# Another scratch ticket thread, for math geeks!

#### StephenCampbell

##### macrumors 65816
Original poster
So, as some of you know, I am an avid scratch player, and rather than quit playing as I was originally planning, I've simply stopped spending more than I can reasonably afford.

Anyways, that's not the topic of this thread. I am trying to calculate what would be the most efficient manner of buying scratch tickets in terms of the ratio of \$20, \$10 and \$5 tickets, or if there should even be a representation of all those three.

This gets extremely complicated and elaborate, and I'm just not sure what the ultimate combination is.

I used to buy these 'rounds' that consisted of a \$20 ticket, two \$10s and four \$5s.... so essentially, \$20 of each type of ticket.

But then I realized that one \$20 ticket has a 1:25 chance of winning \$100, whereas four \$5 tickets collectively have only a 1:248 chance of winning \$100. So if I were to replace the four \$5 tickets with a second \$20, I'd have much greater odds of winning \$100 than if I have the \$20 and four \$5s.

However, a \$20 ticket has a 1:3.51 chance of winning any prize. So with a \$20 ticket there's a 71.5% chance that you lose all your money in one blow.

Whereas with four \$5 tickets, the odds that you don't get any of your money back are actually quite low. An average \$5 ticket has a 1:3.76 chance of winning any prize, so between four tickets you have a 106% likelihood of hitting at least one prize.

So, \$20 ticket gives you much better odds of hitting something big, but also higher odds of losing all your money at once.

And \$10 tickets lie somewhere in between. Between two of them you have the same odds of winning \$100 as you do on one \$20 ticket, but you don't have nearly the odds of winning \$200 that you do with a \$20 ticket. But again, with two tickets your odds of winning Something are greater than your odds on a \$20 ticket, so the game lasts longer, assuming you're not hitting a big prize either way.

Of course, \$5 can only get you \$50,000, whereas \$10 gets you up to \$200,000 and \$20 up to \$1,000,000.

So those of you who are math geeks, what would you do? In what ratios would you purchase the various tickets? One \$5 for every \$10 for every \$20? Or four \$5s for every two \$10s for every one \$20? Or would you only buy \$5 tickets? Or only buy \$20 tickets? Assuming you were going to establish a concept of a 'round' like I did, and always buy a fixed ratio of tickets in batches, how many \$10s and \$5s would you get for each \$20 that you get?

#### And

##### macrumors 6502
Ultimately I would buy none, because I would lose money.

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#### juanm

##### macrumors 68000
If they are a viable business it's because you, as a consumer/client/addict, end up paying more than you get out of it. Simple as that, no need to be a math geek. No matter how you try to twist the numbers, the company making the tickets has probably gone for a definite profit number, and all the different tickets have the same ratio in terms of total batch cost/prizes.

Try to calculate how much you have spent and, how much you've made. If it's too hard, monitor your spendings/wins for a week/month with an excel spreadsheet, and see how much it costs you.
eg:
week 1, \$60 spent, \$40 won
week 2, \$80 spent, \$32 won
week 3, \$40 spent, \$45 won
...

You'll end up with a net amount. Then it's up to you to decide whether the time you lost is worth whatever you made/lost.

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#### Macman45

##### macrumors G5
There is only one winner.....The organisation that run these lotteries. The Uk is now getting flooded with them...not just the "Official" lottery, but we have:

The Postcode Lottery ( zip code based five bucks a week)
The Health care Lottery ( NHS run)

And numerous others.

I played the official lottery for 20 years and apart from a couple of £10 wins and 2 four number prizes, one of £45 and one of £64, I've had nothing.

I now no longer play....I should say that I used the same numbers on each draw.

I've paid enough out for no return.

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#### yg17

##### macrumors G5
Assuming you were going to establish a concept of a 'round' like I did, and always buy a fixed ratio of tickets in batches, how many \$10s and \$5s would you get for each \$20 that you get?
I'd get none, because the house always wins.

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#### alent1234

##### macrumors 603
some years back an MIT professor figured out that you can tell a winning ticket by the serial number on the back. look at your tickets and try to figure out the algorithm

my wife used to "play" these and at least in NYC when you take the winning tickets in they scan them in to the machine to verify the winnings. they don't even look at what you scratched off. hence the serial # on the back is what you need to look at

#### mobilehaathi

##### macrumors G3
So, as some of you know, I am an avid scratch player, and rather than quit playing as I was originally planning, I've simply stopped spending more than I can reasonably afford.
Interesting to see this. You seemed quite adamant last time that this wasn't happening.

As for the topic of this thread,

Anyways, that's not the topic of this thread. I am trying to calculate what would be the most efficient manner of buying scratch tickets in terms of the ratio of \$20, \$10 and \$5 tickets, or if there should even be a representation of all those three.
This is not a well defined question. What do you mean by "efficiency?"

#### maflynn

##### Moderator
Staff member
I am trying to calculate what would be the most efficient manner of buying scratch tickets in terms of the ratio of \$20, \$10 and \$5 tickets, or if there should even be a representation of all those three.

This gets extremely complicated and elaborate, and I'm just not sure what the ultimate combination is.
You mean try to find a system that beats the odds? Not going to happen, there's a reason why governments love lotteries, its one of the easiest and most cost effective ways to get people to hand money over to them.

The odds are consistently stacked against you.

----------

my wife used to "play" these and at least in NYC when you take the winning tickets in they scan them in to the machine to verify the winnings. they don't even look at what you scratched off. hence the serial # on the back is what you need to look at
But you need to buy the ticket to see serial number. Plus you need to find the proper algorithm which means buying lots of tickets to get a winner and then discern the serial number construction.

#### carjakester

##### macrumors 68020
no matter how you put it you will end up losing money. id put the money away in a savings account instead of purchasing tickets to have a chance to win your money back.

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#### yg17

##### macrumors G5
But you need to buy the ticket to see serial number. Plus you need to find the proper algorithm which means buying lots of tickets to get a winner and then discern the serial number construction.
And I doubt there's any pattern to the serial number either - it's probably just a random number and the lottery has a database of which serial numbers are winners and how much. When you scan the ticket, it checks the database to see if it's a winner. I'd be surprised if there was any algorithm to determine winners based off the s/n

#### rdowns

##### macrumors Penryn
You mean try to find a system that beats the odds? Not going to happen, there's a reason why governments love lotteries, its one of the easiest and most cost effective ways to get people to hand money over to them.

QFT. The only thing the government likes more than an ignorant lottery player is a 'smart' one who thinks he can beat the odds.

#### tobefirst

##### macrumors 601
*Sigh*

Yes, the OP is ultimately fighting a losing battle. So is everyone who goes to the casino and drops some money in a slot machine. Even the table "skill" games are setup in a way that the house will always make money. That doesn't mean that they can't be an enjoyable source of entertainment, or that you can't figure out the best way to spend your money to minimize your loss and maybe, for awhile at least (with a run of luck), beat the house.

OP, I'm not a huge math guy, but I believe we'd need to know all of the payouts of each of the tickets you'd like to play in order to figure out how to play best.

#### Raid

##### macrumors 68020
Well like many have said if you'd like to play and get ahead financially you are going to have a bad time... you can figure out the expected value on each ticket using the formula:
Expected Value = Prize Value1 x Prize Odds1 + Prize Value2 x Prize Odds2 + .... + Prize Valuen x Prize Oddsn
where n is the number of different prizes able to be won on the ticket. If the goal is to gain financially then Expected Value should be greater than the cost of the ticket... and that's never going to be the case.

If you are just playing hoping to win something then the formula changes slightly. You can look at it in terms of "cost per win" that means the formula looks like:
Cost Per Win = Cost of ticket x ( Prize Odds1 + Prize Odds2 + .... + Prize Oddsn )
where n is the number of different prizes able to be won on the ticket and assuming the odds of one prize is independent of winning a different prize. Here though you would chose the ticket with the lowest cost per win. However there are other factors that might impact your enjoyment of playing so this is just a simple estimate.

Also check your math again, you are using ratio's instead of percent chances and 4 \$5 tickets with that ratio is not a 106% chance of winning for many, many reasons....

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##### macrumors 6502
An average \$5 ticket has a 1:3.76 chance of winning any prize, so between four tickets you have a 106% likelihood of hitting at least one prize.
I'm sure you can see for yourself that this makes no sense. If a ticket has a 1:3.76 chance of winning a prize, that's about a 73.4% chance that it will win nothing. Thus, for four tickets, you have a 0.734 x 0.734 x 0.734 x 0.734 chance of winning nothing, or about 29%. That gives you about a 71% chance of winning something on at least one ticket.

#### alent1234

##### macrumors 603
And I doubt there's any pattern to the serial number either - it's probably just a random number and the lottery has a database of which serial numbers are winners and how much. When you scan the ticket, it checks the database to see if it's a winner. I'd be surprised if there was any algorithm to determine winners based off the s/n

there is one

the guy from MIT won a lot of money figuring this out. i think he also figured out that the serial number increments by 1 or whatever the pattern is at each location and was able to figure out where to buy the winning tickets

at least this was the case a few years ago. might have changed now

#### Don't panic

##### macrumors 603
i think we all know, including the OP, that it is a losing proposition, so yes, the best 'strategy' is always not to play.
however, what i think he is asking is how to "maximize the wins", which is better expressed as how to "minimize the losses".
it really depends on what you are looking for in your 'wins'.
what do you want to maximize?
i can see 3 desirable outcomes (which one is more desirable is more psychological than mathematical)
1. maximize number of win (you are gratified by the "i won" moment)
2. maximize money won (final return for money invested, this will ALWAYS be a loss in the long run)
3. maximize chances of a single big win

if you seek 2., then you have to calculate the return per dollar in each group of tickets, using ALL the different possible prizes, and their respective odds.
you will find how much, on average, one ticket of each class 'wins' (this will be by definition less than the value of the ticket).
let's say for example (and these are completely made up numbers) that the \$5 ticket on average wins \$ 1.21/ticket (meaning that if you invested \$100,000 of 20,000 tickets, you'd expect \$24,200 in total prizes), the \$10 wins 2.95/ticket and the \$20 wins 4.21/ticket.
if those were the numbers, then the best strategy would be the \$10 tickets, as they'd pay (on average) 29 cents/dollar invested, compared to 24 and 21 for the other two, respectively.

if you seek 3., i'd imagine that you want the \$20 ticket, but it also depends on what you consider the threshold for it to be a 'big prize'. basically you'd do like in 2. but only include the 'big prizes' in the calculations.

----------

there is one

the guy from MIT won a lot of money figuring this out. i think he also figured out that the serial number increments by 1 or whatever the pattern is at each location and was able to figure out where to buy the winning tickets

at least this was the case a few years ago. might have changed now
if there was, the sellers would get all the winning tickets out of their rolls.
and even if they didn't, as a buyer you'd still need to be have access to large number of unplayed ticket to pick from.
if there ever was such a loophole (which honestly sounds like an urban myth) i am pretty sure they would quickly close it. they are actually quite serious about the 'fairness' of these games (among the players, not the states).

edit: i got curious and found this interesting article: http://www.wired.com/magazine/2011/01/ff_lottery/all/
the guy indeed 'broke' one of the games (but never made money out of it), based on the visible part of the design of that specific game, which was flawed. in they article they do mention barcodes, so I guess there might have been flaws in that part that have now been fixed.
it remains that the most likely people to take advantage of the system, if loopholes exist in some specific games, are the retailers, as they can just scan the rolls and pick out the winners

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#### StephenCampbell

##### macrumors 65816
Original poster
Okay let me clarify a few things for those who are unfamiliar with how the game works.

NOBODY knows where winners are after the tickets are printed. If even the people who printed the tickets knew where they were, they'd be able to know which stores to go to to pick out the big winners.

There are specifications set in the printing machines (i.e. print 3 \$200,000 prizes, 250 \$500 prizes, 120,000 \$10 prizes) etc, but when they are printed out nobody knows what's where. I believe there may be a guaranteed minimum number of prizes per roll, but again, nobody would know what those prizes are.

The barcode which knows whether it's a winner or not is underneath the scratch-off surface. That barcode is not scanned when the ticket is sold. The barcode and number on the back of the ticket only indicate which game number it is, and let the lottery know at which location the ticket was sold.

Now, back to the topic. I'm confused about how the odds work for multiple tickets. Ray Brady's explanation makes sense, but at the same time, if the odds are 1:3.76, if you had let's say, 1000 groups of 3.76 tickets each, you would have just about 1000 prizes between those groups, yes? 3.76:3.76 odds means having one prize On Average, correct?

#### Raid

##### macrumors 68020
I'm confused about how the odds work for multiple tickets. Ray Brady's explanation makes sense, but at the same time, if the odds are 1:3.76, if you had let's say, 1000 groups of 3.76 tickets each, you would have just about 1000 prizes between those groups, yes? 3.76:3.76 odds means having one prize On Average, correct?
Ok your assumptions are correct, but by using ratios you run into problems like trying to buy .76 of a ticket! The win to ticket ratio of 1:3.76 is roughly translated to a 26.6% chance of winning. In your example the purchase of 3760 tickets times 26.6% yes would mean you'd expect on average 1,000 prizes.

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#### mobilehaathi

##### macrumors G3
Okay let me clarify a few things for those who are unfamiliar with how the game works.

NOBODY knows where winners are after the tickets are printed. If even the people who printed the tickets knew where they were, they'd be able to know which stores to go to to pick out the big winners.

There are specifications set in the printing machines (i.e. print 3 \$200,000 prizes, 250 \$500 prizes, 120,000 \$10 prizes) etc, but when they are printed out nobody knows what's where. I believe there may be a guaranteed minimum number of prizes per roll, but again, nobody would know what those prizes are.

The barcode which knows whether it's a winner or not is underneath the scratch-off surface. That barcode is not scanned when the ticket is sold. The barcode and number on the back of the ticket only indicate which game number it is, and let the lottery know at which location the ticket was sold.

Now, back to the topic. I'm confused about how the odds work for multiple tickets. Ray Brady's explanation makes sense, but at the same time, if the odds are 1:3.76, if you had let's say, 1000 groups of 3.76 tickets each, you would have just about 1000 prizes between those groups, yes? 3.76:3.76 odds means having one prize On Average, correct?
You still haven't defined what your goal is.

##### macrumors P6
You mean try to find a system that beats the odds? Not going to happen, there's a reason why governments love lotteries, its one of the easiest and most cost effective ways to get people to hand money over to them.

The odds are consistently stacked against you.
Agreed. It can be fun to play on occasion, but you gotta know it is a losing proposition.

#### StephenCampbell

##### macrumors 65816
Original poster
Ok your assumptions are correct, but by using ratios you run into problems like trying to buy .76 of a ticket! The win to ticket ratio of 1:3.76 is roughly translated to a 26.6% chance of winning. In your example the purchase of 3760 tickets times 26.6% yes would mean you'd expect on average 1,000 prizes.

Yes, I was talking about on average. If you can expect 1,000 prizes from 3,760 tickets on average, than you can expect at least one prize from four tickets on average.

My goal is to strike a balance between having less-expensive tickets that will ensure me some of my money back, vs. getting more of the \$10 or \$20 tickets and having a chance at a really big prize.

The thing is, once you're buying sizable amounts of tickets, it might all average out, and the only difference between \$5 and \$20 tickets is that you don't have a chance at anything bigger than \$50,000 with the \$5 ticket. Because if you Do win with the \$20 ticket, you win at least \$20. Your "almost guaranteed win" between four \$5 tickets will often be just \$5.

#### ejb190

##### macrumors 65816
Okay, I was curious. Living in Indiana, I went to the Hoosier Lottery webpage and looked up the scratch tickets. Picking one of the low cost tickets at random, I ended up on Instant Cash 5. Digging a bit deeper, I found the Game Rules. The odds of winning are 1:3.11. Pretty good, but lets take that apart a bit. Overwhelmingly, the prizes consist of free tickets. The odds of winning cash is only 1:6.453. Most of the cash prizes are \$1. The odds of winning anything more then break even is 1:11.543.

But putting the odds aside, there is a much more telling number: Prize payout. Let's say you bought all the tickets - all 2,568,000 at \$1 each. So you win all the prizes - \$1,350,157. That's right - you won every prize and still lost \$1.2 million! The prizes total 52% of the face value of the tickets.

I checked out a number of the high dollar games as well. The highest payout I saw was 75%. And this number was a bit misleading as the prizes over \$1 million were payed out as annuities - meaning the lottery only has to pay out a fraction of the prize and let compound interest do the rest.

The endgame is this, the only way to win money is for someone else to lose it and the lottery is not going to run a game where they lose money. To quote Wargames, "A strange game. The only winning move is not to play. How about a nice game of chess?"

Two lessons you can learn from the lottery. 1) If it seems too good to be true, it probably is. 2) Time and compound interest are your friends. You know those annuities I mentioned above? The lotteries use them for a reason and you can take advantage of the same math. I ran the numbers in a previous thread started by the same OP.

#### mobilehaathi

##### macrumors G3
My goal is to strike a balance between having less-expensive tickets that will ensure me some of my money back, vs. getting more of the \$10 or \$20 tickets and having a chance at a really big prize.
I'm not trying to be difficult, but this is still not well defined. What do you mean by 'strike a balance between' and 'ensure me some of my money back?'

Do you wish to maximize the number of winning tickets 'per round?' Do you wish to minimize net losses? Do you wish to maximize gross gains?

#### Shrink

##### macrumors G3
I'm not trying to be difficult, but this is still not well defined. What do you mean by 'strike a balance between' and 'ensure me some of my money back?'

Do you wish to maximize the number of winning tickets 'per round?' Do you wish to minimize net losses? Do you wish to maximize gross gains?
It's really very simple....

He wants to win on every ticket, make a fortune, and retire to an island in the Caribbean.

So just tell him how to do that, for goodness sake!

#### Don't panic

##### macrumors 603
Yes, I was talking about on average. If you can expect 1,000 prizes from 3,760 tickets on average, than you can expect at least one prize from four tickets on average.

My goal is to strike a balance between having less-expensive tickets that will ensure me some of my money back, vs. getting more of the \$10 or \$20 tickets and having a chance at a really big prize.

The thing is, once you're buying sizable amounts of tickets, it might all average out, and the only difference between \$5 and \$20 tickets is that you don't have a chance at anything bigger than \$50,000 with the \$5 ticket. Because if you Do win with the \$20 ticket, you win at least \$20. Your "almost guaranteed win" between four \$5 tickets will often be just \$5.
again, are you going for maximum number of wins, maximize number of dollars won or going for big prizes?
that affects your ideal strategy the most (again keeping in mind that in term of net money, statistically you certainly lose)

if you are looking at total money 'earned', than you need to calculate the statistical average of your return of investment per dollar spent.
if you get 5 dollars back on the \$5 dollar tickets or 20\$ back on the 20\$ ticket, then it's the same, but what are respective odds of winning that specific prize? that will tell you which one is more advantageous. and you have to get that for all the prizes, including the middle ones (not sure if the odds per each prize are available)