Raffle probability

My daughter had a raffle at school. There were several different prizes, each had a bucket next to it. People, interested in getting a particular prize, would place one or more tickets into the corresponding bucket. Then one ticket was drawn from every bucket, and the prize got awarded to whoever put that ticket in.

My daughter had about 50 tickets. She identified 5 or so prizes she would not mind winning, and then asked me what she should do with her tickets. I suggested that she put ten tickets into each of the 5 buckets. That way, she'd get a shot at winning several prizes.

However, she did not care how many prizes she would win, she just wanted to win anything. I believe, that even with that constraint added, it is still better to distribute tickets evenly between buckets rather than put them all into one, I even made a spreadsheet, demonstrating it for a few fixed values for the numbers of buckets and tickets, but I can't find a way to show that analytically. The problem "feels" like it should have a simple and elegant solution, but I just don't see it :(

Any takers?

Assume, there are N buckets, each already contains n tickets. You have Nk < n tickets of your own, and are trying to maximize the probability of any win. The question is to find an analytical way to tell which strategy is better - should you put k tickets into each of N buckets, or should you just pick one bucket and drop all Nk tickets there?

I think the best strategy would be to put all the tickets into one bucket.

Think about it this way, You have 10 roulette cheques. Are your odds of hitting a number better if you play one number for 10 spins, or 10 different numbers on one spin?

Quote: Ayecarumba

I think the best strategy would be to put all the tickets into one bucket.

I can show, this is not true at least in some cases.

For example, suppose, you have 10 tickets, and 2 buckets, already containing 10 tickets each.
If you put all 10 tickets into one bucket, the probability of win is 1/2. If you go for 5+5, then it is
1-(2/3)^2 = 5/9 > 1/2

Quote: Ayecarumba

I think the best strategy would be to put all the tickets into one bucket.

Think about it this way, You have 10 roulette cheques. Are your odds of hitting a number better if you play one number for 10 spins, or 10 different numbers on one spin?

It is the latter (but the difference isn't tremendously large): 1-(37/38)^10 ~ 8.9/38 < 10/38
I think, this is different though, because the denominator is fixed at 38, whereas in the bucket case, it grows as you put more tickets into the bucket. I.e., the probability of a win grows slower than linearly with the number of tickets you throw at it.

If the number of tickets in each bucket is a large amount, it doesn't matter (I did a test with 100000 tickets and both ways came out with the same odds of winning anything).

Split up the tickets into as many buckets as possible. If there are a lot of tickets (I'd say on the order of 10k) in each bucket and your daughter has one prize she really wants, you're not losing a whole lot of EV on choosing one over the others. On the other hand, if each bucket has 400-500 in them already, splitting your ticket share into as many buckets as possible is (by far) your best choice. Good luck.

Quote: ahiromu

Good luck.

Thanks ... The drawing is actually over (she did win a book, and some kind of a robotic kitten).

The probability of *not* winning anything by throwing Nk tickets into one bucket already containing n is n/(n+Nk), and the probability of losing by spreading between buckets is (n/(n+k))^N. Asymptotically, when n is very large, both of these are close to 1, so you are right that with a huge number of tickets already in it does not matter.

Still, what I am looking for is an (analytical) proof (or at least a reasonably convincing illustration) of the fact that spreading is always better.

Well, starting at an even spread then moving one ticket increases your chance of losing, since

n/(n+k-1) * n/(n+k+1) > n/(n+k) * n/(n+k)

because (n+k-1)*(n+k+1) < (n+k)^2

So, while not being a solid proof, shows that the even spread is better than anything in its neighbourhood. This assumes all buckets have an equal n tickets in them beforehand.

Quote: ahiromu

If the number of tickets in each bucket is a large amount, it doesn't matter (I did a test with 100000 tickets and both ways came out with the same odds of winning anything).

Split up the tickets into as many buckets as possible. If there are a lot of tickets (I'd say on the order of 10k) in each bucket and your daughter has one prize she really wants, you're not losing a whole lot of EV on choosing one over the others. On the other hand, if each bucket has 400-500 in them already, splitting your ticket share into as many buckets as possible is (by far) your best choice. Good luck.

But what if each bucket doesn't contain the same number of tickets before you add yours?

I think your best strategy to win a prize from the available universe, is to "buy" the least popular bucket by putting all your tickets in it. Big fish, small pond.

Quote: Ayecarumba

But what if each bucket doesn't contain the same number of tickets before you add yours?

I think your best strategy to win a prize from the available universe, is to "buy" the least popular bucket by putting all your tickets in it. Big fish, small pond.

For only 2 buckets containing A tickets and B tickets, respectively, the optimal number of tickets for her to place in bucket A would be (50 - A + B)/2, provided that A>0, B>0, A-B<50, and she is allowed to play last. If A-B>=50, then she should put all her tickets in bucket B. And if A=0 or B=0, then she would only have to put one ticket in bucket A or bucket B, respectively.