The Economics of 50/50 Draws

"Boy," you might be saying, "this blog has sure posted a lot this month!" You're right! In fact, March so far has had at least twice as many page views as any other month in the existence of this blog. Figuring that this is about as successful as this is ever going to get, I figured I'll just keep posting while the going is hot.

Recently I attended a fair number of arena curling games. People who attend arena curling games often enjoy things like expensive beer, ridiculously addictive popcorn, curling (sometimes), and the 50/50 draw.

A fun way to think about casino games and lotteries (if you're me) is based on their expected return (ok, it's actually not fun at all).

Take Roulette, for instance. If you pick a solid colour in Roulette you have an 18/38 chance of winning, where winning would double your money. Doubling your money isn't quite good enough to break even, though, because 18/38 is slightly less than half. As a result, for every dollar you spend on Roulette, you'd expect to lose 5.26 cents. This is the house edge for Roulette, and what ensures that the casino always comes out on top.

Other common casino games have house edges like 1.41% (pass line in craps), 1.06% (banker bet in Baccarat), and 0.43% (perfect play in Blackjack without counting cards). Slot machines will often run house edges ranging from 7%-15%.

Lotteries are a little bit different. Lotto 6/49, for instance, runs a house edge of about 30%, Keno gambling is approximately 30%-40% depending on the rules, and I suspect that sports betting using SportSelect can get as high as 30%-50%.

Now that we have a reference point, we can compare 50/50 lotteries to these other games. In a basic 50/50 game, everyone could buy $5 tickets, and one person would win 50% of all the money paid into the lottery. Buying one ticket would give you a 1/n chance of winning, and you would win an amount of money equivalent to n/2 time the price per ticket, at a cost of entry of that same price per ticket. As a result, your expected loss per ticket purchased is 50%. This is way worse than pretty much anything else.

Things get a little bit more complicated though. Many arenas nowadays offer the option of paying $5 for 1 ticket, $10 for 3 tickets, or $20 for 10 tickets. Apart from the obvious differences in price per ticket (making $20 for 10 seem a substantially better deal already), how do these different options translate into house edges?

Fortunately I made another set obscure and hard-to-read ternary plots for ya! Check it out:

Due to me not thinking before colouring, the axes are a bit backwards. Along the bottom is the fraction of sales for the $5/1 combo, along the right hand side is the fraction of sales for the $10/3 combo, and along the left hand side is the fraction of sales for the $20/10 combo. If you've never read one of these, check out this cool website on how to do so.

In general, the house edge depends on what everyone else buys, which makes sense. It's interesting just how large this effect is, though. Buying a single ticket for $5 ranges from a house edge of 50% (if everyone else does too) to 80% (if everyone else buys the best combo).

Buying three tickets for $10 ranges from a house edge of 25%-70%, and buying ten tickets for $20 ranges from a house edge of 50% to a player edge of 25%. The player edge, of course, only occurs if you are one of a very tiny number of people buying the $20/10 combo.

Though I can't find any sources for the actual distribution at sporting events, if we had an even split in sales then we'd be looking at a house edge ranging from 40%-75%, depending on which package is purchased. This is by far the worst set of house odds out of any of the games previously mentioned.

50/50 draws are maybe justified in the sense that the money primarily goes to charities, but as an investment (or even just a source of gambling for fun) they're really probably one of the worst things you can do.

See ya!