## 3 Method
### Method 2: Multiplication Rule of Probability and Chebyshev's Inequality  
Let us assume that we have a random value distributed at the segment \([a,\,b]\) and also a random sample of this distribution. Let us call the significant extrema those that correspond to the leftmost (rightmost) bin in the histogram, and the length of each bin is \(\%\)\(\%\)\(0.27\%\) of the length of the segment \([a,\,b]\). This arrangement arises from the following considerations. For the normal distribution, the part of the distribution outside the \(3\sigma\) interval should contain approximately \(0.27\%\) of the whole distribution. If we take the uniform distribution at segment \([a,\,b]\), these last bins are expected to receive the same \(0.27\%\) of the overall samples. Later, to map the 3D objects, we will only consider the extrema from these last bins. Let us explore another way to show that the number of significant extrema of summation of two spherical functions will not increase compared to the number of significant extrema of each function. This consideration is based on the probability multiplication theorem and does not even require a Gaussian assumption. Since the same number of nodes (namely, 10,242) is used on each of \(n\) spheres, we divide all intervals, \(\left[f_{\min}^{(i)}f_{\max}^{(i)}\right]\) and \(i=1,\ldots,n\), by the same constant number \(N_{\text{bin}}\) of bins. Let \(P(X_{i})\) be the probability of the random variable \(X_{i}\) to be a significant maximum, i.e., to get into the rightmost bin \(\left[f_{\max}^{(i)}-\left(f_{\max}^{(i)}-f_{\min}^{(i)}\right)/N_{\mathrm{bin}},f_{\max}^{(i)}\right]\) or to the leftmost one. Let us reiterate that the length of both bins is \(0.27\%\) of the length of segment \(\left[f_{\min}^{(i)},f_{\max}^{(i)}\right]\). According to the multiplication rule of probability, we have \(P(X_{1}X_{2})=P(X_{1})P_{X_{1}}(X_{2})\), where \(P(X_{1}X_{2})\) is the probability of occurrence of both events, i.e., the probability that the sum \((f_{1}+f_{2})\) is the significant maximum, and it includes the significant maxima for \(f_{1}\) and \(f_{2}\). \(P_{X_{1}}(X_{2})\) is the conditional probability of \(X_{1}\) occurring given that \(X_{2}\) occurs, i.e., the probability of a point on the sphere to be a significant maximum of \(f_{1}\), when \(f_{2}\) is a also significant maximum. In the case of independent random variables \(X_{1}\) and \(X_{2}\), the conditional probability is \(P_{X_{1}}(X_{2})=P(X_{2})\). As we see, in any case, it turns out to be \(P(X_{1}X_{2})\leq P(X_{1})\) since \(P_{X_{1}}(X_{2})\leq 1\). On the other hand, \(P(X_{1}X_{2})=P(X_{2})P_{X_{2}}(X_{1})\) and \(P(X_{1}X_{2})\leq P(X_{2})\), thus  
\[P(X_{1}X_{2})\leq\min\{P(X_{1}),P(X_{2})\}.\]  
By induction, a similar conclusion can be drawn for \(n\) random variables as follows:  
\[P(X_{1}X_{2}\cdots X_{n})\leq\min\{P(X_{1}),\ldots,P(X_{n})\}. \tag{12}\]  
Due to the multiplication rule of probabilities  
\[P(X_{1}X_{2}\cdots X_{n})=P(X_{1})P_{X_{1}}(X_{2})P_{X_{1}X_{2}}(X_{3})\cdots P _{X_{1}\cdots X_{n-1}}(X_{n}).\]  
Formula (12) only shows a non-increase of the number of significant extrema in the sum and does not provide a quantitative assessment. For significant minima, the same considerations are valid. Chebyshev's inequality allows us to quantify the probabilities \(P(X_{1})\), \(\ldots\), \(P(X_{n})\) as being significant extrema in (12). **Chebyshev's Inequality. **_If a random variable \(X\) has expectation \(E[X]\) and variance \(D[X]\), then the inequality holds for any \(\varepsilon>0\)_  
\[P(|X-E[X]|>\varepsilon)\leq\frac{D[X]}{\varepsilon^{2}}.\]  
If we consider the sum \(S_{n}\) from Method 1 as a random variable \(X\) (without any assumptions about the dependence or independence of its terms), then, we estimate the probability that the deviation of this quantity \(S_{n}\) from its expectation \(E[S_{n}]\) will be more than three standard deviations in absolute value.