I recently received the following email from my job at TC3…
2013 “POT OF GOLD” 50/50 RAFFLE
Our 17th year of making some lucky winner $1,000 richer!
$20 per ticket – only 100 tickets sold.
This got me to thinking: Is this 50/50 raffle worth the ticket? Are such raffles ever worth it? Is it even possible for it to be worth it?
Tools to Answer the Question
To answer this question, we need to calculate something called an expected value. This is a tool in statistics that we can use to find the average payout for a situation, given the cost of participating and the probability of winning. To understand expected value, consider the following example.
A simple card game costs $5 to play. You draw a single card from a well-shuffled deck and win a prize based on its value. If the card is the Ace of Hearts, you win $100. If it is a different Ace, you win $20. If it is a different heart, you win $5. If it is a different card entirely, you win nothing.
Imagine playing this game a number of times — 52 times to be exact. If the world perfectly followed probability rules, you would expect to get each individual card exactly once in 52 plays. So you’d win the $100 one time, the $20 three times, the $5 twelve times, and lose 36 times. We need to factor in the cost of playing, though, as well. This gives us the following breakdown of Profit.
- $95 – 1 time
- $20 – 3 times
- $0 – 12 times
- $-5 – 36 times
The expected value of the game is merely the average of all the possible payouts. So we would add one 95, three 20s, 12 zeroes, and -5 thirty-six times: 95 x 1 + 20 x 3 + 12 x 0 + -5 x 36 = -25. Divide this result by 52 to get the average, and we can say that, per game, we’d lose an average of around 48 cents.
We use the expected value to determine whether the game is worth playing. While we might luck out in the short-run and get that Ace of Hearts, if we keep playing the game (theoretically, if we do so an infinite number of times) we would lose money – 48 cents per game in fact. A rational person would decide not to play.
A slightly different – but significantly useful – way to calculate this expected value is to divide by 52 all throughout the calculation: 95 x (1/52) + 20 x (3/52) + 0 x (12/52) + -5 (36/52) = -0.48. What this turns the calculation into is merely a product of each possible outcome for the game and its respective probability. There’s a 1/52 chance we get the Ace of Hearts from the deck, so there’s a 1/52 chance we make a profit of $95.
This means that to find the expected value for any situation, we can simply multiply each outcome by its probability and add up all those products.
Is there an App(lication) for That?
So let’s apply this process to the 50/50 game described above. We pay $20 to play, and have a 1/100 chance of winning half of the pot: $1000. For every 1 time we win $1000, we lose $20 ninety-nine times. Put another way, we have a 1/100 chance of getting a profit of $980, and a 99/100 chance of a profit of -$20. The calculation for the expected value is easy to do:
980 x (1/100) + -20 x (99/100) = -10
So if we played this game every year, we would expect to lose, on average, $10 per year. Keep in mind that it isn’t actually possible to lose $10 playing this game, just as it isn’t possible for an American woman to have had 2.06 kids. This is an average. Still, this means that this is a game stacked against the player, and so it is not a good idea for us to play.
A 50/50 by Any Other Name…
This version isn’t how 50/50 raffles are often played, however. If you’ve ever been to a Cornell Big Red Hockey game, you’ve probably seen people selling tickets, or have heard an announcement about the result. Often, there is not a fixed number of tickets like there is in the example above. People buy a ticket, usually for $1 each, and the randomly chosen winner gets half the total sales that are made. If 438 tickets are sold, then the winner gets $219 and the raffle organizers, usually a charity of some kind, gets the other $219. This presents an interesting problem when calculating the expected value, as your probability of winning and how much you win change depending on how many people play. We can still approach this in the same way.
First, we’ll assume that each ticket costs $1 and there are no special deals for buying more tickets (15 tickets for $10, for example). We’ll also assume that, as in most 50/50 raffles, there is one winner and that person wins half the pot (and that amount isn’t rounded; if the pot is $5, the winner gets $2.50). We will have to introduce the variable x to represent the number of people playing. This means that the pot is $x, and the winner will receive x/2 dollars as a prize (meaning a profit of x/2 – 1 for you). It also means that you have a 1/x probability of winning, assuming you only buy one ticket.
All this being known, we can say the following…
In other words, regardless of how many people are playing, your expected value is -0.5. This is actually a pretty big deal. It doesn’t matter if you’re the only one playing, or if there are 100,000 people playing, you can expect to lose 50 cents for each game. And it doesn’t matter if you buy more tickets. If you buy 2 tickets, you can expect to lose $1 for each game; you can expect to lose $1.50 if you buy 3 tickets, and so on. These 50/50 raffles have the worst guarantee ever: you can only expect to get half your money back.
Is It Ever Worth It?
Can we construct a 50/50 raffle where you can expect to make a profit? Assuming we keep the 50/50 premise — that the one winner gets exactly half the pot — the only thing we can alter that might have an effect is the cost of the tickets. This is easy to figure out. Since we expect to lose half of the money that we spend on a ticket, we need to find a way of getting free tickets. If the raffle has a buy-one-get-one-free deal, like 20 tickets for the price of ten, the money that we’re saving on the extra 10 tickets will cancel out the $0.50 we’re losing on the tickets we do pay for, and so our expected value will become zero. We won’t expect to lose money, but we won’t expect to win money, either.