Say that you have a travel agency. Your job is to book rooms at hotels. Some hotels fill up more quickly than others, and you want to figure out which hotels to book at first so that your net booking rate is as high as it could be the staff you have.
The logic of prioritization is simple: prioritize those hotels where the expected loss if you don’t book now is the largest. The only thing we need to do is find a way to formalize the losses. Going straight to formalization is daunting. A toy example helps.
Imagine that there are two hotels Hotel A and Hotel B where if you call 2-days and 1-day in advance, the chances of successfully booking a room are .8 and .8, and .8 and .5 respectively. You can only make one call a day. So it is Hotel A or Hotel B. Also, assume that failing to book a room at Hotel A and Hotel B costs the same.
If you were making a decision 1-day out on which hotel to call to book, the smart thing would be to choose Hotel A. The probability of making a booking is larger. But ‘larger’ can be formalized in terms of losses. Day 0, the probability goes to 0. So you make .8 units of loss with Hotel A and .5 with Hotel B. So the potential loss from waiting is larger for Hotel A than Hotel B.
If you were asked to choose 2-days out, which one should you choose? In Hotel A, if you forgo 2-days out, your chances of successfully booking a room next day are .8. At Hotel B, the chances are .5. Let’s play out the two scenarios. If we choose to book at Hotel A 2-days out and Hotel B 1-day out, our expected batting average is (.8 + .5)/2. If we choose the opposite, our batting average is (.8 + .8)/2. It makes sense to choose the latter. Framed as expected losses, we go from .8 to .8 or 0 expected loss for Hotel A and .3 expected loss for Hotel B. So we should book Hotel B 2-days out.
Now that we have the intuition, let’s move to 3-days, 2-days, and 1-day out as that generalizes to k-days out nicely. To understand the logic, let’s first work out a 101 probability question. Say that you have two fair coins that you toss independently. What is the chance of getting at least one head? The potential options are HH, HT, TH, and TT. The chance is 3/4. Or 1 minus the chance of getting a TT (or two failures) or 1- .5*.5.
The 3-days out example is next. See below for the table. If you miss the chance of calling Hotel A 3-days out, the expected loss is the decline in success in booking 2-days or 1-day out. Assume that the probabilities 2-days out and 1-day our are independent and it becomes something similar to the example about coins. The probability of successfully booking 2-days and 1-days out is thus 1 – the probability of failure. Calculate expected losses for each and now you have a way to which Hotel to call on Day 3.
| | 3-day | 2-day | 1-day |
| Hotel A | .9 | .9 | .4 |
| Hotel B | .9 | .9 | .9 |
In our example, the number for Hotel A and Hotel B come to 1 – (1/10)*(6/10) and 1 – (1/10)*(1/10) respectively. Based on that, we should call Hotel A 3-days out before we call Hotel B.