Lottery scratch

Sorry for thwe rudimentary(I assmue) math question from a noob that is mathematically challenged but...

If you have a lottery scratch game that has the odds of winning are 1:3 and you know there are 100k winning tickets, do you know how many losing tickets there are in that game?

Thanks

Quote: udrunkenmonkey

Sorry for thwe rudimentary(I assmue) math question from a noob that is mathematically challenged but...

If you have a lottery scratch game that has the odds of winning are 1:3 and you know there are 100k winning tickets, do you know how many losing tickets there are in that game?

I will say can not be determined from the information provided.

Your initial instinct is to multiple 100k winning tickets by 3 to have 300k losing tickets forcing there to be an equality. However, in most if not all lottery games of chance, the "odds of winning" are based on the expected value and probability. In your example, you don't state the prize amounts, costs of tickets, or even number of tickets. Ever look at many state lotteries where you have to say pick 6 numbers from 59. The probability of hitting the jackpot is well over several hundred million, and yet the "odds of winning" are stated to be substantially lower. Lower demonination prizes impact the "odds of winning", yet the true probability of the jackpot remains the same. Same logic applies to instant scratch offs. You see how the odds on most websites are aprox 1 in 4.5, 1 in 6, etc. Even if you know the # of winning tickets, you still can't determine the number of losing tickets without further information.

By 1:3, do you actually mean "1 in 3", or "1 to 3". "1:3" literally means 1 to 3, or a 25% chance of winning (1 winner to every 3 losers), but typically, lottery odds are given in a "1 in 3" format.

Assuming it is really 1 to 3, than that would imply there are 400k tickets printed, of which 100k are winners. But how many are losers?

-Assuming no "free ticket" prizes, than there would 300k losers.
-If there are free ticket prizes, and they are considered winners, than the answer is still 300k losers.
-However, if there are free ticket prizes and they are neither considered losers nor winners, then the question cannot be answered.

Most lottery websites will list the details of each game, including the total number of tickets distributed, and the breakdown of the prize amounts. Subtract the total number of "winners" from the total distributed, and you will have the number of losers.

Note that the scratch off tickets I have seen list both an, "odds of winning a cash prize" and "odds of winning any prize" on each ticket. If the 1:3 is "odds of winning any prize", then it does include free tickets.

Thanks to all who responded. Sorry for my delay with additional info.

The lottery scratch game cost $20 per ticket. From the back of one ticket:
"on the average, overall odds of winning any prize are 1 in 3.09"

The prizes are as follows:

prize ammount total prizes

$20 79,018

$25 158,036

$50 42,371

$100 36,696

$200 1,979

$500 759

$1000 499

$5000 10

$10000 4

$25000 4

$1000000 2

There are no "free ticket" prizes.
I don't know how many total tickets (both winners and losers) otherwise it would simply be subtraction of all the winners from the total to determine the number of losers.
Is the total number of winners known and the overall odds of 1 in 3.09 enough info to determine the number of losing tickets?

Thanks everyone

One of the prizes is $20 - for a ticket that costs $20. While this qualifies as a 'free ticket', it's obvious based on how it's written that the lottery commission counts them as wins.

Note that the 1 in 3.09 odds is OVERALL odds of winning ANYTHING.

Quote: udrunkenmonkey

Is the total number of winners known and the overall odds of 1 in 3.09 enough info to determine the number of losing tickets?

It seems to me that simple math should be sufficient, but prior posts indicated that it's not....

I know, its comical that the lottery commission counts the $20 prize as winning. I don't think I've won anything unless its more than my initial bet.

The $20 win is the same as a free ticket in the sense that it is the cost of a ticket, but you have the option of getting your $20 bucks back in cash in case you needed to buy a pepperoni stick, coke and a half gallon of gas.

It seems to me that simple math should be sufficient like one of the earlier responses suggested, I'm just looking for validation from a true mathematician. I obviously am not....


Thanks

I think 1 IN 3 means that there are twice as many losers than winners.

Given that there are 319,378 "winners", and the odds of winning "any prize" are 1 in 3.09, there are approximately 667,501 losing tickets.

This assumes that all tickets are sold. Note that with scratchers, all the winners may already be sold, leaving all the losers. The actual odds of winning depend on the number of unsold winners that remain in the pool. There is no guarantee that the distribution and sequence of winners will be even.

The Wizard responded to a similar lottery question (quoted below) on his Wizard of Odds website: here

"My wife and I bought a $20 raffle ticket for the Indiana lottery. My understanding of this game is that the drawing for the winning prizes (which number 777) will be held on August 16, 2007, regardless of the number of actual tickets sold and with the absolute maximum available number of tickets being 325,000. As of today, only 60,000 tickets have been sold. Would it be a good gamble to buy a few additional tickets? What would our odds of winning a prize be? - David B. from Evansville, IN
According to the Indiana Lottery web site, there is a total of $3,270,000 in prize money given to 325,000 ticket holders. That would make each ticket worth $10.615 each on average, assuming the series sold out. At a cost of $20 each, the return is 50.31%. If only 60,000 tickets were sold then each would be worth $54.50, for a return of 272.50%. The breakeven point is 163,500 tickets sold. If you believe that fewer than that will be sold, then buying tickets becomes a good bet, putting aside tax and the utility of money implications. September 30, 2007"

I would say there are 960,000 total tickets. Previous $20 tickets in that state had 960,000 tickets, and all tickets printed are a multiple of 120,000 (atleast all of them that I checked.)