Everyone loves a comp. They’re not always easy to come by, with one prominent exception: free drinks for play at the bar. You know how it works—load a machine with $10 and play max coins and you drink free. Some places impose a few more restrictions (or require a $20 buy-in), but for the most part that’s the way it works. What’s so good about it? Let’s break down the numbers.
To make things easy, I’ll assume that you play video poker at an average skill level and that a cocktail goes for $4.50. So the choice is: A) pay $4.50 for the cocktail, or B) buy a $10 roll of quarters and play video poker. The result is obvious if you pay. How about if you play?
The cost for playing depends on the pay schedule on the game at hand. Understanding pay schedules is a discussion of its own, so let’s simplify again and assume you’re playing a machine with a 3 percent casino edge, which is a fairly common scenario. While the concept is a bit slippery for some to grasp, a 3 percent edge means that the machine is mathematically entitled to 3 percent of the money you play through it—no more and no less.
That means that for each max-coin bet you make on a 25-cent machine ($1.25), the house edge earns 3.75 cents. Round up to 4 cents for convenience and start counting ’em up. Play a hand, tally minus-4 cents. Play two, minus-8 cents. And so on. Of course, you’re winning and losing in units of $1.25 as you play, but this is the mathematical way to accurately calculate this proposition.
In reality, all you’re expected to do for your drink is play the roll through one time. That’s eight plays, which is an accrued expected loss of 40 cents. Would you rather pay 40 cents or $4.50? The economics are clear.
You say it’s cheesy to buy a roll and play it just once? Fine. Play it through twice. Hell, play it through 10 times. You still come out ahead by 50 cents. Understand that you’ll lose your $10 in many sessions. But there’ll also be sessions when you cash a profit. Add them all together and you win.