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Deep Thoughts: What Would A Monty Hall Problem For Animal Lovers Look Like?

October 28th, 2013 · 4 Comments
Fun With Math · Game Theory

I first learned about the Monty Hall problem my sophomore year of college, and then, for some reason that I can’t even begin to remember, it was also discussed at the beginning of my graduate micro theory course. In other words, I’ve been annoyed by this problem for a good long while now.

The problem itself is pretty simple and based off of one of the games in the Let’s Make a Deal Game show, originally hosted by Monty Hall and now starring the guy who plays Barney’s cute gay brother on How I Met Your Mother. Anyway, the basic idea is that the contestant is presented with three doors, one of which has a car or something behind it and two of which have goats behind them. The contestant is told to pick one and then a door with a goat is opened, at which point the contestant can either stay with his original choice or switch to the other unopened door.

The math behind the problem isn’t what annoys me- it’s pretty simple to see that switching doors is a better strategy to get the car. Think about it- if you switch, the only way you lose if you had originally picked the right door, which would happen with probability 1/3. This leaves you with a 2/3 probability of getting the car. On the other hand, if you don’t switch, you only win the car if you originally picked the door with the car, which happens with probability 1/3. Instead, my beef with the problem is that it never considers the possibility that someone would actually prefer one of the goats:

Hear me out- preferring the goat isn’t nearly as absurd as it might initially seem. First of all, I currently own a perfectly usable car (except for that stupid check engine light) but have zero goats. Therefore, the principle of diminishing marginal utility suggests that, even though cars are generally more useful that goats on average, it’s entirely possible that a first goat might bring me more utility than a second car. I mean, where would I park a second car? (Please, asking where I would park a first goat is just silly.) Furthermore, I would have to pay taxes on much of the value of the “free” duplicate car, but I somehow doubt that the IRS would really try to argue that a goat would put me over the gift allowance. Lastly, I mean, COME ON…

To be fair, I don’t think that the game show uses pygmy goats, but I would argue that this would make the game more interesting, if for no other reason than it would add a layer of uncertainty regarding what the contestant was aiming to win. (Or, more realistically, maybe the contestants on the show don’t have the same utility function than I do, but at least the pygmy goats would make the show cuter.) Of course, we can’t ignore the obvious follow-up question: What strategy would maximize the goat-winning probability in the game?

To figure this out, we can use reasoning very similar to the logic employed above. If one doesn’t switch, then he will win whatever he initially chose, which will be a goat 2/3 of the time. If one does switch, the only way switching will result in a goat is if the person had originally chosen the car, which will happen 1/3 of the time. Therefore, staying with one’s original choice is the optimal goat-acquisition strategy. (It shouldn’t be surprising that the optimal goat strategy is the opposite of the optimal car strategy, but it’s nice to have the numbers to think about.)

It’s worth noting that the above analysis was conducted under the assumption that the contestant doesn’t care which goat he wins. Assuming that the contestant is shown the winnable items ahead of time, it seems reasonable that the contestant would have a preference for one goat over the other. Is it possible to incorporate this preference into the contestant’s strategy?

Unfortunately not- since a door with a goat is always opened following the contestant’s initial choice, the contestant knows that he is from that stage forward playing for either the car or other other goat, and he doesn’t have enough information to control which goat is initially exposed. That said, there is some opportunity for strategizing if the contestant preferred one of the goats to the car but preferred the car to the other goat.

In related news, I fail to understand why I have never successfully gotten through a game-show audition process. (Fun fact: A friend hooked me up with an invite to audition for Deal or No Deal a while back, and my “this is the only game show where being smart doesn’t help” schtick was working pretty well until they asked what I would do with the money. Pro tip: answers to this question that invoke the permanent income hypothesis tend not to be deemed television friendly.)

Tags: Fun With Math · Game Theory

4 responses so far ↓

  • 1 Michael // Oct 28, 2013 at 4:46 pm

    I’d argue that the car is still the optimal goat-receiving strategy as well. Pygmy goats typically run between $25 to $300. The resale value (net taxes) of any car put on a game show is greater than $300. Thus by playing to win the car, with probability 2/3, you win the car which can quickly be sold (think, Carmax) and turned into money. At that point, a pygmy goat can be purchased with the leftover money going to other utility-giving goods. With probability 1/3 you win the goat anyway. Thus for a sufficiently patient individual, playing for the car is weakly dominant.

  • 2 Dave Frye // Oct 29, 2013 at 7:55 am

    Regarding your probability math, once a door is opened you need to apply the conditional probabilities of the two remaining doors. Each has an equal chance (they were both 1/3 to start with) of housing the desired prize. With only two doors remaining, you have a 50% probability of obtaining the prize with a choice of either door.

  • 3 econgirl // Oct 29, 2013 at 6:09 pm

    @ Michael: Wait- are you trying to tell me that goods and services can be exchanged for money, which can then be exchanged for other goods and services? I will believe your logic if you get me a price quote on a pygmy goat. 🙂

    @ Dave: You should check out the link that I put on the Monty Hall problem at the beginning of the post, or check out Naked Statistics by Charles Wheelan, since he does a nice job of thoroughly explaining the math behind the problem.

  • 4 Dave Frye // Oct 29, 2013 at 7:33 pm

    I remain unconvinced but I will run the simulation to model it myself. The prior probabilities are as stated; however the a priori probabilities consider. That there are only two possibilities and we have no reason to believe that one is more likely than another.

    I will let you know when I have a chance to model this problem.

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