**Update** See the comments, but thanks to readers’ arguments I believe I was wrong in an essential calculation. I’ve reworked the text to correct my error, which was small in most cases—*except in one important place*. But in making this error, it reveals that the New York Lottery has one gamble which is near enough a sure thing (at least to break even), a very odd situation.

**Update 2** See HaroldW’s comments: the tickets aren’t the same!

I received a call from Noelia de la Cruz at *Business Insider* for an article she was writing about scratch-off tickets at the New York Lottery, “We’ve Figured Out What New York Scratch Ticket Has The Best Payout.”

The easiest calculation is the amount you will probably lose. This is the ticket price. So if you buy a $1 ticket you’ll probably lose $1. If you buy a $30 ticket (yes, they have one), you’ll probably lose $30. And so on for the other ticket face values.

In other words, you’ll probably lose. Unfortunately, this statement is logically equivalent to “you might win.” And somehow this is all some ticket buyers ever hear. The mere possibility of winning, no matter how improbable, is all the encouragement that these people need.

It should come as no surprise that the odds are against you. But if you are going to gamble, wouldn’t you like to gamble on the game which has a best payout? Here’s how.

The New York Lottery has nearly four dozen different “games”, i.e. gambles, you can play. These are ever in flux. While we were preparing this article, two gambles went out of circulation and two more came on line. The NY Lottery website lists the current gambles. Each ticket has an explanation of the odds of winning various payouts plus the odds of winning at least one of these payouts.

For example, the $1 Take Five ticket allows you to win $1 which has odds 1 in 8.82, $2 at odds 1 in 23.17, and so on up to winning $5,555 at odds 1 in 529,200. You can even win a free ticket which has odds 1 in 2:84. So when you buy a ticket, you might win nothing, or you might win $1, or $2, and so on. You therefore have a chance of winning nothing or at least one prize.

*That you can win a free ticket means something special.* Stay tuned for what.

The formula you need to convert odds to probability of winning is this:

probability = 1/odds.

The probability of winning $1 on this ticket is thus

probability = 1/8.82 = 0.113,

which is just over 10%. The probability of winning the “jackpot” is 0.0000019, which is about 2 in a million.

The webpage lists the probability you win *something*, for this ticket: “Overall Chances of Winning: 1 in 4.64”. Thus, the probability is 0.216.

But the lottery erred in printing this number. This is the probability of winning money for *just this ticket*. But since you can win a *free* ticket (with probability 0.26), you might win money on this free ticket, too. The chance of that is 0.216, as well. Of course, this free ticket might win you another free ticket, which gives you another chance to win money—or another free ticket! And so on to infinity.

The overall chance of winning something is then 0.291 and not 0.216 (this is odds of 1 in 3.43). Which means you have a 71% of losing.

The ticket with the highest chance of winning is $30 Win $1 Million A Year for Life: 0.341. The $1 Take Five has the second highest. The lowest, at 0.195, is the $5 Cashwords ticket.

There is another calculation relevant to gamblers and the lottery: the payout. Now, scratch-off tickets are a kind of modified raffle. The Lottery knows exactly how much money they are going to give away (on a rolling basis, anyway) and just how much they will take in (if all tickets are sold). These two numbers, modified by circulation statistics, are all that is needed to calculate the expected payout to set the odds of winning each prize.

Since the pool of money is fixed by the Lottery, the expected payout is known, i.e. not random, to them. But it still is to the gambler, because of course he doesn’t know how much has already been won and how much is left in the pool. Therefore, at least as a rough guide, we can calculate the “expected payout” for each ticket. Think of this as the rate of return.

The calculation is simple: for each ticket, multiply the probability of winning each amount by that amount, then sum these together. For the Take Five, this is $1 x 0.113 + $2 x 0.043 + … + $5,555 x 0.0000019 = $0.65.

**Here is where** the mistake reveals something strange. We still have to account for the free tickets, so the final expected payout is $0.9995, that is, $1. I.e., you will *break even* on this game (on average). *The New York Lottery is thus losing money on this ticket*, after considering their light bills, shares to retailers, and so forth.

It was this strange finding that convinced me my original calculations were correct. I could not believe the lottery would create a gamble where they effectively lose money. But two readers have convinced me that this is so.

The second highest expected payout is $0.93 (per dollar) for the $30 Win $1 Million A Year for Life ticket (I assumed that in all “win for life” games you lived 40 years more). Most payouts are in the 60 – 70 cent range (per dollar gambled). Meaning the Lottery takes about 35 cents gross profit per ticket.

The lowest payout is the $2 Money Match gamble at $0.62 (per dollar, or $1.24 for the ticket). Interestingly, this ticket has the most—37!—different ways to win ($1, $2, $3, and so on). I’ll leave it as a homework problem to show that as the number of ways to win increases, the expected payout will *decrease*.

Categories: Fun, Statistics

What is missing from the analysis is the amount the winner of a large sum would have to pay in both state and federal taxes. This reduces the value of the large prizes substantially.

It’s true that winners must pay taxes, but we don’t need to consider it here. For example, if two different employers were offering you a job and Company A offered to pay you more than Company B, then all things equal, you should accept Company A’s offer even though you will end up paying more income taxes.

In the UK, the National Lottery is also colloquially known as the Stupidity Tax.

And then there is this.

http://www.wired.com/magazine/2011/01/ff_lottery/all/1

Wm

I believe you’re misinterpreting the odds posted. Odds of “8 to 1” work the way you say: it means that you lose 8 for each 1 you win, i.e. you lose 8 out of 9 and win 1 out of 9.

Odds of “1 in 8.82”, as they use here, literally mean you win 1 out of 8.82, not 1 out of 9.82; i.e. probability of winning is 0.1134, not 0.1018

The probability of winning cash directly on an “instant take 5” is therefore 0.2154, or “1 in 4.64”. (doing it your way, it would come out as 0.200) As you say, this doesn’t take into account the free tickets. Since 1 in 2.84, you get a free ticket, you can multiply by 1/(1-(1/2.84)), so the chance you win something comes out as 0.332, or 1 in 3.01. Possibly they’re not allowed to treat the free ticket as a win at all.

The expected return (not counting free tickets) comes out as 0.6475. With the free tickets, the expectation comes out as 0.9995, which seems extremely high to me — possibly the taxes referred to by Bob are where the real profit margin is in all this.

Anomaly UK,

It is high because you have made a mistake. Here are the basic formulae;

odds = probability / ( 1 – probability)

probability = odds / ( 1 + odds)

You sometimes have to subtract the later from one depending on whether the odds are “for” or “against” the event in question.

I’m with AnomalyUK on the interpretation of the posted chances. Say the probability of an event is 0.1. The usual phrasing is either “the odds are 1:9 against” or “the chances are 1 in 10”. In this case, the New York state lottery uses the latter formulation.

You can confirm this interpretation by looking at their webpage concerning the odds of winning the Megamillions lottery, http://nylottery.ny.gov/wps/wcm/connect/NYSL+Content+Library/NYSL+Internet+Site/Home/Jackpot+Games/MEGA+MILLIONS/MegaMillions+-+Chances+of+Winning

In this case, the probabilities (assuming unbiased choosing of balls!) are straightforward to calculate and agree with the interpretation of “1 in X” as being equivalent to a probability of 1/X. [As opposed to the interpretation as 1:X odds against, which corresponds to a probability of 1/(X+1).]

This doesn’t guarantee that the same interpretation is applied to scratch tickets, but it seems improbable (!) that they’d vary within the same organization.

Read it again, slower. “1 in 10”, means that out of 10 tickets, 1 wins. One. In. Ten.

“Ten to one” means that out of 11 tickets, 1 wins. Ten losses to one win. Ten. To. One. Unless it’s “Ten to one on”, meaning that out of 11 tickets, 10 win. (written 10/1 and 1/10 respectively).

As a general rule, it is very dangerous to quote formulae without understanding what they mean or why you’re using them. The formulae you quote work on ratios of wins to losses. There is no way that “1 in 8.82” would be a ratio of wins to losses — it is very clearly a ratio of wins to trials.

Odds can be expressed in several different ways, all called “odds”. Wikipedia calls these ones “Decimal odds”. http://en.wikipedia.org/wiki/Fixed-odds_betting#Decimal_odds

Anomaly UK,

Read it all, slow or fast. You’re welcome to go to the NY Lottery website, as suggested, where you can read things like (for the $1 Take Five) “Overall Chances of Winning: 1 in 4.64” and the chance of winning $1 is “1 in: 8.82,” etc. Go. To. The. Website. And. Read.

You are not paid out $8.82 (or $0.82) on a dollar bet when you win a dollar. You win a dollar. But since you paid a dollar, you come out even. These are not fractional odds.

As a rule, it is very dangerous to claim misquotations without verifying the source.

Briggs, in the post you call “1 in 8.82” etc. “odds”, and on the site, correctly quoted in your comment above, they are described as “chances of winning”. I think this inconsistency was described as misinterpretation rather than a misquote, and since the numbers given are indeed not really any sort of odds, fractional or decimal, but probabilities, it is in fact a misinterpretation to ‘convert’ them to probabilities as though they were odds.

Note that the interpretation of “1 in 8.82” as a probability is confirmed not only by the language used but also by the consistency of the arithmetic both in the scratchie case and the Megamillions page referred to by Harold. The probabilities 1/8.82, 1/23.17 etc. do add up to an overall probability of winning cash directly from one ticket of 1/4.64. (It amazes me that they do not take the opportunity to also take into account any returns from any resulting free tickets.)

To be clearer, obviously the numbers you took as odds are not probabilities, but you took “1 in 8.82” to mean odds of 8.82/1, whereas it is intended to mean a probability of 1/8.82.

I had studied the sources very carefully before I made my first comment, and, like Jonathan D, I had done the calculations which show that the New York Lottery interprets the “chances of winning” the way I said, as probabilities — and not the way you said, as odds. I invite you to demonstrate a probability of winning of either 1/4.64 or 1/(4.64+1) by any method other than the one Jonathan and I have used.

You are correct that “these are not fractional odds”. As I demonstrated with the wikipedia link, they are decimal odds. However ” probability = odds / ( 1 + odds) ” is a formula for working with fractional odds, which is why it gives the wrong result in this case. That is what I meant by knowing what your formulae mean.

What is the chance of rolling a six on a six-sided die? It is 1 in 6. What is that as a probability? It is 1 divided by 6. It is not 1 divided by 7.

Expressed as fractional odds, that would be “5/1” – five to one against, as per your formula. It is unusual (though by no means unheard of) to express winning chances that way. It is more common here in Britain, where fractional odds are advertised by bookmakers on every high street, than in the USA. Even here, though, that would always be written “5/1”, never “1 in 5” or “1:5”.

Jonathan D, Anonymous,

Hmm. Well, you guys have convinced me that your interpretation is correct.

Three things kept me from agreeing: my stubborness, Anonymous’s cattiness, and that the $1 Take Five ticket (when I did the original calculation) had an expected winning just under a dollar. I could not believe that the lottery would allow a gamble on which they effectively

lose(after considering the share they have to pay for the sellers, etc.).The expected payout for the Take Five, on the “1 in” interpretation is $1, or near enough to it ($0.9995). So I figured the Lottery

musthave been using regular odds, which puts the expected payout at $0.84 (as in the text).But one error does not fix another, so what I’ll do is to rework the text with this interpretation in mind, highlighting the strangeness of the Take Five. This may be today or tomorrow.

Thanks to you both. For your help, if ever I see you, I’ll buy you a Take Five ticket.

There could well be an error in the reported figures, but I wonder how many people bother claiming free tickets, and whether the Lottery are happy to bet on that.

Jonathan,

It’s possible. Before today, I would have said the probability of misprints was negligibly low. But after this, I’m not so sure. I mean, they are losing money on the Take Five ticket. Not much, to be sure. But a definite loss.

Perhaps it is meant as a “loss leader”, a gamble which gives citizens the false hope that they are winning frequently, so that they might as well buy other tickets, too. Since this is a public entity, this practice hardly seems kosher.

It didn’t make sense to me that the payout on the Take 5 scratch ticket (counting free tickets) is just about 1.00. I went back to the Lottery site and discovered the missing piece of information. The most likely positive result (chances of 1 in 2.84) is described there as a free “Take 5” — but it’s *not* another scratch card! It’s a free quick pick in the Take 5 drawing game. One can quickly determine that the expected payout for that game is approx. $0.558. [Presuming that if one wins a free Take 5 quick pick in the drawing, it is valued at the expected payout — or equivalently, that one plays the Take 5 drawing until a decisive result is obtained.]

Using $0.558 as the value of the “free Take 5” prize, the Take 5 scratch game’s expected payout is computed as approximately $0.844.

Good to know that the Lottery is not losing money on it. Incidentally, according to this source, New York State Lottery pays out an average of 61 cents for every dollar wagered. Not that it’s relevant to this game, but it hints that many NYSL games do not pay out at a rate as high as this particular scratch card.

Well there you go – our problem was failing to distinguish between “Take Five” and “Instant Take 5”!

It is perhaps more interesting to distinguish between this Take Five and this one. 😉

HaroldW,

You’re a sweetheart! Thanks very much.

Don’t forget you get to keep that nickel you used to scratch the ticket too, always a winner!

This is an interesting discussion. I have quite a bit experience in calculating payouts. Let me know if you agree if the following is correct:

*) With a free game you are partially exposed to luck/odds twice. Meaning if e.g. a payout-percentage is 90%, then playing twice the payout-percentage is still 90%, but your personal payout reduces to 81%. Meaning, although you win a free ticket worth $1, it is something you could call “Schrödingers odds” – it is $1 at the same time as it is not.

*) A free ticket is financially $1. In the view of one game-round you have won $1. But, you have to play it again. The trouble is here that in order to have the correct “payout” within one game-round, you need to think into infinity. Because if you only look at one batch of tickets, the player is partially exposed to the payout-percentage twice.

*) So, imo, the currrent Take 5 ticket has “Overall odds” of winning anything at all of 1.77. It has “Odds excl. Freeplay” of 4.69. It has a payout-percentage of 97.1% within one (theoretical) game-round. But, if you as a person buy all the 546,440 tickets in the batch, the free tickets get mixed up as you then look at more than one game-round. Imagine there is no free-play, but instead if you win $1, a man puts a gun to your head and asks you to play again. The payout-percentage is still 97.1%, but because you essentially played again you lost more and thus the lottery made more money.

Robert

I play a $20 scratch off game in Maryland, the odds on the card say about 1:3 (really 1:2.97), the minimum you can win is $20. I was thinking about the odds on playing multiple games.

If I play three games, I calculate the odds of winning on all 3 games 1/27 = 3%. The odds on winning nothing as 8/27 =~ 30%. To calculate the odds of winning at least one prize, I simply subtracted the odds of winning nothing from the 100%, and came up with =~70%, this is known as “Least Once”. To calculate the odds of winning only two prizes was a bit more tricky, I came up with 3/27. Any flaws in my calculations?

The thing I want to know is what ticket number is most common for the jackpot winners to be printed on for the scratchoffs? Is it ticket number 1? 50?