## 7 Discussion  
At this point, two related questions arise: Does the paradoxical situation under consideration arise: (1) when the chatbot is completely reliable, i.e., its overlook probability \(a_{i}=0\), and (2) when the chatbot is not completely reliable, that is, its overlook probability is strictly positive and can be arbitrarily small?  
Let us answer the first question first. A simple counterexample clearly shows that the answer here is: _there is no paradox_. For completeness, we provide proof below.  
Consider the following numerical example, the idea of which was suggested to us by [PERSON]. For an absolutely reliable chatbot 1, we set \(a_{i}^{1}=0\); for chatbot 2, we set \(a_{i}^{2}=0.1\); the known satisfaction levels are \(r_{i}=10\) for each chatbot and any location, and the prior probabilities that the target information is hidden at location \(i\) are \(p_{i}=0.5\), \(i=1\), \(2\); \(d=0.1\). Then, the absolutely reliable chatbot 1 will inevitably find the target information at the right location, either during the first search with probability \(\
u_{i}\) or during the second search (after an unsuccessful first search) with probability \(\
u_{i}\), and the search process ends. Assuming \(d=0.1\), the overall objective function \(F^{1}(\mathbf{\pi}^{\star})\) for the first chatbot is: \(p_{i}\cdot 10+p_{i}\cdot 10\cdot e^{-0.1}\), that is, about 9.6.  
As for the second, unreliable chatbot, since \(a_{i}^{2}>0\), the iterative search in these two locations can have an infinite number of rounds, and the expected total satisfaction on an infinite horizon will obey expression (1). Elementary arithmetic operations on Formula (1) using the data for chatbot 2 and the sum of an infinite geometrical progression show that the expected satisfaction \(F^{2}(\mathbf{\pi}^{\star})\) of the second chatbot does not exceed 4.9 points, and in fact, it will even be much less for a very small \(a_{i}^{2}>0\). We see that the efficiency of an absolutely reliable chatbot is greater, which means _there is no paradox_ in this case.  
Now let us answer the second question. At first glance, and here too, the answer should not be paradoxical, since the objective function defined in expression (1) is continuous in the variable \(a_{i}\), and, therefore, one would expect to obtain the same answer as in the case where the probability \(a_{i}=0\). However, this answer is incorrect--in fact, it is only true for the special case where the chatbot is absolutely reliable and retrieves the target information during the first round of the algorithm. In this case, the process is finite and terminates after one scan of each location. However, if the chatbot is unreliable and thus does not detect the target in the first round, then, starting from the second round of the multi-round iterative search, the objective function (1) is discontinuous in the discrete variable \(u_{i}\) (i.e., the number of unsuccessful scans of location \(i\)), and the proof based off the continuity of the objective function is no longer valid. The correct answer to the second question is that if the AI chatbot does not extract the target information in the first round of search, then during the iterative search procedure described in Section 5.2, the less reliable chatbot is more efficient--for any positive overlook probability \(a_{i}\), even as small as you wish. This claim is rigorously proved in Proposition 2 of this section.