October 3, 2005 3:50 PM Subscribe

If a company is holding a series of 24 raffles where they give away a prize, and you have 24 raffle tickets which can be used in any number of raffles you choose, do you have better odds entering one ticket per draw, or using them all in one draw? This is assuming that the number of other participants in the raffle is roughly the same from draw-to-draw.

posted by Jairus to Grab Bag (17 answers total)

posted by Jairus to Grab Bag (17 answers total)

I get distribute them:

If 5 other people enter each draw:

Probability you won’t win any if you enter them all in one = 5/29 = .1724

Chance you won’t win any if you distribute them: (5/6)^24 = .012

if 1000 other people enter each:

Chance you won’t win if you enter them all in one = 1000/1024= .9765

Chance you don’t win if you distribute them = (1000/1001)^24 = .9762

So whether small or large numbers of people enter each draw, the chance that you won't win anything are smaller if you distribute the tickets. The smaller the number of people who enter, the greater the discrepancy (and of course the greater your chances of winning, either way).

posted by duck at 4:01 PM on October 3, 2005

If 5 other people enter each draw:

Probability you won’t win any if you enter them all in one = 5/29 = .1724

Chance you won’t win any if you distribute them: (5/6)^24 = .012

if 1000 other people enter each:

Chance you won’t win if you enter them all in one = 1000/1024= .9765

Chance you don’t win if you distribute them = (1000/1001)^24 = .9762

So whether small or large numbers of people enter each draw, the chance that you won't win anything are smaller if you distribute the tickets. The smaller the number of people who enter, the greater the discrepancy (and of course the greater your chances of winning, either way).

posted by duck at 4:01 PM on October 3, 2005

If you get one ticket in each raffle, your chance of winning a prize in each one is (say) *x*. Out of all 24 raffles, your chance of winning any prize at all is 24*x*.

If you get 24 tickets in a single raffle, your chance of winning that particular prize is 24*x*. So if there's one prize you want the most, go with this method, you're 24 times more likely to win *that* prize than in the previous example.

Your chances of winning a prize at all are the same with each method, but the latter method doesn't have any possibility of winning multiple prizes. It depends on what you're hoping to win.

posted by knave at 4:04 PM on October 3, 2005

If you get 24 tickets in a single raffle, your chance of winning that particular prize is 24

Your chances of winning a prize at all are the same with each method, but the latter method doesn't have any possibility of winning multiple prizes. It depends on what you're hoping to win.

posted by knave at 4:04 PM on October 3, 2005

Hm, **duck** makes a good point. My math basically assumes you're in a very large pool, such that buying 24 tickets is pretty much multiplying your odds by 24. But that's not even close to true in a small pool, like 5 people (assuming 1 ticket each).

posted by knave at 4:08 PM on October 3, 2005

posted by knave at 4:08 PM on October 3, 2005

Kirth Gerson is right.

Suppose x is the number of participants in each raffle (all the same) and n_i is the number of tickets you buy for raffle i). Then the probability you don't win anything is the (1-n_1/x)(1-n_2/x)...(1-n_24/x). You want to MINIMIZE the probability you don't win anything.

You minimize this, it turns out, by letting n_1 = 24 and n_i=0 for all i not equal to 24. The easiest way to see this is to think about just two raffles: in that case, (1-n_1/x)(1-n_2/x) = 1-(n_1+n_2)/x+n_1*n_2/(x*x), which is bigger than (1-(n_1+n_2)/x). The same argument works for more raffles.

On the other hand, duck's is also right. His argument looks at the case where the value of x depends on what YOU decide to do - in that case it's quite a bit more complicated, but you can write out a formula like the one I wrote above, and minimize it, and see that you're doing best if you spread your tickets out.

Finally, as can be seen from duck's calculation with "1000", once there are lots more people entering the raffles than you have tickets, it basically doesn't matter. And you *could* do similar calculations to figure out how to maximize the probability that you win, say, at least 3 prizes - you'll get the answer that you're best off putting all your tickets in 3 raffles (if you do a Kirth-style calculation) or distributing them all equally (if you do a duck-style calculation). Again, when there are a lot of contestants it won't really matter what you do.

posted by louigi at 4:09 PM on October 3, 2005

Suppose x is the number of participants in each raffle (all the same) and n_i is the number of tickets you buy for raffle i). Then the probability you don't win anything is the (1-n_1/x)(1-n_2/x)...(1-n_24/x). You want to MINIMIZE the probability you don't win anything.

You minimize this, it turns out, by letting n_1 = 24 and n_i=0 for all i not equal to 24. The easiest way to see this is to think about just two raffles: in that case, (1-n_1/x)(1-n_2/x) = 1-(n_1+n_2)/x+n_1*n_2/(x*x), which is bigger than (1-(n_1+n_2)/x). The same argument works for more raffles.

On the other hand, duck's is also right. His argument looks at the case where the value of x depends on what YOU decide to do - in that case it's quite a bit more complicated, but you can write out a formula like the one I wrote above, and minimize it, and see that you're doing best if you spread your tickets out.

Finally, as can be seen from duck's calculation with "1000", once there are lots more people entering the raffles than you have tickets, it basically doesn't matter. And you *could* do similar calculations to figure out how to maximize the probability that you win, say, at least 3 prizes - you'll get the answer that you're best off putting all your tickets in 3 raffles (if you do a Kirth-style calculation) or distributing them all equally (if you do a duck-style calculation). Again, when there are a lot of contestants it won't really matter what you do.

posted by louigi at 4:09 PM on October 3, 2005

The pool size would be reasonably large, 1500-2000. You'd have your best luck putting all the tickets in one raffle then, yes?

posted by Jairus at 4:10 PM on October 3, 2005

posted by Jairus at 4:10 PM on October 3, 2005

No. Assuming we're talking about something real-worldish, I would go with distributing them. The reason is that if 2000 other people are going to enter raffle A, they're going to do so whether you put one ticket there or whether you put 24 tickets there. (which is my scenario).

If the scenario is that there will be 1500 tickets in your raffle whether that includes your 24 tickets or not (so 1476 other people will enter a ticket if you put all yours there, or 1499 will enter a ticket if you only put one there), then you'd be better of going with one single raffle. This scenario would require that your putting your 24 tickets in there somehow induced others to

And with that number of contestents, it really won't make much of a difference regardless.

posted by duck at 4:22 PM on October 3, 2005

Agreed that it depends on what you want to win. If you want to win multiple prizes, I say distribute them amongst the raffles with the least amount of tickets. If you want to win the best prize, put them all in **that** bucket. If you want to have fun, distribute them all over... then you at least get to feel like you have a chance each and every time they do a raffle.

posted by whatisish at 4:30 PM on October 3, 2005

posted by whatisish at 4:30 PM on October 3, 2005

To further duck's argument, the chances of you winning also include you winning one or more prizes. In the all-in-one case its just one prize. In the distributed case, its 1-24 prizes.

posted by vacapinta at 4:32 PM on October 3, 2005

posted by vacapinta at 4:32 PM on October 3, 2005

Depends also on whether the value of the prizes varies by raffle, what the definition of 'roughly' is and if it's random or a function of your entry or a function of the prize if variable, whether there's a cap on the # of raffle tickets in a given raffle, and whether you can win more than one prize.

Assuming that the prize value (v) stays constant through all raffles, that there are exactly the same # (n) of non-your entries per draw, and that there's no cap and you can win more than one prize, your expected value is

ALL_IN_ONE_BASKET: ((n+24) choose 1) * v

vs

SPREAD_THE_WEALTH: ((n+1) choose 1) * v) + ... (23 other instances)

for n = 1000, assuming v = 1,

ALL_IN_ONE_BASKET = 24/1024 = 0.0234375000

SPREAD_THE_WEALTH = (v/1001) + (v/1001) + ... = 0.023976024

So spreading the wealth is a superior strategy. Especially as the number of other entrants gets low. If n = 1 (there's only one other guy playing one ticket per raffle), then the ALL_IN_ONE strategy gives you 24 chances out of 25 to win one prize. But the SPREAD_THE_WEALTH strategy gives you 24 chances of 1 in 2, which is just ridiculously better.

posted by felix at 4:34 PM on October 3, 2005

Assuming that the prize value (v) stays constant through all raffles, that there are exactly the same # (n) of non-your entries per draw, and that there's no cap and you can win more than one prize, your expected value is

ALL_IN_ONE_BASKET: ((n+24) choose 1) * v

vs

SPREAD_THE_WEALTH: ((n+1) choose 1) * v) + ... (23 other instances)

for n = 1000, assuming v = 1,

ALL_IN_ONE_BASKET = 24/1024 = 0.0234375000

SPREAD_THE_WEALTH = (v/1001) + (v/1001) + ... = 0.023976024

So spreading the wealth is a superior strategy. Especially as the number of other entrants gets low. If n = 1 (there's only one other guy playing one ticket per raffle), then the ALL_IN_ONE strategy gives you 24 chances out of 25 to win one prize. But the SPREAD_THE_WEALTH strategy gives you 24 chances of 1 in 2, which is just ridiculously better.

posted by felix at 4:34 PM on October 3, 2005

If you know that people can choose a raffle, but you don't know who has put what tickets in what raffle, then put all your tickets in raffle no 13. This assumes that people will not be thinking the same as you. If this is the case, then I'd use a random number generator to pick two numbers (e.g. raffles 6 and 16) and put half your tickets in each.

When the lottery started in the UK, they did a thing about which combinations of numbers were picked most often. It was astonishing to see how many people picked numbers 1..6 because (probably) those people assumed that everyone else would assume the chance was that 1..6 would never come up.

posted by seanyboy at 4:39 PM on October 3, 2005

When the lottery started in the UK, they did a thing about which combinations of numbers were picked most often. It was astonishing to see how many people picked numbers 1..6 because (probably) those people assumed that everyone else would assume the chance was that 1..6 would never come up.

posted by seanyboy at 4:39 PM on October 3, 2005

What I'm saying here is that psychology, not statistics will play the most important role here.

posted by seanyboy at 4:40 PM on October 3, 2005

posted by seanyboy at 4:40 PM on October 3, 2005

For the case of 2000 people, spread 'em out.

U:\>perl t1.pl

total winnings for spread: 12029

total winnings for basket: 514

U:\>perl t1.pl

total winnings for spread: 12077

total winnings for basket: 506

U:\>perl t1.pl

total winnings for spread: 12134

total winnings for basket: 498

posted by felix at 4:54 PM on October 3, 2005

U:\>perl t1.pl

total winnings for spread: 12029

total winnings for basket: 514

U:\>perl t1.pl

total winnings for spread: 12077

total winnings for basket: 506

U:\>perl t1.pl

total winnings for spread: 12134

total winnings for basket: 498

#!/usr/bin/perl srand($$); $iterations = 1000000; $tick = 24; $raffle = 24; $entrants = 2000; $totalwinnings = 0; foreach $i (1...$iterations) { $winnings = 0; foreach $raf (1..$raffle) { if (int(rand($entrants + 1)) == 0) { $winnings++; } } $totalwinnings += $winnings; } print "total winnings for spread: $totalwinnings\n"; $totalwinnings = 0; foreach $i (1...$iterations) { $winnings = 0; if (int(rand($entrants + 24)) == 0) { $winnings++; } $totalwinnings += $winnings; } print "total winnings for basket: $totalwinnings\n";

posted by felix at 4:54 PM on October 3, 2005

Your best odds of winning a prize are: figure out which prize is desired least by your colleagues, and enter all your tickets for that prize.

This is akin to lottery draws - if you pick high numbers (over 31), your odds of winning are the same (assuming the pick is random), but your odds of *sharing the prize if you win* are much reduced. Many people play birthdays and important dates for their lottery tickets. This means that lots more people have bets on the numbers from 1-31 than on the numbers 32 and up (and very many on the numbers from 1-12). So if you win $10 million with the numbers 1, 2, 6, you'll probably have to split it with ten people. But if you win it with the numbers 59, 63, 65, you probably won't.

posted by jellicle at 4:59 PM on October 3, 2005

This is akin to lottery draws - if you pick high numbers (over 31), your odds of winning are the same (assuming the pick is random), but your odds of *sharing the prize if you win* are much reduced. Many people play birthdays and important dates for their lottery tickets. This means that lots more people have bets on the numbers from 1-31 than on the numbers 32 and up (and very many on the numbers from 1-12). So if you win $10 million with the numbers 1, 2, 6, you'll probably have to split it with ten people. But if you win it with the numbers 59, 63, 65, you probably won't.

posted by jellicle at 4:59 PM on October 3, 2005

I was just thinking what jellicle said before I read it: Didn't quite understand the scenario, since if raffle participants can choose where to put their tickets, will the amount of participants vary in each draw? Or is it just you who has 24 tickets to play freely?

posted by keijo at 10:35 PM on October 3, 2005

posted by keijo at 10:35 PM on October 3, 2005

This thread is closed to new comments.

posted by Kirth Gerson at 3:53 PM on October 3, 2005