A Mathematician’s First Camping Trip

My son has reached the age where being a scout means being outdoorsy. Last weekend, he and I went on our first camping excursion with his den.

Scouting urges to BE PREPARED, and I didn’t want to be the one family that was not. But there was a lot of equipment I didn’t have (most notably a tent, but also, as I learned when I arrived and saw the scout-mates’ set-ups, grills, axes and sledgehammers, sleeping mats, specialized coolers, etc.). I surely didn’t want to drop a lot of money buying gear until I was reasonably sure we’d get a good amount of use of it.

Thus I arrived at the classical dilemma: should I rent, or should I buy?

Buying the camping gear could easily cost upward of \$500. Renting was much more reasonable, only about \$50. If I knew in advance that we would go camping more than 10 times, buying would be the more economical option. On the other hand, if I would only camp fewer than 10 times, then my best move would be to rent each time. Without an oracle to consult, though, I have to risk wasting some money. But can I at least minimize the risk?

What I need is a strategy – some method of making my decision. The only real choice I have is when to buy (after 6 trips? after 1 trip? after 0 trips? never?), so let’s suppose I will buy after $t$ trips. The amount of money I’ll spend in the process is $50t+500$ (the first term accounts for the rental fees, while the second is the cost of purchase).

Let’s compare this amount of money to what I would have spent had I been clairvoyant. We’ll call the number of camping trips I take in my life $n$.

  • If I $n\geq 10$, then I was going to spend at least \$500 either way, my optimal expense is just the \$500 cost of buying. The ratio of actual to optimal spending is $\frac{50t+500}{500}$, which simplifies algebraically to $\frac{t+10}{10}$.
  • If $n\leq 10$, then renting every time is the more affordable option, so $\$50t$ is my optimal expenditure. The ratio in this case of actual to optimal is $\frac{50t+500}{50t}$, which simplifies as $\frac{t+10}{t}$.

Here’s the tricky part: I don’t know the value of $n$. For example, if I buy after 5 trips, then:

  • I may go camping 20 times, in which case my ratio of actual-to-optimal expense is $\dfrac{750}{500}=1.5$ (which isn’t that bad), or
  • I may never go camping again, spending $\frac{750}{250}=3$ times as much as I needed to (which is kind of bad).

Since I don’t know what the future holds, I have to protect myself against the worst possible case. And no matter what, one of the ratio fractions is at least $2$:

  • If $t\geq 10$, then $\frac{t+10}{10}\geq \frac{10+10}{10}=2$.
  • If $t\leq 10$, then $10\geq t$, so $\frac{t+10}{t}\geq \frac{t+t}{t}=2$.

So, my strategy always risks my spending at least $2$ times as much money as I needed to. That’s hardly reassuring, but I suppose it’s not avoidable.

While I’d rather not have to pay double, this analysis did help me make my decision for this camping trip: renting was better. In fact, I’ll continue to rent for the first 10 trips and only buy thereafter; doing so makes my spending ratio only 2 (and no higher). In general, I would want to rent until those fees equaled the cost of buying, and then buy after that.

After our trip, my son asked if we could get a personalized “I $\heartsuit$ CAMPING” license plate. So, it looks like I may end up buying a tent after all.