DrLotto
Posted by DrLotto
Nov 06, 2019

How to calculate Florida Scratch-off ticket value from their new website.

We can now figure out how much money a Florida Scratch-off ticket is worth.

Thanks to Florida Lotteryís new website, we can now figure out how much money each Florida Scratch-off ticket is worth. They now publish remaining prizes for each scratchoff.

For example, you can see that $50,000 Holiday Luck has 6 of 6 remaining grand prizes (at the time of this post). And if you look at the $2 prize, you can see that of the original 1798751 $2 prize tickets, only 1624725 remain. That means the percentage of tickets that have been sold is probably about 1 ó (1624725 / 1798751), or about 90%.

Since there are fewer overall tickets remaining but all of the grand prizes still remain, then the odds of getting a grand prize are higher now than they actually should be!

The original odds of getting a grand prize were 1 in 3,000,000.

With 6 grand prizes, that means the total number of tickets printed is 6 * 3,000,000 or about 18,000,000.

But using our deductive skills above, we determined that only about 90% of the tickets remain. That means only 18,000,000 * 0.90 tickets remain, or about 16,200,000.

And since there are still 6 grand prizes, and only 16,200,000 tickets, then the *true* odds of getting a grand prize are 16,200,000 / 6, or about 1 in 2,700,000!

Thatís still not very likely! But, if youíre going to be playing scratchers anyways, you might as well do everything you can to maximize your odds of hitting it big!

The math I just showed you lets you quickly and easily figure out which games have grand prizes that are more/less likely. But determining the exact value of a scratch ticket is a little more difficult.

To do that, estimate the total number of remaining tickets, the same way we did above. In the case of $50,000 Holiday Luck, we got 16,200,000.

Then, multiply that number by the price of the ticket. In this case, itís $2, so $32,400,000. Thatís the price it would cost to buy every remaining ticket.

Next, add up the number of each remaining prize multiplied by the value of that prize. So, for $50,000 Holiday Luck, it would be (6 * $50,000 + 24 * $10,000 + 555 * $1,000 + Ö).

If you do that for each prize, you should get about $22,680,000. (AgainÖ thatís the value Iím getting at the time of this writing. You will get something else, but the idea is the same.)

Now, we know how much it would cost to buy every remaining ticket, $32,400,000, and we know how much we would win if we bought every remaining ticket, $22,680,000.

So this definitely isnít a very good ticket! Florida will take in $32,400,000 in sales and only pay out $22,680,000. But thatís the point of the lottery, right? At least the money they make will go to good causes, like education.

But what this means for you is that on average youíll only get back a small percentage of the money you gamble. That percentage is calculated as $22,680,000 / $32,400,000, or about 70%!

If you buy $100 worth of tickets, you can expect to only get back $70 on average. Thatís a $30 loss.

Some other tickets are worse. Some are better.

You can see the [link redacted by Moderator] for every game at [link redacted by Moderator] where we do the math for you so that you can easily know which is the best scratch off to buy!

[Edit by Mod: Thanks for the explanation, but promoting your website here is spam]

DrLotto
Posted by DrLotto
Mar 05, 2019

Lawsuit reveals scratchoffs not random

There's a lawsuit going on in Texas right now. Players are claiming that the rules of the Fun 5 game were unclear and they should be receiving payouts which the state is denying.

The lawsuit is a little interesting. But the documents that have been released as part of the discovery process is what I find most exciting.

Check out the game parameters that are part of the contract that the state has with GTECH, the company contracted to print the tickets.

Quote:

No consecutive string of non-winning tickets will be greater than <redacted> which is 2 times the overall odds for this game rounded to the nearest whole number.



The overall odds for the game have already been published so it's easy to deduce the number that was redacted. Now we know that lottery retailers can wait until they see ~6 tickets in a row sold as losers and they are guaranteed to have a winner next in line.

Comments

OnceDear
OnceDear Mar 06, 2019

Woah at item 3. That blows random out of the water completely and would allow store staff or other vultures an enormous advantage.

Is this game still available?

Mission146
Mission146 Mar 06, 2019

There were a few WV Lottery games kind of like this several years ago. I worked in a, ďSmoke Shop,Ē part of a grocery store that sold them.



The problems were identifying which scratchers for which this no more than x consecutive losers was the case, then thereís actually the matter of being present for that number to be sold AND scratched in front of you. Any ticket that leaves without a known result screws the pooch and restarts the count.



Even if you information traded with a co-worker, you would go most days (just speaking to this ONE store) without the target number of that particular scratcher even being sold, much less having the results revealed. For my part, I got to buy maybe one guaranteed winner a month and most of those were breakeven prizes.

Ahigh
Ahigh Mar 14, 2019

Texas is retarded when it comes to gambling laws.



Period.

DrLotto
DrLotto Jul 27, 2019

I think they can be beat even by people who don't work at a retailer. [Link redacted my moderator]



At the very least, you can increase your odds by only buying from new packs and then stopping once you get a prize over a certain amount. Or if the first few are losers, continuing to buy until you get a winner since you know each consecutive ticket is more likely to be a winner.