# What's the optimal amount to spend on this raffle?

July 24, 2009 9:04 AM Subscribe

There's a monthly raffle at work for 5 parking spaces out back. There are about 30 people in my office. There's no limit to the amount of tickets you can buy. You can only win once. Otherwise, parking costs about $4 a day, or about $80 a month. Raffle proceeds go to charity. What's the optimal amount to spend on this?

I should tell you also that this question is purely academic because the raffle has been discontinued. But I still want to know what you all think because work is boring as all get out.

What's the best way to figure this out? One friend said she'd use a stats program, another said Matlab. I was thinking maybe a nice line graph with asymptotes and bright colors and such?

I figure the absolute most you would want to spend on this raffle would be $80, only because it goes to charity. Also, I'm assuming no one else in the office ever bought more than 2 or 3 tickets.

I should tell you also that this question is purely academic because the raffle has been discontinued. But I still want to know what you all think because work is boring as all get out.

What's the best way to figure this out? One friend said she'd use a stats program, another said Matlab. I was thinking maybe a nice line graph with asymptotes and bright colors and such?

I figure the absolute most you would want to spend on this raffle would be $80, only because it goes to charity. Also, I'm assuming no one else in the office ever bought more than 2 or 3 tickets.

This is really unsolvable, because you can't control how many tickets other people buy.

However, if you assume that other people will buy only one, and you are willing to buy up to 15 (15*$5=$75, which is better than 80), then:

Your odds of winning = tickets you bought / total tickets bought

Your "utility" = odds of winning / money spent.

If you buy one ticket, your odds of winning are about 3%, and your utility is 0.0067; if you buy 15 tickets, your odds of winning are about 34% but your utility drops to 0.0045.

posted by adamrice at 9:29 AM on July 24, 2009

However, if you assume that other people will buy only one, and you are willing to buy up to 15 (15*$5=$75, which is better than 80), then:

Your odds of winning = tickets you bought / total tickets bought

Your "utility" = odds of winning / money spent.

If you buy one ticket, your odds of winning are about 3%, and your utility is 0.0067; if you buy 15 tickets, your odds of winning are about 34% but your utility drops to 0.0045.

posted by adamrice at 9:29 AM on July 24, 2009

When you say everyone only bought 2 or 3 tix, how many people are we talking about? I would look at it one of two ways. First, it is for charity so I would pick a number I was willing to give to Charity and buy that amount worth of tickets. Or, I would figure out my expected positive return. Sort of like buy lottery tickets. If I had a 1 in 10 chance with 1 ticket, my expected return would be $8 so I would do it. More than 1 in 16 I would not do it. Less than 1 in 16 I would probably enter.

posted by JohnnyGunn at 9:29 AM on July 24, 2009

posted by JohnnyGunn at 9:29 AM on July 24, 2009

How much are you willing to lose (er... donate to charity) for the chance of winning a space? That's the amount.

posted by The Deej at 9:32 AM on July 24, 2009 [1 favorite]

posted by The Deej at 9:32 AM on July 24, 2009 [1 favorite]

You want to keep buying tickets until buying a ticket increases your probability of winning by less than 1/16 = 0.0625. (That's $5, the price of a ticket, divided by $80, the value of the prize.)

So let's say the number of tickets bought by other people is n, and you buy k tickets. Let C(n, k) denote the binomial coefficient. There are C(n+k, 5) ways that the winning tickets can be drawn, of which C(n, 5) have you losing. So your chances of losing are C(n,5)/C(n+k, 5).

If n = 60 (that is, sixty other tickets are sold), your chances of winning are 0.08197 if you buy one ticket; 0.15600 if you buy two; 0.22299 if you buy three; 0.28369 if you buy four. Taking the differences, it's worth buying the third ticket but not the fourth.

But it turns out that if everybody does this and buys

So there's some game theory in your problem that I don't feel like thinking about.

posted by madcaptenor at 9:34 AM on July 24, 2009 [3 favorites]

So let's say the number of tickets bought by other people is n, and you buy k tickets. Let C(n, k) denote the binomial coefficient. There are C(n+k, 5) ways that the winning tickets can be drawn, of which C(n, 5) have you losing. So your chances of losing are C(n,5)/C(n+k, 5).

If n = 60 (that is, sixty other tickets are sold), your chances of winning are 0.08197 if you buy one ticket; 0.15600 if you buy two; 0.22299 if you buy three; 0.28369 if you buy four. Taking the differences, it's worth buying the third ticket but not the fourth.

But it turns out that if everybody does this and buys

*three*tickets, and we set n = 90 (you said thirty people in the office), it's not worth buying tickets at all! Then your chances of winning on a single ticket are 0.05495.So there's some game theory in your problem that I don't feel like thinking about.

posted by madcaptenor at 9:34 AM on July 24, 2009 [3 favorites]

Yes there IS some game theory in there, and I feel like thinking about it as I have nothing better to do than data entry.

There are some known unknowns here, namely a)maybe not everyone will buy a ticket, and b)how many tickets will each person buy?

That all the money goes to charity softens the blow a bit if I end up losing. It could also provide an incentive for me to manipulate my coworkers into buying more tickets.

madcaptenor, I'm not so hot with calculus so forgive me if you addressed this in the paragraph above, but I should clarify that each person can only win once. So there will be five separate winners.

posted by boghead at 10:11 AM on July 24, 2009

There are some known unknowns here, namely a)maybe not everyone will buy a ticket, and b)how many tickets will each person buy?

That all the money goes to charity softens the blow a bit if I end up losing. It could also provide an incentive for me to manipulate my coworkers into buying more tickets.

madcaptenor, I'm not so hot with calculus so forgive me if you addressed this in the paragraph above, but I should clarify that each person can only win once. So there will be five separate winners.

posted by boghead at 10:11 AM on July 24, 2009

For the short answer, see the final paragraph.

The optimal strategy depends on everyone else's strategy. You don't know their strategies. One way to handle that is to assume everyone else uses your strategy as well. But in that case, the optimal strategy is to buy one ticket. Everyone else will buy the same number as you, so your chance of winning is constant and you might as well minimize cost.

[I'm going to use fake-o probabilities here. We'll pretend that the chance of winning is [your entries] over [total entries] times five. It's really way more complicated than that, with "this choose that" and "no replacement" and even "that guy just won so now it's actually (this minus his) choose that," but I don't like probability that much. Fake-o probabilities are close enough for large enough values of [total entries] and for certain definitions of "enough."]

One ticket is $5. In the everyone-buys-one game, you have a fake-o 5 in 30 chance of winning. Value of the prize is $80. Expected winnings (averaged over infinite months, say) are 5/30*80 = $13.33. $5 is less than $13.33, so you come out ahead there.

But that's not going to happen unless you get together with your coworkers and game the system, colluding to each buy only one ticket (this *would* be mutually beneficial to everyone, but all it takes is one jerk buying five tickets...). So then we get into the murky area of other people buying unknown numbers of tickets.

We can still do some analysis, because all that really matters is the total number of tickets bought other than your own. For any such total, we can figure out the optimal number of tickets for you to buy.

Let's say everyone else buys about 5 tickets, so there are 150 tickets purchased. You're going to buy X on top of that. Value of the prize is $80. Expected winnings are your chance of winning (fake-o: 5 times X out of 150+X) times the value of the prize: 5*X/(150+X)*80. The total expected value (winnings - cost) is then 5*X/(150+X)*80 - 5*X. Oops, that's negative for any positive value of X. In the long run, you'll lose money consistently over just paying for parking.

So what if everyone else only buys 2 tickets, on average? 5*X/(60+X)*80 - 5*X has a maximum at about 9. So if each of your coworkers buys 2 tickets, on average, and you buy 9, you'll come out ahead. By about $7 per month. And that's the best you can do if that many tickets are sold on average.

[On preview, madcaptenor's math says I'm wrong, but my fake-o math is unassailable, I say! Plus, I think he or she didn't account for the "you can only win once" clause, which makes it even more messy. Uh, so we're both wrong. But I'm more wrong.]

And. If anyone else figured out your clever scheme, then they'd start wanting to buy 9 tickets, too. Then everything is thrown off, nobody comes out ahead, and you're all screwed. So tell them you're only buying two.

Ah! But you derive some benefit, some feel-good-ness from giving to charity! Basically, some tickets are "free." How many dollars would you donate to charity without this raffle going on? Consider that a given. If you would donate $20 per month, then you might as well put that in the raffle, and the tickets are in some sense free. Then your expected value, for C free tickets due to charitable donation, X regular tickets, and T total tickets sold to others is:

5*(C+X)/(T+C+X)*80 - 5*X.

Your chances have gone up, but your cost stays the same at 5*X. Plug in your value for C, whatever your best guess is for T, and find the maximum of the curve. I've just been using an online graphic calculator and eyeballing it, though there are probably easier tools than that one.

Anyway, your question was what's the best way to figure this out, right? To get the to the point, I claim that the best way is with some fake math; some graphs with pretty colors, interactive zoom, and maxima marked with overflowing treasure chest icons that make cha-ching noises when you mouseover them; and possibly a few over-the-top assumptions that besmirch your coworkers' intelligence and character.

posted by whatnotever at 10:21 AM on July 24, 2009 [1 favorite]

The optimal strategy depends on everyone else's strategy. You don't know their strategies. One way to handle that is to assume everyone else uses your strategy as well. But in that case, the optimal strategy is to buy one ticket. Everyone else will buy the same number as you, so your chance of winning is constant and you might as well minimize cost.

[I'm going to use fake-o probabilities here. We'll pretend that the chance of winning is [your entries] over [total entries] times five. It's really way more complicated than that, with "this choose that" and "no replacement" and even "that guy just won so now it's actually (this minus his) choose that," but I don't like probability that much. Fake-o probabilities are close enough for large enough values of [total entries] and for certain definitions of "enough."]

One ticket is $5. In the everyone-buys-one game, you have a fake-o 5 in 30 chance of winning. Value of the prize is $80. Expected winnings (averaged over infinite months, say) are 5/30*80 = $13.33. $5 is less than $13.33, so you come out ahead there.

But that's not going to happen unless you get together with your coworkers and game the system, colluding to each buy only one ticket (this *would* be mutually beneficial to everyone, but all it takes is one jerk buying five tickets...). So then we get into the murky area of other people buying unknown numbers of tickets.

We can still do some analysis, because all that really matters is the total number of tickets bought other than your own. For any such total, we can figure out the optimal number of tickets for you to buy.

Let's say everyone else buys about 5 tickets, so there are 150 tickets purchased. You're going to buy X on top of that. Value of the prize is $80. Expected winnings are your chance of winning (fake-o: 5 times X out of 150+X) times the value of the prize: 5*X/(150+X)*80. The total expected value (winnings - cost) is then 5*X/(150+X)*80 - 5*X. Oops, that's negative for any positive value of X. In the long run, you'll lose money consistently over just paying for parking.

So what if everyone else only buys 2 tickets, on average? 5*X/(60+X)*80 - 5*X has a maximum at about 9. So if each of your coworkers buys 2 tickets, on average, and you buy 9, you'll come out ahead. By about $7 per month. And that's the best you can do if that many tickets are sold on average.

[On preview, madcaptenor's math says I'm wrong, but my fake-o math is unassailable, I say! Plus, I think he or she didn't account for the "you can only win once" clause, which makes it even more messy. Uh, so we're both wrong. But I'm more wrong.]

And. If anyone else figured out your clever scheme, then they'd start wanting to buy 9 tickets, too. Then everything is thrown off, nobody comes out ahead, and you're all screwed. So tell them you're only buying two.

Ah! But you derive some benefit, some feel-good-ness from giving to charity! Basically, some tickets are "free." How many dollars would you donate to charity without this raffle going on? Consider that a given. If you would donate $20 per month, then you might as well put that in the raffle, and the tickets are in some sense free. Then your expected value, for C free tickets due to charitable donation, X regular tickets, and T total tickets sold to others is:

5*(C+X)/(T+C+X)*80 - 5*X.

Your chances have gone up, but your cost stays the same at 5*X. Plug in your value for C, whatever your best guess is for T, and find the maximum of the curve. I've just been using an online graphic calculator and eyeballing it, though there are probably easier tools than that one.

Anyway, your question was what's the best way to figure this out, right? To get the to the point, I claim that the best way is with some fake math; some graphs with pretty colors, interactive zoom, and maxima marked with overflowing treasure chest icons that make cha-ching noises when you mouseover them; and possibly a few over-the-top assumptions that besmirch your coworkers' intelligence and character.

posted by whatnotever at 10:21 AM on July 24, 2009 [1 favorite]

Better people have done the math and walls of text, so I'll be short and sweet: given that your 29 coworkers buy three tickets each, your profit is a line that starts with a negative y and a negative slope (-0.15668), as well as a negative derivative, so further tickets will only bring you more quickly down that slippery slope (pun intended).

But hey, it's for charity, right? So dump in five bucks, and you'll only lose fifteen cents while giving five dollars to charity. (It's the one who rents these parking spaces that really loses.)

posted by ThatRandomGuy at 10:34 AM on July 24, 2009

But hey, it's for charity, right? So dump in five bucks, and you'll only lose fifteen cents while giving five dollars to charity. (It's the one who rents these parking spaces that really loses.)

**So the real question is...how much do you like this charity?**posted by ThatRandomGuy at 10:34 AM on July 24, 2009

This reminds me of a betting trap that a friend likes to spring on people. "I have a dollar. The winning bid gets the dollar. The catch is that the loser has to pay their bid, too." If he runs this on two people who are strongly motivated to "win" (or are strongly motivated to make the other player lose), they'll bid way, way past $1. The only winner is my friend, who pockets the (large) difference.

Unless you know whether you're bidding against people who are really motivated to win, I'd wait for more information. That might mean sitting out the first round, or going in for $5 if that's what it takes to find out how other people bid.

posted by dws at 10:37 AM on July 24, 2009

Unless you know whether you're bidding against people who are really motivated to win, I'd wait for more information. That might mean sitting out the first round, or going in for $5 if that's what it takes to find out how other people bid.

posted by dws at 10:37 AM on July 24, 2009

5/X + 4/(X-1) + 3/(X-2) + 2/(X-3) + 1/(X-4) chances of winning $80.

Lets call the sum of these odds Y1/X.

Price of ticket: $5

Thinking of strictly financial gain, ignoring benefits to charity:

Simple case: You would want better than 1 in 16 odds on your money if you buy 1 ticket. (6.25% chance you'd win.

The equation at the top will change to Yn/X, if you buy n tickets, where Yn is a function of n. You would want better than n in 16 odds on your money if you buy n tickets.

If you know what X is (by asking everyone who buys tickets), you will be able to calculate how many tickets to buy. Compare the differences between Yn/X and n/16. Pick the one that gives you the biggest difference.

This all assumes you can determine what X is or make a good assumption. Whether or not you make a good bet that benefits yourself depends on X.

posted by ajackson at 11:07 AM on July 24, 2009

Lets call the sum of these odds Y1/X.

Price of ticket: $5

Thinking of strictly financial gain, ignoring benefits to charity:

Simple case: You would want better than 1 in 16 odds on your money if you buy 1 ticket. (6.25% chance you'd win.

The equation at the top will change to Yn/X, if you buy n tickets, where Yn is a function of n. You would want better than n in 16 odds on your money if you buy n tickets.

If you know what X is (by asking everyone who buys tickets), you will be able to calculate how many tickets to buy. Compare the differences between Yn/X and n/16. Pick the one that gives you the biggest difference.

This all assumes you can determine what X is or make a good assumption. Whether or not you make a good bet that benefits yourself depends on X.

posted by ajackson at 11:07 AM on July 24, 2009

Haha, X being the number of tickets bought.

You could easily use MatLAB to spit out a graph for you--you would need to input the 16 Yn/X equations and then plot Yn/X - n/X as a function of n and X--the peaks will give you your answers.

posted by ajackson at 11:15 AM on July 24, 2009

You could easily use MatLAB to spit out a graph for you--you would need to input the 16 Yn/X equations and then plot Yn/X - n/X as a function of n and X--the peaks will give you your answers.

posted by ajackson at 11:15 AM on July 24, 2009

Do they release the amount of $$$ raised for charity from the past raffles? That info would tell you what to expect the ### of tickets sold to be and allow for an optimal calculation (assuming the correct formula is used).

posted by Bort at 11:30 AM on July 24, 2009

posted by Bort at 11:30 AM on July 24, 2009

Since each person can win only once, you're guaranteed to win the 30th month of the raffle, right?. So, one ticket for $5.00 that month should do it. If you don't mind waiting for two and a half years, that is.

I'm just saying.

posted by Dolley at 4:04 PM on July 24, 2009

I'm just saying.

posted by Dolley at 4:04 PM on July 24, 2009

It's once per month, not once per the life of the lottery -- as in, there are five spaces, and no one person gets more than one of them.

posted by jacquilynne at 4:26 PM on July 24, 2009

posted by jacquilynne at 4:26 PM on July 24, 2009

I believe that without having a good guess as to the total number of tickets sold, and thus your probability of winning, there isn't a good way to model this sensibly.

If you're going to park in back no matter what but are looking to subsidize that cost by winning, $5/ticket isn't going to give you anything approaching betting odds for any reasonable distribution or number of tickets sold as demonstrated by ThatRandomGuy. That is you're going to have to buy a great many more tickets than the sixth-most-tickets-purchased person to be relatively certain of winning a spot such that your total cost is going to approach $80 anyway. Not to mention, if you don't win you're out the cost of your tickets plus the $80/month.

For any sensible distribution and number of tickets sold, I don't think the limiting factor is so much how many tickets others are buying as it is the value of the prize relative to the cost per ticket. In order for buying tickets to begin to pan out for you the company is going to be contributing considerably less to charity than the value of the parking spaces so why not sell them at the full rate and just drop $80/spot/month in charitable contributions?

None of this considers any potential non-economic value you might derive from either gambling at work or making charitable contributions though. Basically you're free to graph what ever curves you feel look good and then create a PowerPoint presentation around your inputs and findings to support whatever conclusions you feel appropriate. Bonus points for using Microsoft ClipArt. This is more fun than data entry in the same way that drafting this answer was more fun than fixing tricksy MSI bugs on a Friday afternoon...which was the point of this question, no?

posted by Fezboy! at 4:57 PM on July 24, 2009

If you're going to park in back no matter what but are looking to subsidize that cost by winning, $5/ticket isn't going to give you anything approaching betting odds for any reasonable distribution or number of tickets sold as demonstrated by ThatRandomGuy. That is you're going to have to buy a great many more tickets than the sixth-most-tickets-purchased person to be relatively certain of winning a spot such that your total cost is going to approach $80 anyway. Not to mention, if you don't win you're out the cost of your tickets plus the $80/month.

For any sensible distribution and number of tickets sold, I don't think the limiting factor is so much how many tickets others are buying as it is the value of the prize relative to the cost per ticket. In order for buying tickets to begin to pan out for you the company is going to be contributing considerably less to charity than the value of the parking spaces so why not sell them at the full rate and just drop $80/spot/month in charitable contributions?

None of this considers any potential non-economic value you might derive from either gambling at work or making charitable contributions though. Basically you're free to graph what ever curves you feel look good and then create a PowerPoint presentation around your inputs and findings to support whatever conclusions you feel appropriate. Bonus points for using Microsoft ClipArt. This is more fun than data entry in the same way that drafting this answer was more fun than fixing tricksy MSI bugs on a Friday afternoon...which was the point of this question, no?

posted by Fezboy! at 4:57 PM on July 24, 2009

Thank you all so much for your answers. I'll come back and read this tomorrow when I'm not drunk. I should also mention that while there are five winners in each raffle drawing, the first winner gets to pick the charity.

A few final thoughts:

Looking at how much money has been raised for charity in previous months is a good idea.

Non-raffle parking is $80 per month at minimum and a solid 5-7 minutes walk further away from raffle parking.

I'm a graphic designer, so I like buttons and pretty colors. Whatnotever, it's good to know I could use a Nissa Maxima with a Cha-Ching! hover pseudostate. Or a line graph.

I walk to work from home. Before this raffle was discontinued by the bosses in May, I was considering entering and then auctioning off my winning ticket for the same charity. Ah, good times at Bank of Awfulness.

Thanks again Mefis.

posted by boghead at 10:56 PM on July 24, 2009

A few final thoughts:

Looking at how much money has been raised for charity in previous months is a good idea.

Non-raffle parking is $80 per month at minimum and a solid 5-7 minutes walk further away from raffle parking.

I'm a graphic designer, so I like buttons and pretty colors. Whatnotever, it's good to know I could use a Nissa Maxima with a Cha-Ching! hover pseudostate. Or a line graph.

I walk to work from home. Before this raffle was discontinued by the bosses in May, I was considering entering and then auctioning off my winning ticket for the same charity. Ah, good times at Bank of Awfulness.

Thanks again Mefis.

posted by boghead at 10:56 PM on July 24, 2009

Oh also, I registered my name and then decided NOT to donate the 5 dollars because I was poor. Then four months later I did. Why? Because of this raffle. Because I love you guys.

posted by boghead at 11:18 PM on July 24, 2009

posted by boghead at 11:18 PM on July 24, 2009

After thinking quite a bit and playing around in excel, here are my conclusions.

The number of potential buyers doesn't factor into the calculation(s) (which I find interesting and somewhat counter intuitive). My thinking:

There are 5 spots worth $80/spot (assume same value as other parking - but throw in a premium for being close to the building if you'd like), so the spots are worth $400. Now, the tickets will be giving away the spots, so the value of the tickets should be equal to the spots, so tickets value = $400. The price of a ticket is $5, so if 80 tickets are sold ($400/$5), then the tickets have the value they cost, and it is a wash. If less than 80 tickets are sold, then you should buy at least 1 (up to about 5, from my calculations, depending on how many tickets are sold) and if there are 80 or more tickets sold, then you are losing money to buy one.

The crucial number is still how many tickets were sold during previous raffles, but now I think we know what the magic number to compare it to is: 80.

Assume you give parking in one of the 5 raffled spots to be worth $20/month. Then the 5 spots are worth (to you) $100/spot * 5 = 500. Tickets are $5, so 100 is the magic number of tickets sold.

[Note that with these numbers, since the cost of the ticket is equal to the number of spots for sale, that the optimal tickets sold (the magic/important number) is equal to the value of a spot - this is just a coincidence due to the equality of ticket value and number of spots. If these changed, then the magic number would not correspond to spot value. See?]

posted by Bort at 9:13 AM on July 25, 2009

The number of potential buyers doesn't factor into the calculation(s) (which I find interesting and somewhat counter intuitive). My thinking:

There are 5 spots worth $80/spot (assume same value as other parking - but throw in a premium for being close to the building if you'd like), so the spots are worth $400. Now, the tickets will be giving away the spots, so the value of the tickets should be equal to the spots, so tickets value = $400. The price of a ticket is $5, so if 80 tickets are sold ($400/$5), then the tickets have the value they cost, and it is a wash. If less than 80 tickets are sold, then you should buy at least 1 (up to about 5, from my calculations, depending on how many tickets are sold) and if there are 80 or more tickets sold, then you are losing money to buy one.

The crucial number is still how many tickets were sold during previous raffles, but now I think we know what the magic number to compare it to is: 80.

Assume you give parking in one of the 5 raffled spots to be worth $20/month. Then the 5 spots are worth (to you) $100/spot * 5 = 500. Tickets are $5, so 100 is the magic number of tickets sold.

[Note that with these numbers, since the cost of the ticket is equal to the number of spots for sale, that the optimal tickets sold (the magic/important number) is equal to the value of a spot - this is just a coincidence due to the equality of ticket value and number of spots. If these changed, then the magic number would not correspond to spot value. See?]

posted by Bort at 9:13 AM on July 25, 2009

This thread is closed to new comments.

posted by amtho at 9:09 AM on July 24, 2009