Using the presented Bayesian approach for each customer we obtain a distribution for she/he to belong to Low, Middle or High class.
Suppose that there are three different mails prepared for Low, Middle and High class correspondingly. And we have a client with probabilities $(0, 0.68, 0.32)$.
For this client a simple strategy will always send High class mail, and Bayesian strategy with probability $0.68$ will send Middle class mail and with probability $0.32$ you will send High class mail.
Suppose that you gain $1$ point if you send correct mail to client and $-1$ point if you send incorrect mail to client.
Then for the first strategy, the mean gain is $0.68 \cdot 1 + 0.32 \cdot (-1) = 0.36$. If you adopt Bayesian strategy you get $0.68 (0.68 - 0.32) + 0.32 (0.32 - 0.68) \approx 0.13$. So in this case it is better to send always High class mails.
While, suppose that your mail includes $5$ different goods, and customer is unsatisfied if her/his mail has no goods related to her/his class. Suppose that again gain is $1$ if customer is satisfied (at least one good item) and $-1$ otherwise. In this case for the first strategy the mean gain is again $0.36$. However, if you select each good from 5 at random according to the posterior probability than you average gain is $0.68 (1 (1 - 0.32^5) - 1 \cdot 0.32^5) + 0.32 (1 (1 - 0.68^5) - 1 \cdot 0.68^5) = 0.9$. As $0.9 > 0.36$ using Bayesian approach here we get better mean gain.
The presented example is not very good, as it oversimplifies things. To get optimal decision you need to use industry specific loss function. Also note that you get better estimate of probabilities if you have a customer with more than one purchase, and Bayesian approach allows you to take this thing in account.