The game theory of free drinks

A couple weeks ago, a New York bar had an interesting promotion:

From 11pm to midnight every 5th drink each bartender serves will be free. That means if the person before you orders four drinks, you win. Doesn’t matter if it’s top shelf: gratis. (You still have to tip though, you cheap bastards.)

All attempts to game the system are welcome. [via Chapman/Chapman]

How would you maximize your free drinks?

I have yet to work out an optimal strategy, but here are a few of my thoughts.

Case 1: you know the last person’s order

You could eavesdrop on orders to try to game the system. Let’s say you can hear the order of the person ahead of you. You know how many drinks ordered and whether any of them is free.

In this situation, there is a way to improve your odds. You wait until someone orders k < 5 drinks, none of which are free, and then make your order at least 5 – k drinks. If the person ahead of you paid full price for two drinks, for instance, then you can order three and guarantee yourself a free drink. This improves your discount from 20 percent (one out of five) to 33 percent (one out of three).

You could further exploit the position by choosing expensive, top-shelf liquors.

Case 2: position is unknown

In a busy bar, you may not be able to know your position. The bartender might fulfill a bunch of orders at the other side of the bar–some drinks being free–but you can’t tell because people have tabs and they are tipping anyway. Your drink might be the 5th (free), 4th, 3rd, 2nd, or 1st in the queue.

In this case, it’s advantageous to make large orders with friends. If you just order one of two drinks, you may not get a free drink at all. If you order 5, however, you can be sure you’ll get a free drink. If you’re lucky, you could even get two drinks free by ordering six (if your drinks are the sequence 5-6-7-8-9-10, then the first and last would be free).

Of course, everyone would be thinking along similar lines. I would think most people would order five drinks at a time, eliminating any positional advantage for the next order.

Notice this encourages more people to order drinks–seems like a good game for the bar!

What are your thoughts?

How can you maximize your free drinks in the two cases of knowing and not knowing your position?

Have you ever been in a bar with a free drink promotion like this?

What other free drinks schedules could generate revenues? Dan points out a random free drink schedule could generate more drink orders, but it might anger drunk people and not be worth it.



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  • Scott

    I think we’re missing some information here. If I place a single order of 5 drinks of differing costs, which of those is considered to be free? Given what information we have, and the fact that, regardless of simultaneous orders, the bartender must make them sequentially, it would be at the discretion of the bartender, who could make the least expensive drink the free one. This minimizes any potential gain.

    For any given order of size X, your number of free drinks equals INT((X+n)/5), where INT is the integer function (taking only the integer portion of a fraction or decimal) and ‘n’ is how many drinks ago the last free drink was (from 0-4).

    Assuming ‘n’ is unknown and has an equal chance of being any of its possible values (from 0-4) your odds of getting at least one free drink is X/5. Obviously ordering 0 drinks will net you no free drinks and ordering 5 or more will be guaranteed to get you at least 1 free drink.

    Based on this I’m not sure it’s advantageous to place larger than normal orders, at least not compared to the same number of drinks spaced over multiple orders.

    Consider 2 orders of 4 drinks as opposed to 1 order of 8:

    2 orders of 4: 4% chance of no free drink, 32% chance of 1 free drink, 64% chance of 2 free drinks.
    1 order of 8: 0% chance of no free drink, 30% chance of 1 free drink, 60% chance of 2 free drinks.

    In both cases, you will net an average of 1.6 free drinks.

  • http://wjspaniel.wordpress.com William Spaniel

    The problem is that ordering n < 5 leaves you open to being abused by someone who better understands the system.

    Assuming all anyone cares about is being a cheapskate, ordering five drinks at a time seems to be the only sensible equilibrium. The best alternative I can think of is attempting to "tip" (re: bribe) the bartender for intelligence. This will work, but it will be extremely costly. In fact, the bartender should be able to extract all of the rent in this situation, leaving you with an extremely costly free drink.

    And if you order five drinks at a time, I wouldn't worry about the bartender putting in the cheapest drink as the fifth. There's a huge agency problem here. As long as the bartender doesn't actually own the bar, he has all the incentive in the world to make the most expensive drink free. If the $4 drink is free, how much are you going to tip? I'd guess significantly less than if he put the $20 drink as free, especially since the interaction might repeat later in the night.

    All told, if everyone is playing optimally, the bartender comes out a big winner, the patrons come in a close second, and the owner loses big. But that is if everyone is playing optimally. Note that the promotion begins at 11 pm, by which point most of the patrons won't be thinking straight.

  • G. Pearson

    Am I missing something? If you order 5 drinks, there is zero chance of two free drinks. The example you gave, 5-6-7-8-9-10, is six drinks.

    Am I wrong? Did you mean ordering 6 drinks, but typed 5?

  • http://www.mindyourdecisions.com/blog/ Presh Talwalkar

    Thanks all for the comments! And you’re right G. Pearson about needing six drinks to get two free–thanks for the correction.

  • http://www.xploit.dk Jörn Martin Hajek

    “That means if the person before you orders four drinks, you win.”

    They obviously got that one wrong – the person before you will get the free drink four times out of five.

    As for a solution: If the bar is really so busy that no one except the barkeeper can keep track of it, this is equivalent to a 20% discount on all drinks, with some variance added.

    If everybody can keep track of everything, everybody will always try to get the fifth drink. If a drink is worth over 80% of the normal price, the first person will buy all drinks in the bar (possibly excluding the last modulo 5 drinks). If a drink is worth exactly 80% of the normal price, then people should be indifferent to not buying them, and buying them in batches of 5.

    Realistically, drinks have diminishing returns. So if everybody is well-informed and acts rationally (which may be a silly assumption, given the amount of drinks consumed) everybody will buy the drinks in batches of 5. Should a group need less than 5 drinks (the last drink may actually have a negative value when consumed), they will strike a deal with another group.

  • Shiny

    Is it illegal to try and strike a deal with the bartender? I can ask him to give a signal if there is a free drink up for grabs if I ordered 2 or 3 drinks. Then tip the bartender generously for it. May be you shouldn’t do this more than twice, to keep both the bartender and yourself safe.

  • Aaron Zielinski

    Presh,

    In the condition of “Case is Unknown,” doesn’t it simply become an expected value proposition?

    If you buy one, you have no idea whether your position is
    1 (pay)
    2 (pay)
    3 (pay)
    4 (pay)
    5 (free)
    Expect to pay: 4/5 = 0.8 each

    If you buy two, you have no idea whether your positions are
    1, 2 (pay 2)
    2, 3 (pay 2)
    3, 4 (pay 2)
    4, 5 (pay 1)
    5, 1 (pay 1)
    Expect to pay: 8/10 = 0.8 each

    If you buy three, you have no idea whether your positions are
    1, 2, 3 (pay 3)
    2, 3, 4 (pay 3)
    3, 4, 5 (pay 2)
    4, 5, 1 (pay 2)
    5, 1, 2 (pay 2)
    Expect to pay: 12/15 = 0.8 each

    And so on.

    You can see that regardless of how many you purchase, your expected value is always going to be 20% off of the full price. Thus, you should not change your consumption patterns at all and simply enjoy an expected 20% reduction in your night’s expenses.

  • http://www.mindyourdecisions.com/blog/ Presh Talwalkar

    Good point Aaron, but there is a difference in how you order. While the expected value is the same, the variance is decreased when you order more drinks since you’re guaranteed a free drink if you order 5 or more. (I wonder: is there a nice formula for variance of the savings?)

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