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# Are the odds (ever) in my favor in this raffle?April 19, 2012 10:39 AM   Subscribe

Are the odds in this raffle significantly better than last year?

An organization I am part of holds an annual raffle.
This year, the price to participate has gone up quite a bit, but the rules have also changed. I'm trying to figure out if it is a better or worse deal but using vaguely remembered probability formulas isn't helping.

Last Year :
2 big prizes to win.
18,000 tickets sold.
One prize per ticket(meaning once your ticket is pulled, it's out of the drum).
They claim your odds of winning a big prize were 2 in 18,000.

This year:
12 big prizes to win.
Maximum of 10,000 tickets to be sold.
Multiple prizes per ticket (meaning all tickets go back in the drum for each prize)
They claim your odds of winning a big prize are 12 in 10,000 for a roughly 11 times greater chance of winning a big prize.

Last year, there were a total of 26 prizes to be won.
This year, there are 62 total prizes.

Now, here's the rub.
Last year, \$10/ticket.
This year, \$100/ticket.

I'm trying to work out the following:
1) If the odds are actually as they say (12 in 10,000/2 in 18000)
(They seem not, especially the 18,000, for reasons I can't explain)
2) If the odds are increased, are they increased enough to cover a 10x jump in price.
3) What the overall odds are of winning any prize compared to last year.

Any stats people out there willing to help?
Showing your work would be great so I can see where I went wrong but if you plug it into Matlab, that's fine too
posted by madajb to Science & Nature (17 answers total) 1 user marked this as a favorite

Last year, there were three possibilities: (1) You win the first prize, (2) you don't win the first but you win the second prize, (3) you win no prize.

The odds of winning no prize would be (17,999/18,000)*(17,998/17,999) or 17,998/18,000 [ie, they don't pull your ticket the first time, and they don't pull it the second time]. So the odds of winning any prize would be 1 - (17998/18000) or 2/18000 (or .0111%).

This year, your chance of not winning any prize is (9999/10,000)12 (they don't pull your ticket 12 times), so your chance of winning at least 1 prize is one minus that, or 0.1199% (not exactly 12 in 10000 but close enough).

So your chance of winning at least 1 prize is about 10x higher than last year. But it's still pretty small.
posted by muddgirl at 10:52 AM on April 19, 2012 [2 favorites]

Some information that would be helpful - what are the values of the 62 prizes? One way to approach this is the expected return, and you need to know how much they are offering in terms of prizes to compute that.
posted by procrastination at 10:52 AM on April 19, 2012 [1 favorite]

I do agree with procrastination that this doesn't speak to expected return - you might be 10x more likely to win a prize, but the expected return might be lower depending on the cost of the prizes.
posted by muddgirl at 10:53 AM on April 19, 2012

Multiple prizes per ticket (meaning all tickets go back in the drum for each prize)

I just want to point out that this actually decreases your odd of getting at least one prize which may seem counter-intuitive.
posted by vacapinta at 10:56 AM on April 19, 2012 [2 favorites]

You can't really answer without knowing how many tickets are actually sold -- it seems quite likely that if they only sold 18k tickets at \$10 last year, they won't be able to sell 10k tickets at \$100.
posted by Perplexity at 10:57 AM on April 19, 2012 [3 favorites]

I think we'd need something other than "number of prizes to be won" when you're also introducing a price increase per ticket.

On the face of it though, let's see. Can we assume the "big prizes" have the same value this year as last year?

Last year was 1 in 9000 for a "big prize"
This year is 1 in 833 for a "big prize"

Disregarding "small prizes" for the time being, it appears that there is indeed a slight edge this year over last year, considering the ten-fold price increase (833:1 vs. 900:1)
posted by ShutterBun at 10:58 AM on April 19, 2012

meaning once your ticket is pulled, it's out of the drum

Once a ticket is pulled, there are 17999 tickets left (assuming all tickets are sold). Assuming that you didn't win the first time around, the odds for the second pull on the first raffle must be 1/17999. The only way to win both pulls is to buy two tickets, which increases your odds to 1/18000+1/17999, which is close to but not exactly 2/18000.

For the second raffle, the tickets are put back into the barrel. Your odds of having one winning ticket are the same on each pull: 1/10000. (Assuming all tickets are sold.) If you buy twelve unique tickets, your odds are 1/10000+...+1/10000=12/10000.

So you'll have to buy multiple and more expensive tickets to get better odds. I imagine the increased expense on the wager covers their increased losses on prizes.
posted by Blazecock Pileon at 10:58 AM on April 19, 2012

The prizes this year range from about \$18,000 at the top end to about \$3-400 on the low end.

Last year, the prizes ranged from about the same on the top to about \$1000 on the low end.
posted by madajb at 10:58 AM on April 19, 2012

Oh, if there are small prizes in both raffles that aren't counted in your question, and tickets are not returned to the drum, your chance of winning a 'big prize' is indeed very slightly lower than 2/18,000 (because your ticket has a small chance to be pulled before that drawing), but your chance of winning any prize at all is slightly higher. We'd have to know the total number of prizes in each raffle to calculate this precisely.
posted by muddgirl at 11:03 AM on April 19, 2012

Assuming all of the tickets are/were sold, here's your chances of one ticket winning:
```
Last year:      This year:
any prize       0.14%           0.62%
big prize       0.011%          0.12%```
I'm going to assume that the prizes have the same value as last year.
Just talking about the big prizes, you have a slightly better return this year than last year. But for other prizes, your expected return is about half of what it was last year. If the big prize is a lot bigger than the other prizes (e.g., a car vs a \$50 gift certificate at the Cheesecake Factory) then that decrease doesn't really matter, and the odds are effectively the same.

But the small prizes have to be a lot smaller than the big ones for the expected return to have actually increased. If the small prizes have decreased in value (which it sounds like is the case) then the expected return is really less.

The easiest way to calculate all this is to just add up the value of all the prizes, and divide it by the number of tickets. That's your expected return on one ticket.

If the raffle doesn't sell all 10k tickets though the story is different, and each ticket is better off. However it's not very likely that buying a raffle ticket is ever going to be a good investment (though I suppose charity raffles with donated prizes can afford to give away more than they take in. I don't know how often this actually happens though). The main reason to do it is to support the organization running the raffle.
posted by aubilenon at 11:04 AM on April 19, 2012

We'd have to know the total number of prizes in each raffle to calculate this precisely.

(I did put this in the original post, sorry if it wasn't clear)

Last year, there were a total of 26 prizes to be won.
This year, there are 62 total prizes.
posted by madajb at 11:05 AM on April 19, 2012

Gosh, I totally missed that the number of prizes was in the question. My answers are all wrong.
posted by muddgirl at 11:05 AM on April 19, 2012

Wait, muddgirl, is that correct? It's my understanding that last year, if you won a "minor" prize, your ticket was out of the running and you could not win one of the two "major" prizes, whereas this year you still have a chance to win a major prize even if you also win a minor prize.

But that doesn't actually affect your odds of winning the "major" prize. Imagine if last year they drew all the major prizes first, then drew the minor prizes. I'm sure they didn't do it that way, because it would be really anticlimactic, but it *didn't actually matter* for the sake of the odds what order the prizes were drawn.
posted by mskyle at 11:06 AM on April 19, 2012

Addendum: if they put your ticket back in the drum after you win a *major* prize, then yes, there is a slightly lower chance of winning a major prize, but you also have a small chance of winning two major prizes. I'm *guessing* that this is not the case, but I don't know.
posted by mskyle at 11:14 AM on April 19, 2012

I'm trying to figure out if it is a better or worse deal but using vaguely remembered probability formulas isn't helping.

I think this is actually a much simpler question. It really doesn't matter how many prizes there are, if tickets are returned after being pulled, etc. All you want to know is the expected value of a ticket relative to its cost. You just need to tally up or estimate the total value of the prizes to be won and divide this by the total cost of all tickets sold (you may need to estimate if you don't expect all 10,000 will be sold). Do the same for last year, compare the two and you'll have your answer.
posted by ssg at 11:32 AM on April 19, 2012 [2 favorites]

Addendum: if they put your ticket back in the drum after you win a *major* prize, then yes, there is a slightly lower chance of winning a major prize, but you also have a small chance of winning two major prizes. I'm *guessing* that this is not the case, but I don't know.

As far as I know, each ticket is in the drawing for all 62 prizes. Multiple prizes are allowed for each ticket.
Order is not specified, but based on past contests, it'll probably go in rough cash equivalent order.
So, a bunch of minor awards, then up a level, then the most expensive.
posted by madajb at 3:52 PM on April 19, 2012

Also, just looked in the official rules, this years drawing is subject to a minimum sale of 7500 tickets.
posted by madajb at 4:06 PM on April 19, 2012

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