:55 PM
With the jackpot now up to $225 million, Stefan, can you run your numbers again? The quick and dirty calculation now gives a value for each $1 ticket of $1.46. And there's something on the order of 40 million tickets now being sold per drawing. So what's the odds that if you win, you'll have to share the jackpot with someone else? About 40 in 175? So that should be taken into account. And the chances of sharing with two other people, and so on and so forth. But I reckon at this point the expected value of each lottery ticket is definitely positive, no, before taxes?
Of course the interesting thing is how that can or should translate into behaviour. If the expected value gets high enough, is there a point at which buying a lottery ticket becomes rational? I have a feeling the answer is no: after all, winning a $225 million jackpot is not twice as nice as winning a $112.5 million jackpot, but the calculation assumes it is.
Posted by: Felix on November 6, 2005 08:18 PM
I'm going to keep on plugging away at this. EV is now $1.67 per $1 ticket, but is $262 million really that much more attractive than $225 million? It would certainly seem that way, given the steady increase in tickets sold per drawing as the jackpot increases.
Posted by: Felix on November 9, 2005 10:05 PM