Partially Revealed Scratch Off Lottery Ticket

The California Lottery has a $3 scratch off game where a player will reveal 25 symbols in the play area. If three or more are "diamonds", the player will win a prize ranging in value from $3 for three diamonds, to $20k for 13 diamonds:



The question is, "What is the new value (if any) of the ticket, if the first two of the 25 spots scratched reveal diamonds?" (the other 23 spots will be revealed after the resale of the ticket).

Here is the complete payout table:

Considering the fixed odds and all that, I would think the value of said scratch-off ticket would remain constant.

I would bet just about any amount of money that all of the non-winning tickets have 2 diamonds on them. To the extent that all tickets will have at least two diamonds on them, you haven't added any value by pre-locating two (of, hopefully, many) diamonds on your ticket. The most likely scenario, by far, for your ticket is to reveal blanks in all of the remaining spaces.

Let's think of this a different way. Suppose that every losing ticket had 2 diamonds (reasonable assumption I think). What are the odds that upon scratching two squares you reveal those two diamonds? If you assume the 2 diamonds are randomly distributed, that is actually quite a long shot... 2/(25*24). Which implies a 299/300 chance the ticket is a winner.

Now, it may be more tickets have the only 2 diamonds in the 1st and 2nd position, and thus less than a 299/300 chance it is a winner, but I think it's probably a winner.

However, the big question is are the diamonds randomly distributed around the scratch area, or do they put diamonds near the beginning on purpose? If the diamonds are random, the ticket is valuable and that value can be calculated. If not, the ticket is probably worth $3-5.

they are not randomly distributed. the designs are fixed. there will be some variations. but if you bought a bunch you would see copies. im sure theyre designed to provide excitement for the purchaser even if they are losers. kindof like slot machines displaying nearwinners.

I think this is an extremely interesting question. So interesting that I almost did a blog post about it last week.

I think your odds have to change as you scratch off the ticket. If your overall odds of winning are 1 in 4, they are only 1 in 4 before you have scratched off the first box. If the first box is a diamond, your odds have to have gotten better. If the second box you scratch is a diamond, they go up again. Think about it, I would definitely pay $3 for a ticket that has two diamonds randomly scratched with only 2 scratches.

I don't think there is anyway to actually know what the odds are if you've scratched 24 boxes, and have two diamonds. Your overall odds are 1 in 4, but I think there has to be a less than 1 in 4 chance you will get a diamond on the last scratch. So the answer has to be, that the odds on a ticket change as you scratch it.

Quote: FinsRule

I think this is an extremely interesting question. So interesting that I almost did a blog post about it last week.

I think your odds have to change as you scratch off the ticket. If your overall odds of winning are 1 in 4, they are only 1 in 4 before you have scratched off the first box. If the first box is a diamond, your odds have to have gotten better. If the second box you scratch is a diamond, they go up again. Think about it, I would definitely pay $3 for a ticket that has two diamonds randomly scratched with only 2 scratches.

I don't think there is anyway to actually know what the odds are if you've scratched 24 boxes, and have two diamonds. Your overall odds are 1 in 4, but I think there has to be a less than 1 in 4 chance you will get a diamond on the last scratch. So the answer has to be, that the odds on a ticket change as you scratch it.

I agree. If you witness the unscratched ticket being randomly scratched to reveal two diamonds in the first two random boxes, then I think that ticket is worth MUCH more than 3 dollars. It's almost like the Monty Hall problem, inside out. What are the odds that you picked the ONLY 2 winning doors out of 25? Actually they are very very low. In fact, the odds are very low that if you did that, there are only 3 or 4 or 5 diamonds. It's more likely you have a very high number of diamonds.

Now, if someone had knowledge and intentionally revealed two diamonds, it would be meaningless. Again, it's Monty Hall, in reverse.

Which got me thinking...imagine the same Monty Hall problem with 2 cars and one goat. You pick, then he reveals one of the cars, and asks you to switch. This time, you don't want to, because your odds were better BEFORE the reveal.

Sorry, triple post when i was sending this one.

Sorry, triple post when i was sending this one.

WARNING!!!!!!!!!!!!!!!! LATEST SCAM!!!!!!!!!! DON'T BE A VICTOM!!!!!!!!!!!!!

Here is the latest scam with scratch off lottery tickets! It's being done by crooked cashiers, store owners etc etc...

When you have a winning scratch off rub off lottery ticket and you bring it to the store to cash it in if you watch them person who validates it through the lottery machine if you didnt completly scratch off the ticket all the way at the bottom the clerk will scratch more off and the reason being below the actual game part is a series of numbers and after the clerk scans the bar code on the back of the ticket he needs to enter these numbers into the machine to validate your winning ticket to pay you, now here is how the scam works the dishonest clerk gently rubs away the bottom part of the scratch off ticket and he makes it look like natural scratches from being pulled through the lottery box like just a few lines but enough for him to get the validating number he needs but where to the unsuspecting buyer it wouldnt raise any suspision since the actual game still is covered and cant be seen, and then once he has this number he runs it through the machine and if the ticket comes up as a winner he buys it and if it doesnt he leaves it in the rack for a victom to buy the loosing ticket! Be carefull if you see a ticket with some small scratches on it don't chance it DONT BUY IT! Help Get this warning out! Tell everyone you know or feel free to cut copy and paste this message in an email and send it to everyone in your address book! Lets stop these theives! It's hard enough to actually win with out sitting down at a game with a stacked deck!

I agree that if you saw someone scratch two squares randomly and they revealed 2 diamonds that the ticket has value. Tough to calculate the value without making unrealistic assumptions, but it is almost certainly a winner

Quote: dwheatley

...but it is almost certainly a winner

I completely disagree.

If I understand it correctly, the card has 25 spaces, and the spaces either have a diamond or nothing, and three diamonds is the minimum to win.

That being the case, I would venture to say that ALL of the non-winning tickets have two diamonds, and that the person selling this ticket was simply lucky to find them immediately.

Therefore, finding two diamonds in the first two scratches does not increase the odds of finding at least one more.

This is the case we have been arguing against. We can assume they all have 2 diamonds, no problem. But what are the odds of uncovering those 2 diamonds in your first two scratches? If you scratch two squares AT RANDOM (different from OP's suggestion that it was the 1st two spots), and uncover 2 diamonds, you have a winning ticket, or were really 'lucky' in your first two scratches. The odds of uncovering the only 2 diamonds in 2 random scratches out of 25 is 1 in 300.

Now, if the 1st two spots are diamonds, we can be less sure. The tickets could be 'programmed' for near misses of that type. But if you randomly scratch off 2 diamonds out of 2 scratches, I'd give you at least $5 for the ticket.

I agree with the other posters that each losing ticket has two diamonds. Assuming that the diamonds are randomly distributed and using Bayes Theorem, P(A|B) = P(B|A)* P(A)/P(B) (see http://en.wikipedia.org/wiki/Bayes_theorem), I get $11.60 for the value of a ticket with only two spots revealed both being diamonds.

Here's an example of the calculation. The probability of having a total of three diamonds given that the two spots revealed are both diamonds is P(3|2) = P(2|3)*P(3)/P(2) = 0.01 * 0.10 / 0.00885129 = 0.1130

By the way, I get $1.86 as the value of a $3.00 unscratched ticket.

Thank you all for your input. From the other tickets I have seen, the pattern is randomized. The two spots on the ticket in question were not the first two, nor in sequence, but one was an upper corner.

Quote: Ayecarumba

Thank you all for your input. From the other tickets I have seen, the pattern is randomized. The two spots on the ticket in question were not the first two, nor in sequence, but one was an upper corner.

Just so we're clear here...You DO have to tell us how this turns out. You realize that, right?:-)

Hehe, yes. I will reveal how it actually turns out. Stay tuned.

Quote: DJTeddyBear

I completely disagree.

If I understand it correctly, the card has 25 spaces, and the spaces either have a diamond or nothing, and three diamonds is the minimum to win.

That being the case, I would venture to say that ALL of the non-winning tickets have two diamonds, and that the person selling this ticket was simply lucky to find them immediately.

Therefore, finding two diamonds in the first two scratches does not increase the odds of finding at least one more.

You missed the point though. We weren't saying that the lottery or someone without outside knowledge revealed two meaningless Diamonds. In my post, I specifically said that they had to be random spaces. I can't believe, after all of our mathematical posts, you'd devolve into the "lucky enough to find them immediately" flawed logic. The odds that there are only two diamonds, and you find both immediately, is 300 to 1. If there were three, the odds would be 100 to 1 that you could go 2 for 2. The much, much, MUCH more likely scenario is that there are many more than two diamonds.

Quote: cclub79

The much, much, MUCH more likely scenario is that there are many more than two diamonds.

I think I would have refrained from using the "much" and "many" characterizations. I suspect that the calculations performed by ChesterDog are probably pretty good, if he used the proper overall probabilities for the various number of diamonds.

Quote: Doc

I think I would have refrained from using the "much" and "many" characterizations. I suspect that the calculations performed by ChesterDog are probably pretty good, if he used the proper overall probabilities for the various number of diamonds.

Good point! Here's how I calculated the overall probability of scratching off two spots and getting two diamonds for each type of ticket. The first column (n) is the type of ticket by number of diamonds. The second column is the number of each type of ticket printed. The third column is the odds of getting each type of ticket expressed as "1 in". The fourth column (P(n)) is the probability of each type of ticket. The fifth column (P(2r|n)) is the probability of revealing two diamonds given that the ticket has n diamonds; it's calculated in Excel by permut(n,2)/permut(25,2) And the last column (P(n)*P(2r|n)) is the probability of buying a type of ticket and revealing two diamonds. The sum of the last column, 0.00885129, is the probability of revealing two diamonds on a ticket with two scratches; this number is used later as the denominator in P(n|2r) = P(2r|n)*P(n)/P(2r).

n	tickets	1 in	P(n)	P(2r|n)	P(n)*P(2r|n)
2	11,418,524	1.4	0.7320	0.0033	0.0024
3	1,560,000	10.0	0.1000	0.0100	0.0010
4	1,560,000	10.0	0.1000	0.0200	0.0020
5	592,800	26.3	0.0380	0.0333	0.0013
6	218,400	71.4	0.0140	0.0500	0.0007
7	124,800	125.0	0.0080	0.0700	0.0006
8	65,000	240.0	0.0042	0.0933	0.0004
9	45,500	342.9	0.0029	0.1200	0.0004
10	14,300	1,090.9	0.0009	0.1500	0.0001
11	520	30,000.0	0.0000	0.1833	0.0000
12	130	120,000.0	0.0000	0.2200	0.0000
13	26	600,000.0	0.0000	0.2600	0.0000
Sum	15,600,000		1.0000		0.0089

Quote: ChesterDog

I agree with the other posters that each losing ticket has two diamonds. Assuming that the diamonds are randomly distributed and using Bayes Theorem, P(A|B) = P(B|A)* P(A)/P(B) (see http://en.wikipedia.org/wiki/Bayes_theorem), I get $11.60 for the value of a ticket with only two spots revealed both being diamonds.

Here's an example of the calculation. The probability of having a total of three diamonds given that the two spots revealed are both diamonds is P(3|2) = P(2|3)*P(3)/P(2) = 0.01 * 0.10 / 0.00885129 = 0.1130

By the way, I get $1.86 as the value of a $3.00 unscratched ticket.

How much rounding are you doing? I get $11.73 for the value of a ticket with only two spots revealed both being diamonds. $3.64 for 1 diamond and 1 other. $1.47 for no diamonds and 2 others. $1.97 for an unscratched ticket.

Quote: rdw4potus

I would bet just about any amount of money that all of the non-winning tickets have 2 diamonds on them. To the extent that all tickets will have at least two diamonds on them, you haven't added any value by pre-locating two (of, hopefully, many) diamonds on your ticket. The most likely scenario, by far, for your ticket is to reveal blanks in all of the remaining spaces.

Well, not all the tickets have at least two diamonds. Some have one, and I did come across one without any. The symbols on the tickets I have seen are are randomly distributed.

The outcome... The ticket with the two initial diamonds was a winner.



Unfortunately, there was only one more diamond amongst the remaining 23, for a $3 break-even return. Had fun, "dreaming big" though. Thanks everyone for your input.