Review statement of Problem 1.

 

Solution: Assume that all the balls in the vessel are distinguished from one another by numbers in such a way that the white balls have numbers \(1, 2, 3, \dots, a\) and the black have numbers \(a+1, a+2, \dots, a+b\).

The numbers on the chosen balls must form some set of \(\alpha + \beta\) numbers from all \(a+b\) numbers \(1, 2, 3, \dots, a+b\).

The number of different sets of \(\alpha + \beta\) numbers that can be formed from \(a+b\) numbers is equal to \[\frac{(a+b)(a+b-1)(a+b-2)\cdots (a + b - \alpha - \beta +1)}{1\cdot 2 \cdot 3 \cdot\cdots \cdot (\alpha + \beta)}.\]

Corresponding to this, we can distinguish \[\frac{(a+b)(a+b-1)(a+b - 2) \cdots (a+b - \alpha - \beta +1)}{1\cdot 2 \cdot 3 \cdot\cdots \cdot (\alpha + \beta)}\] equally likely cases, each of which consists of the appearance of a specific \(\alpha + \beta\) numbers.

From all these exhaustive and disjoint cases, the appearance of \(\alpha\) white and \(\beta\) black balls is favored only for those in which any set of \(\alpha\) numbers from the group \(1, 2, 3,\dots, a\), together with any \(\beta\) numbers from the group \(a+1, a+2, \dots, a+b\), appear.

The number of distinct sets of \(\alpha\) numbers that can be formed from \(a\) numbers is equal to \(\frac{a(a-1) \cdots (a- \alpha + 1)}{1\cdot 2\cdot\cdots \cdot \alpha}\) and the number of distinct sets of \(\beta\) numbers that can be formed from \(b\) numbers is equal to \(\frac{b(b-1) \cdots (b- \beta + 1)}{1\cdot 2\cdot\cdots \cdot \beta}\).

Therefore, the number of distinct sets of \(\alpha + \beta\) numbers that can be formed from \(a+b\) numbers is expressed by the product \[ \frac{a(a-1) \cdots (a- \alpha + 1)}{1\cdot 2\cdot\cdots \cdot \alpha} \cdot \frac{b(b-1) \cdots (b- \beta + 1)}{1\cdot 2\cdot\cdots \cdot \beta}.\]

Then the number of cases considered that favor the appearance of \(\alpha\) white and \(\beta\) black balls is expressed by the indicated product.

And, consequently, the desired probability that among the chosen \(\alpha + \beta\) balls, there will be \(\alpha\) white and \(\beta\) black is expressed by the ratio \[\frac{\frac{a(a-1) \cdots (a- \alpha + 1)}{1\cdot 2\cdot\cdots \cdot \alpha} \cdot \frac{b(b-1) \cdots (b- \beta + 1)}{1\cdot 2\cdot\cdots \cdot \beta}} {\frac{(a+b)(a+b-1)(a+b-2)\cdots (a + b - \alpha - \beta +1)}{1\cdot 2 \cdot 3 \cdot\cdots \cdot (\alpha + \beta)}},\] which, after simple transformations, reduces to⁴ \[\frac{1\cdot 2 \cdot 3 \cdot\cdots \cdot (\alpha + \beta)}{1\cdot 2 \cdot\cdots \cdot \alpha \cdot 1\cdot 2 \cdot\cdots \cdot \beta} \cdot \frac{a(a-1) \cdots(a- \alpha + 1) \cdot b(b-1) \cdots (b- \beta + 1)}{(a+b)(a+b - 1)(a+b - 2) \cdots (a + b - \alpha - \beta +1)}.\]

 

Continue to Markov's presentation of a numerical example for Problem 1.

Skip to statement of Problem 2.

 

 

Review statement of Problem 1.

 

\(a=3,\  b=4,\  \alpha =2,\  \beta= 2 \).

Let us assume the white balls are set by the numbers 1, 2, 3 and the black by the numbers 4, 5, 6, 7.

The numbers on the chosen four balls can be represented by any of the following \(\frac{7 \cdot 6 \cdot 5 \cdot 4}{1\cdot 2 \cdot 3 \cdot4} = 35 \) sets:

1, 2, 3, 4	1, 2, 3, 5	1, 2, 3, 6	1, 2, 3, 7	1, 2, 4, 5	1, 2, 4, 6	1, 2, 4, 7
1, 2, 5, 6	1, 2, 5, 7	1, 2, 6, 7	1, 3, 4, 5	1, 3, 4, 6	1, 3, 4, 7	1, 3, 5, 6
1, 3, 5, 7	1, 3, 6, 7	1, 4, 5, 6	1, 4, 5, 7	1, 4, 6, 7	1, 5, 6, 7	2, 3, 4, 5
2, 3, 4, 6	2, 3, 4, 7	2, 3, 5, 6	2, 3, 5, 7	2, 3, 6, 7	2, 4, 5, 6	2, 4, 5, 7
2, 4, 6, 7	2, 5, 6, 7	3, 4, 5, 6	3, 4, 5, 7	3, 4, 6, 7	3, 5, 6, 7	4, 5, 6, 7

If two white and two black balls are chosen, then their numbers form one of the following \(\frac{3\cdot 2}{1\cdot2} \times \frac{4\cdot3}{1\cdot2} = 18\) sets:

1, 2, 4, 5	1, 2, 4, 6	1, 2, 4, 7	1, 2, 5, 6	1, 2, 5, 7	1, 2, 6, 7
1, 3, 4, 5	1, 3, 4, 6	1, 3, 4, 7	1, 3, 5, 6	1, 3, 5, 7	1, 3, 6, 7
2, 3, 4, 5	2, 3, 4, 6	2, 3, 4, 7	2, 3, 5, 6	2, 3, 5, 7	2, 3, 6, 7

In this way, we have 35 equally likely cases, 18 of which favor the considered event; consequently, the desired probability that there will be two of each color among the four chosen balls is \(\frac{18}{35}\).⁵

 

 

Review statement of Problem 2.

 

Solution: This problem can be considered as a special case of the previous, with \(a = \alpha\).

Namely, there are \(i\) tickets, whose numbers are chosen in advance, that are similar to the white balls, and the remaining tickets are similar to the black balls.

This similarity immediately reveals that the solution to the posed problem is obtained from the previous through substituting all numbers \(a,\, b,\, \alpha,\, \beta\) by the corresponding numbers \(i,\, n-i,\, i,\, m-i\).

Returning on that basis to the expression \[\frac{1\cdot 2 \cdot 3 \cdot\cdots \cdot (\alpha + \beta)}{1\cdot 2 \cdot\cdots \cdot \alpha \cdot 1\cdot 2 \cdot\cdots \cdot \beta} \cdot \frac{a(a-1) \cdots(a- \alpha + 1) \cdot b(b-1) \cdots (b- \beta + 1)}{(a+b)(a+b - 1) \cdots (a + b - \alpha - \beta +1)}\] found earlier, and making the indicated substitutions, we find the value of the desired probability in the form of the product \[\frac{1\cdot 2 \cdot 3 \cdot\cdots \cdot m}{1\cdot 2 \cdot\cdots \cdot i \cdot 1\cdot 2 \cdot\cdots \cdot (m-i)} \, \frac{i(i-1) \cdots 1 \cdot (n-i)(n-i-1) \cdots (n-m + 1)}{n(n - 1) \cdots (n-m +1)}\] which, after cancellation, reduces to⁶ \[ \frac{m (m-1) \cdots (m-i+1)}{n(n-1) \cdots (n-i+1)}.\]

Thus, the desired probability that among the chosen \(m\) numbers all of the \(i\) numbers indicated in advance appear is expressed by the fraction \[\frac{m (m-1) \cdots (m-i+1)}{n(n-1) \cdots (n-i+1)}.\]

 

Here, as with the first problem, Markov includes a second solution, which we omit.

 

Continue to an application of Problem 2.

Skip to statement of Problem 3.

 

Review statement of Problem 2.

 

For example, let us turn our attention to a lottery which, in the past, was played in France and many regions of Germany.⁷

Figure 5. Two billettes from French lotteries, one dated 27 April 1911 (upper image),
the other 12 October 1909 (lower image). Offered for auction at eBay, September–October 2023.

It consists of 90 numbers and, in each drawing, five numbers appeared. By the rules of the lottery, it was possible to place one or another amount on any numbers, or on any set of two, three, four or, finally, five numbers, which were called, respectively, l'extrait simple, l'ambe, le terne, le quaterne, and le quine.

If the five published numbers were found in the set on which the player placed an amount, then the administrator of the lottery gave an agreed-upon sum, found by a specific ratio to the amount of the bet, to that player.

These ratios are:

for l'extrait simple............................................................................................................ 15,

for l'ambe....................................................................................................................... 270,

for le terne.................................................................................................................... 5500,

for le quaterne........................................................................................................... 75 000,

for le quine............................................................................................................ 1 000 000.

In order to calculate the probabilities that l'extrait simple, l'ambe, le terne, le quaterne, and le quine appear, we should set \(n = 90\) and \(m = 5\) in the expression \(\frac{m (m-1) \cdots (m-i+1)}{n(n-1) \cdots (n-i+1)}\) we found, and give \(i\) the values \(1, 2, 3, 4, 5\), consecutively.

In this way, we find that the probability of the appearance of

l'extrait simple is equal to ................................................................................... \(\dfrac{5}{90} = \dfrac{1}{18}\),

l'ambe......................................................................................................... \(\dfrac{5 \cdot 4}{90 \cdot 89} = \dfrac{2}{801}\),

le terne........................................................................................... \(\dfrac{5 \cdot 4 \cdot 3}{90 \cdot 89 \cdot 88} = \dfrac{1}{11\,748}\),

le quaterne .......................................................................... \(\dfrac{5 \cdot 4 \cdot 3 \cdot 2}{90 \cdot 89 \cdot 88 \cdot 87} = \dfrac{1}{511\,038}\),

le quine................................................................................ \(\dfrac{1}{511\,038} \cdot\dfrac{1}{86} = \dfrac{1}{43\,949\,268}\).

Therefore, if the player's bet is \(M\), then the mathematical expectation⁸ of his profit from participating in the lottery

is expressed by:

in the case of l'extrait simple........................................................... \(\left(\frac{15}{18} - 1\right) M = -\frac{1}{6} M\),

in the case of l'ambe.................................................................... \(\left(\frac{540}{801} -1\right) M = -\frac{29}{89}M\),

in the case of le terne........................................................... \(\left(\frac{5500}{11\,748} - 1 \right)M = -\frac{1562}{2987}M\),

and so on.
 
In all cases, as we see, the mathematical expectation is a negative number; consequently, the lottery in question represented a game that was far from harmless.

This result corresponds to the fact that the lottery yielded a significant profit for its organizers.

 

Continue to statement of Problem 3.

 

 

Review statement of Problem 3.

 

Solution:   For brevity, let \[\left\lbrace \frac{p(p-1) \cdots (p-m+1)}{1\cdot2\cdot\cdots\cdot m}\right\rbrace^k = Z_p ,\] for any number \(p\).

On each draw, the number of selected tickets can represent any set of \(m\) numbers from all \(n\) numbers \(1, 2, \dots, n\).

Corresponding to this, on one draw we distinguish \(\frac{n(n-1)\cdots(n-m+1)}{1\cdot2\cdot\cdots\cdot m}\) equally likely cases, and on \(k\) draws, we distinguish  \(\left\lbrace \frac{n(n-1)\cdots(n-m+1)}{1\cdot2\cdot\cdots\cdot m}\right \rbrace^k = Z_n\) equally likely cases.⁹

Each of the preceding exhaustive and disjoint cases consists of the appearance of \(k\) specific sets of \(m\) numbers on the \(k\) draws considered.

Establishing in this way those cases that we will consider, and showing the total number of them, we proceed to determine the probabilities of the events referred to in the problem by calculating the number of cases favorable to them.

If \(i\) specific numbers \(\alpha_1, \alpha_2, \dots, \alpha_i\) do not appear, then for one draw, instead of  \(\frac{n(n-1)\cdots(n-m+1)}{1\cdot2\cdot\cdots\cdot m}\), there remain \(\frac{(n-i)(n-i-1)\cdots(n-i-m+1)}{1\cdot2\cdot\cdots\cdot m}\) cases, and for \(k\) draws, we have  \(\left\lbrace\frac{(n-i)(n-i-1)\cdots(n-i-m+1)}{1\cdot2\cdot\cdots\cdot m}\right\rbrace^k = Z_{n-i}\) cases, instead of \(\left\lbrace \frac{n(n-1)\cdots(n-m+1)}{1\cdot2\cdot\cdots\cdot m}\right \rbrace^k = Z_n\).

Thus, the probability that, for the \(k\) draws considered, \(i\) specific numbers do not appear is given by the fraction \[\frac{Z_{n-i}}{Z_n} = \left\lbrace\frac{(n-i)(n-i-1)\cdots(n-i-m+1)}{n(n-1)\cdots(n-m+1)}\right\rbrace^k.\]

Then the number of cases for which the \(i\) specific numbers \(\alpha_1, \alpha_2, \dots, \alpha_i\) do not appear and one specific number \(\beta_1\) does appear can be expressed by the difference \(\Delta Z_{n-i-1} = Z_{n-i} - Z_{n-i-1}\), where according to what we just said, \(Z_{n-i}\) represents the number of cases in which the numbers \(\alpha_1, \alpha_2, \dots, \alpha_i\) do not appear, and \(Z_{n-i-1}\) the number of those cases in which, besides the numbers \(\alpha_1, \alpha_2, \dots, \alpha_i\), the number \(\beta_1\) also does not appear.

In a similar manner, the number of cases in which the numbers \(\alpha_1, \alpha_2, \dots, \alpha_i\) do not appear and two specific numbers do appear can be expressed by the second difference \(\Delta^2 Z_{n-i-2} = \Delta Z_{n-i-1} - \Delta Z_{n-i-2}\), where \(\Delta Z_{n-i-1}\) represents the number of all cases in which the number \(\beta_1\) does appear and the numbers \(\alpha_1, \alpha_2, \dots, \alpha_i\) do not appear, and \(\Delta Z_{n-i-2}\) the number of those cases in which, besides \(\alpha_1, \alpha_2, \dots, \alpha_i\), the number \(\beta_2\) also does not appear.

In view of the possibility of continuing similar reasoning, it is not difficult to conclude that, in general, the number of cases in which \(i\) specific numbers do not appear and another \(l\) specific numbers do appear can be represented by the \(l\)^th order difference \(\Delta^l Z_{n-i-l}\), which is equal to: \[Z_{n-i} - \frac{l}{1}Z_{n-i-1} + \frac{l(l-1)}{1\cdot2} Z_{n-i-2} - \cdots \pm Z_{n-i-l}.\]

Then the probability that, in the \(k\) draws considered, \(i\) specific numbers do not appear and another \(l\) specific numbers do appear is equal to¹⁰ \[\frac{\Delta^l Z_{n-i-l}}{Z_n}.\]

Continue to Markov's solution of Problem 3, Parts 3, 4, and 5.

Skip to statement of Problem 4.

 

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[9] In other words, \(Z_p = \binom{p}{m}^k\) for any \(p\), so that \(Z_n = \binom{n}{m}^k\).

[10] For example, suppose \(n = 10, m = 3,\) and \(k = 5\). The total number of outcomes is \(Z_{10}=\binom{10}{3}^5=120^5\). For question (1), let \(i = 2\), so that we consider the number of selections without two numbers \(\alpha_1\) and \(\alpha_2\) in any of the 5 draws. There are \(Z_8=\binom{8}{3}^5=56^5\) such selections. Hence, the probability that \(\alpha_1\) and \(\alpha_2\) do not appear is \(\frac{Z_8}{Z_{10}}=\frac{56^5}{120^5} \approx 0.02213\). For question (2), the number of selections without \(\alpha_1\) and \(\alpha_2\) and containing another number \(\beta_1\) is: \[\Delta Z_7 = Z_8 - Z_7 = \binom{8}{3}^5-\binom{7}{3}^5 = 498~209~901.\] Of these, \[\Delta Z_6 = Z_7 - Z_6 = \binom{7}{3}^5 - \binom{6}{3}^5 = 49~321~875\] do not contain some other number \(\beta_2\). So, the number of selections without \(\alpha_1\) and \(\alpha_2\) and containing \(\beta_1\) and \(\beta_2\) is: \[\Delta^2 Z_6 = \Delta Z_7 - \Delta Z_6 = Z_8 - 2 Z_7 + Z_6 = 448~888~026,\] and the probability of this event is \(\frac{\Delta^2 Z_6}{Z_{10}} = \frac{448~888~026}{120^5} \approx 0.01804\). Similarly, \(\frac{\Delta^3 Z_5}{Z_{10}} = \frac{Z_8 - 3Z_7 + 3Z_6 - Z_5}{120^5} \approx 0.01618\).

 

 

Review statement of Problem 3.

 

The other probabilities referred to in the problem represent three particular cases of the probability just found and, therefore, can be gotten from the expression \(\frac{\Delta^l Z_{n-i-l}}{Z_n}\) for particular assumptions about \(i\) and \(l\): \[\mathrm{1)}\  i=0, \ \ \ \ \mathrm{2)}\  i=n-l,\ \ \ \ \mathrm{3)}\  i=0, l=n.\]

Letting \(i = 0\), we get the following expression for the probability of the appearance of \(l\) specific numbers: \[\frac{\Delta^l Z_{n-l}}{Z_n} = 1 - \frac{l}{1}\left(\frac{n-m}{n}\right)^k + \frac{l(l-1)}{1\cdot2} \left(\frac{n-m}{n}\right)^k \left(\frac{n-m-1}{n-1}\right)^k - \cdots \]

Letting \(i = n - l\), we find that the probability of the appearance of \(l\) specific numbers and the nonappearance of the remaining is equal to:

\[\begin{align} \frac{\Delta^l Z_0}{Z_n} &= \frac{Z_l}{Z_n} - \frac{l}{1}\frac{Z_{l-1}}{Z_n} + \frac{l(l-1)}{1 \cdot 2} \frac{Z_{l-2}}{Z_n} - \cdots \\ &= \left(\frac{n-m}{n}\right)^k \left(\frac{n-m-1}{n-1}\right)^k \cdots \left(\frac{l-m+1}{l+1}\right)^k \left\lbrace 1 - \frac{l}{1} \left(\frac{l-m}{l}\right)^k + \cdots \right\rbrace. \end{align}\]
 Finally, the probability of the appearance of all \(n\) numbers is equal to:
\[\frac{\Delta^n Z_0}{Z_n} = 1 - \frac{n}{1}\left(\frac{n-m}{n}\right)^k + \frac{n(n-1)}{1\cdot2} \left(\frac{n-m}{n}\right)^k \left(\frac{n-m-1}{n-1}\right)^k - \cdots.\]

Stopping at the last formula and noting that it requires tedious calculation for large values of \(n\), we derive two approximate formulas for it.

For getting the first approximate formula, assume that all numbers \(\left(\frac{n-m-1}{n-1}\right)^k\), \(\left(\frac{n-m-2}{n-2}\right)^k, \dots\), are equal to \(\left(\frac{n-m}{n}\right)^k\) which, for brevity, we denote by the letter \(t\).

Under this assumption, the indicated formula immediately gives us: \[\frac{\Delta^n Z_0}{Z_n} \approx (1-t)^n,\] where \(t = \left(\frac{n-m}{n}\right)^k\).

For the second approximation, we note that for small values of \(i\), the ratio \(\left(\frac{n-m-i}{n-i}\right)^k : \left(\frac{n-m}{n}\right)^k \), which is equal to \(\left\lbrace 1 - \frac{im}{(n-i)(n-m)}\right\rbrace^k\), differs little from \(1 - \frac{kim}{n(n-m)}\), and the product \[\left(1 - \frac{km}{n(n-m)} \right) \left(1 - \frac{2km}{n(n-m)} \right) \cdots \left(1 - \frac{ikm}{n(n-m)} \right) \] differs little from \[1-\frac{km(1+2+\cdots+i)}{n(n-m)} = 1 - \frac{kmi(i+1)}{2n(n-m)}.\]

On this basis, we accept \(t^{i+1}\left(1 - \frac{kmi(i+1)}{2n(n-m)}\right)\) as the approximate value of each product \[\left(\frac{n-m}{n}\right)^k \left(\frac{n-m-1}{n-1}\right)^k \cdots \left(\frac{n-m-i}{n-i}\right)^k.\]

Substituting this approximate expression in place of the exact one in the formula, we get \[\frac{\Delta^n Z_0}{Z_n} \approx (1-t)^n - \frac{kmt^2(n-1)}{2(n-m)}(1-t)^{n-2}\approx (1-t)^n\left\lbrace1-\frac{kmt^2}{2}\right\rbrace,\] since we assume the numbers \(\frac{n-1}{n-m}\) and \(\frac{1}{(1-t)^2}\) are close to 1.

We apply our approximate formula to finding the number of draws under the hypothesis that the probability of the appearance of all numbers was approximately equal to a given number \(\frac{1}{C}\).

The first approximation formula gives \((1-t)^n \approx \frac{1}{C}\) from which we deduce \(n \log(1-t) \approx -nt \approx -\log C\); but \(t = \left(\frac{n-m}{n}\right)^k\) and, thus, \(\log t = k \log\left(1- \frac{m}{n}\right) \approx - \frac{km}{n}\).¹¹

Comparing the approximate equations \(-nt \approx -\log C\) and \(\log t \approx -\frac{km}{n}\), we find \[k \approx \frac{n(\log n- \log \log C)}{m} .\]

In the subsequent calculation, let \(\frac{\log C}{n} = t_0\) and \(\frac{n(\log n- \log \log C)}{m} = k_0\), so that \(t_0\) and \(k_0\) are the approximate values of the numbers \(t\) and \(k\).

The second approximate expression of the probability gives \((1-t)^n \left(1-\frac{kmt^2}{2}\right) \approx \frac{1}{C}\), from which, by performing the approximate calculations, we deduce
\[\log C \approx nt + \frac{nt^2}{2} + \frac{kmt^2}{2} \approx nt  + \frac{(n+k_0 m) t_0^2}{2},\]
\[t \approx \frac{\log C}{n} \left[ 1-\frac{(n+k_0 m)t_0^2}{2 \log C}\right],\]
and then \[-\log t \approx \log n- \log \log C + \frac{(n + k_0 m) t_0^2}{2 \log C}.\]

On the other hand, we have \[ -\log t = -k \log\left(1-\frac{k}{m}\right) \approx \frac{km}{n} + \frac{k_0 m^2}{2n^2}.\]

Finally, equating one approximate expression for $\log t$ to the other, we arrive at the approximate equation \[ \frac{km}{n} + \frac{k_0 m^2}{2n^2} \approx \log n - \log \log C + \frac{(n + k_0 m) t_0^2}{2 \log C},\] from which we easily deduce
\[\begin{align} k &\approx \frac{n}{m} \left\lbrace \log n - \log \log C + \frac{k_0 m}{2 n^2}(\log C - m) + \frac{1}{2n} \log C\right\rbrace \\
    &\approx \frac{1}{m} \left\lbrace (\log n - \log \log C) \left( n + \frac{1}{2} \log C - \frac{m}{2} \right) + \frac{1}{2} \log C\right\rbrace.
\end{align}\]

For example, let us assume \(n= 90,\ m=5,\ C=2\).
 
Then

\(\log n = 4.4998\dots\),	\(\log C = 0.69314\dots\),
\(\log \log C = -0.3665\dots\),	\(n + \frac{1}{2} \log C - \frac{m}{2} = 87.84657\dots\).

and, performing a simple calculation by the previous approximate formula, we get \[k = \frac{4.8663 \times 87.8466 + 0.346}{5} \approx 85.5.\]

According to this result, it can be verified that the probability of the appearance of all 90 numbers in 85 draws is somewhat less than one-half, and for 86 draws is already bigger than one-half.

 

Continue to statement of Problem 4.

 

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[11] Here, Markov used “log” to denote natural (base \(e\)) logarithms. Later, he used the same notation for common (base 10) logarithms!

The Taylor series for \(\ln(1-t)\) is \(-t -\frac{t^2}{2} - \frac{t^3}{3} - \cdots \approx -t\), for small values of t. (The series converges for \(-1 \leq t < 1\).)  Later, he included the quadratic term for a better approximation.

 

 

Review statement of Problem 4.

 

Solution 1: First of all, we note that the game can be won by player \(L\) in various numbers of rounds not less than \(l\) and not more than \(l + m - 1\).

Therefore, by the Theorem of Addition of Probabilities,¹³ we can represent the desired probability \((L)\) in the form of a sum \((L)_l + (L)_{l+1} + \cdots + (L)_{l+i} + \cdots + (L)_{l+m-1}\), where \((L)_{l+i}\) denotes the total probability that the game is finished in \(l+i\) rounds won by player \(L\).

And in order for the game to be won by player \(L\) in \(l+i\) rounds, that player must win the \((l+i)\)^th round and must win exactly  \(l-1\) of the previous \(l+i-1\) rounds.

Hence, by the Theorem of Multiplication of Probabilities,¹⁴ the value of \((L)_{l+i}\) must be equal to the product of the probability that player \(L\) wins the \((l+i)\)^th round and the probability that player L wins exactly \(l-1\) out of \(l+i-1\) rounds.

The last probability, of course, coincides with the probability that in \(l+i-1\) independent experiments, an event whose probability for each experiment is \(p\), will appear exactly \(l-1\) times.

The probability that player \(L\) wins the \((l+i)\)^th round is equal to \(p\), as is the probability of winning any round.

Then¹⁵ \[(L)_{l+i} = p \frac{1\cdot 2\cdot \cdots\cdot (l+i-1)}{1\cdot 2\cdot \cdots\cdot i \cdot 1\cdot 2\cdot \cdots\cdot (l-1)} p^{l-1} q^i = \frac{l (l+1)\cdot \cdots\cdot (l+i-1)}{1\cdot 2\cdot \cdots\cdot i} p^l q^i,\] and finally, \[(L) = p^l\left\lbrace 1 + \frac{l}{1}q + \frac{l(l+1)}{1\cdot 2}q^2 + \cdots + \frac{l(l+1)\cdots(l+m-2)}{1\cdot 2\cdot \cdots\cdot(m-1)} q^{m-1}\right\rbrace.\]
In a similar way, we find \[(M) = q^m\left\lbrace 1 + \frac{m}{1}p + \frac{m(m+1)}{1\cdot 2}p^2 + \cdots + \frac{m(m+1)\cdots(m+l-2)}{1\cdot 2\cdot \cdots\cdot(l-1)} p^{l-1}\right\rbrace.\]

However, it is sufficient to calculate one of these quantities, since the sum \((L)+ (M)\) must reduce to 1.

 

Continue to Markov's second solution of Problem 4.

Skip to Markov's numerical example for Problem 4.

Skip to statement of Problem 8.

 

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[13] This “theorem,” presented in Chapter I, says (in modern notation): If \(A\) and \(B\) are disjoint, then \(P(A \cup B) = P(A) + P(B)\).  We would now consider this an axiom of probability theory. Since Markov considered only experiments with finite, equiprobable sample spaces, he could “prove” this by a simple counting argument.

[14] This “theorem,” also presented in Chapter I, says (in modern notation): \(P(A \cap B) = P(A |B) \cdot P(B)\). Nowadays, we would consider this as a definition of conditional probability, a term Markov never used. Again, he “proved” it using a simple counting argument. He then defined the concept of independent events.

[15] The conclusion \((L)_{l+i} = \binom{l+i-1}{i}p^l q^i\) is a variation of the negative binomial distribution; namely, if \(X\) represents the number of Bernoulli trials needed to attain \(r\) successes, then \(P(X=n) = \binom{n-1}{r-1}p^r q^{n-r},\ \ n\geq r\), where \(p\) is the probability of success on each trial and \(q=1-p\).

 

 

Review statement of Problem 4.

 

Solution 2: Noting that the game must end in no more than \(l+m-1\) rounds, we assume that a player does not stop immediately upon achieving one of the appropriate numbers of winning rounds but continues to play until exactly \(l+m-1\) rounds are played.

Under these assumptions, the probability that player \(L\) wins the game is equal to the probability of winning, by that same player \(L\), no fewer than \(l\) rounds out of all \(l+m-1\) rounds.

In this case, if the game is won by player \(L\), then the number of rounds won by him reaches the value \(l\) and the subsequent rounds can only increase this number, or leave it unchanged.

And conversely, if player \(L\) wins no fewer than \(l\) rounds out of \(l+m-1\) rounds, then the number of rounds won by player \(M\) will be less than \(m\); hence, in that case, it follows that player \(L\) wins \(l\) rounds before player \(M\) manages to win \(m\) rounds and, in this way, the game will be won by player \(L\).

On the other hand, the probability that player \(L\) wins no fewer than \(l\) rounds out of \(l+m-1\) rounds coincides with the probability that, in \(l+m-1\) independent experiments, an event whose probability for each experiment is \(p\) appears no fewer than \(l\) times.

The previous probability is expressed by the known sum of products \[\frac{1\cdot 2\cdot \cdots\cdot (l+m-1)}{1\cdot 2\cdot \cdots\cdot (l+i) \cdot 1\cdot 2\cdot \cdots\cdot (m-i+1)} p^{l+i} q^{m-i+1}\]
where \(i= 0,1,2,\dots,m-1\).

Then \[(L) = \frac{(l+m-1)\cdot\cdots\cdot m}{1\cdot2 \cdot\cdots\cdot l} p^l q^{m-1} \left\lbrace 1 + \frac{m-1}{l+1} \frac{p}{q} +  \frac{m-1}{l+1}\frac{m-2}{l+2}\frac{p^2}{q^2} + \cdots \right\rbrace;\] also, we find \[(M) = \frac{(l+m-1)\cdot\cdots\cdot l}{1\cdot2 \cdot\cdots\cdot m} p^{l-1} q^{m} \left\lbrace 1 + \frac{l-1}{m+1}\frac{q}{p} +  \frac{l-1}{m+1} \frac{l-2}{m+2} \frac{q^2}{p^2} + \cdots \right\rbrace.\]

It is not hard to see that these new expressions for \((L)\) and \((M)\) are equal to those found previously.¹⁶

 

Continue to Markov's numerical example for Problem 4.

Skip to statement of Problem 8.

 

 

 

Review statement of Problem 4.

 

Numerical examples:

\(\mathrm{1)}\ \ p = q = \dfrac{1}{2},\ l=1,\ m=2\).

\[ \begin{align} (L) &= p(1+q) = 2pq\left(1+ \dfrac{1}{2}\dfrac{p}{q}\right) = \dfrac{3}{4}, \\
(M) &= q^2 = \dfrac{1}{4}. \end{align} \]

\(\mathrm{2)}\ \ p=\dfrac{2}{5},\ q= \dfrac{3}{5},\ l=2,\ m=3\).

\[ \begin{align} (L) &= p^2\left\lbrace 1+ 2q + 3q^2\right\rbrace = 6p^2q^2\left\lbrace1 + \dfrac{2}{3}\dfrac{p}{q} + \dfrac{1}{6}\dfrac{p^2}{q^2}\right\rbrace = \dfrac{328}{625}. \\ (M) &= q^3 \left\lbrace 1+ 3p \right\rbrace = 4q^3  p\left\lbrace 1 + \dfrac{1}{4}\dfrac{q}{p}\right\rbrace  = \dfrac{297}{625}. \end{align} \]

 

Continue to Markov's statement of Problem 8.

 

 

Review statement of Problem 8.

 

Solution:  Letting \(n= 1,2,3,\dots\) consecutively, we arrive at the conclusion that, for any value of \(n\), the probability of the equation \(X_1 + X_2 + \cdots + X_n = \alpha\), where \(\alpha\) is a given number, can be determined as the coefficient of \(t^\alpha\) in the expansion of the expression \[\left\lbrace\frac{t + t^2 + \cdots + t^m}{m} \right\rbrace ^n\] in powers of the arbitrary number \(t\).¹⁷

On the other hand, we have \[\left\lbrace\frac{t + t^2 + \cdots + t^m}{m} \right\rbrace ^n = \frac{t^n}{m^n} \frac{(1-t^m)^n}{(1-t)^n} = \] \[ = \frac{t^n}{m^n} \left[ 1 - n t^m + \frac{n(n-1)}{1\cdot 2} t^{2m} - \cdots\right] \left[1 + nt + \frac{n(n+1)}{1\cdot 2} t^2 + \frac{n(n+1)(n+2)}{1\cdot 2\cdot 3} t^3 + \cdots \right]. \]

Therefore, denoting the probability of the equation \(X_1 + X_2 + \cdots + X_n = \alpha\) by the symbol \(P_\alpha\), we can establish the formula: \[m^n P_\alpha = \frac{n(n+1) \cdots (\alpha-1)}{1 \cdot 2 \cdot \cdots \cdot (\alpha-n)} - \frac{n}{1} \frac{n(n+1) \cdots (\alpha-m-1)}{1 \cdot 2 \cdot \cdots \cdot (\alpha-n- m)} + \frac{n(n-1)}{1\cdot 2} \frac{n(n+1) \cdots (\alpha-2m-1)}{1 \cdot 2 \cdot \cdots \cdot (\alpha-n-2m)} - \cdots,\] which represents an easy way of calculating \(P_\alpha\) for small values of \(\alpha\).  

It is also not hard to show the equation \(P_{\alpha} = P_{n(m+1) - \alpha}\), which allows us to substitute the number \(\alpha\) for the difference \(n(m+1) - \alpha\) and, in this way, gives the possibility of decreasing \(\alpha\), if \(\alpha > \frac{n(m+1)}{2}\).

 

Continue to Markov's numerical example for Problem 8.

Skip to the second part of Markov's solution of Problem 8.

Skip to the third part of Markov's solution of Problem 8.

Skip to Markov's analysis of repeated independent events.

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[17] Another instance of the use of probability generating functions. This conclusion is correct because he has assumed that, for each \(n\), \(P(X_n = i) = \frac{1}{m}\), for \(i = 1,2,\dots m\).

 

 

Review statement of Problem 8.

 

For example, for \(m = 6\) and \(n = 3\), we find:

\[216 P_{18} = 216 P_3 = 1, \ \ \ 216 P_{17} = 216 P_4 = 3, \]
\[216 P_{16}= 216 P_5 = \frac{3\cdot 4}{1 \cdot 2}  = 6, \ \ \ 216 P_{15}= 216 P_6 = \frac{3\cdot 4\cdot 5}{1 \cdot 2\cdot 3} = 10,\]
\[ 216 P_{14}= 216 P_7 = \frac{3\cdot 4\cdot 5\cdot 6}{1 \cdot 2\cdot 3 \cdot 4} = 15,\]
\[ 216 P_{13}= 216 P_8 = \frac{3\cdot 4\cdot 5\cdot 6 \cdot 7}{1 \cdot 2\cdot 3\cdot 4 \cdot 5} = 21,\]
\[ 216 P_{12}= 216 P_9 = \frac{3\cdot 4\cdot 5\cdot 6 \cdot 7 \cdot 8}{1 \cdot 2\cdot 3\cdot 4 \cdot 5 \cdot 6} - 3 = 25,\]
\[ 216 P_{11}= 216 P_{10} = \frac{3\cdot 4\cdot 5\cdot 6 \cdot 7 \cdot 8 \cdot 9}{1 \cdot 2\cdot 3\cdot 4 \cdot 5 \cdot 6 \cdot 7} - 3 \cdot 3  = 27,\]

Three ordinary six-sided dice, whose sides have numbers 1, 2, 3, 4, 5, 6 can serve to illustrate this example.

If these three dice are thown on a surface and if \(X_1, X_2, X_3\) denote the numbers on the upper sides, then \(\{1, 2, 3, 4, 5, 6\}\) will be the exhaustive and equally likely values of \(X_1\), as well as of \(X_2\) and \(X_3\).

Corresponding to these, the numbers \(P_3, P_4, P_5, \dots P_{18}\) represent the probabilities of different assumptions on the sum of the numbers,¹⁸ revealed on three ordinary throws of the dice.

And the equation \(P_3 + P_4 + P_5 + \cdots + P_{10} = P_{11} + P_{12} + P_{13} + \cdots + P_{18}\) shows the equal probability of the assumptions that this sum does not exceed 10 and, conversely, that it is bigger than 10.

Figure 7. Three dice made of bone or ivory, ca 9th–10th century CE, excavated from Nishapur, Iran.
Metropolitan Museum of Art, Rogers Fund, 1938, public domain.

 

Continue to the second part of Markov's solution of Problem 8.

Skip to the third part of Markov's solution of Problem 8.

Skip to Markov's analysis of repeated independent events.

 

 

 

Review statement of Problem 8.

 

For large values of \(n\), the exact calculation of \(P_\alpha\) requires tiring computations and can hardly represent great interest.

Then the question arises of finding an approximate expression for the probability with the possibility that it is simple and close to exact.

Assuming only \(n\) is large, but \(m\) is not, and considering not the probability of the actual value of the sum \(X_1 + X_2 + \cdots +X_n\) but the probability that this sum lies within given limits, we can return to the general approximate calculations in Chapter III.

To apply these, we should find the mathematical expectation of the first and second powers of the considered variables.

Since the mathematical expectation of any of these variables \(X_1, X_2, \dots X_n\) is equal to \(\frac{1+2 + \cdots + m}{m} = \frac{m+1}{2}\), and the mathematical expectation of their squares is equal to \(\frac{1^2 + 2^2 + \cdots + m^2}{m} = \frac{(m+1)(2m+1)}{6}\), then the difference between the mathematical expectations of their squares and the square of their mathematical expectations¹⁹ reduces to \[\frac{(m+1)(2m+1)}{6} - \left(\frac{m+1}{2}\right) ^2 = \frac{m^2 - 1}{12},\] and therefore, the results of the third chapter give, for the probability of the inequalities \[n \frac{m+1}{2} - \tau \sqrt{n \frac{m^2 - 1}{6}} < X_1 + X_2 + \cdots + X_n < n \frac{m+1}{2}  + \tau \sqrt{n \frac{m^2 - 1}{6}} ,\] an approximate expression in the form of the known integral²⁰ \[\frac{2}{\sqrt{\pi}} \int_0^\tau e^{-z^2}\, dz.\]

We will take advantage of the particular example to show other results of this approximate expression of probability, which can be applied in the general case.

And before anything, we note that, in the expansion of any entire function \(F(t)\) in powers of \(t\), the coefficient of \(t^\alpha\) can be represented in the form of the integral \[\frac{1}{2\pi} \int_{-\pi}^{\pi}  F(e^{\phi\sqrt{-1}} )e^{-\alpha\phi \sqrt{-1}}\,d\phi,\] because \(\int_{-\pi}^\pi d\phi = 2\pi\), and, for any integer \(k\) different from 0, we have \(\int_{-\pi}^\pi e^{k\phi \sqrt{-1}} \, d\phi = 0\).²¹

Therefore,²² \[\begin{align} P_\alpha&= \frac{1}{2\pi} \int_{-\pi}^\pi \frac{ e^{(n-\alpha) \phi \sqrt{-1}}\left(1-e^{m \phi \sqrt{-1}}\right)^n}{m^n\left(1-e^{\phi \sqrt{-1}}\right)^n} \, d\phi \\
    &=\frac{1}{2\pi} \int_{-\pi}^\pi \frac{ e^{ (n \frac{m+1}{2} -\alpha) \phi \sqrt{-1}} \left(e^{\frac{m}{2} \phi \sqrt{-1}} - e^{-\frac{m}{2} \phi \sqrt{-1}}\right)^n}{m^n \left(e^{\frac{1}{2} \phi \sqrt{-1}} - e^{-\frac{1}{2} \phi \sqrt{-1}}\right)^n}\,d\phi \\
    &=\frac{1}{2\pi} \int_{-\pi}^\pi e^{(n \frac{m+1}{2} - \alpha) \phi \sqrt{-1}} \left( \frac{\sin \frac{m}{2} \phi}{m \sin \frac{\phi}{2}}\right)^n\, d\phi \\
    &= \frac{1}{\pi} \int_0^\pi \cos \left(n \frac{m+1}{2} - \alpha\right) \phi\  \left( \frac{\sin \frac{m}{2} \phi}{m \sin \frac{\phi}{2}}\right)^n\, d \phi.
\end{align}\]

 

Continue to the third part of Markov's solution of Problem 8.

Skip to Markov's analysis of repeated independent events.

 

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[20] Markov never defines the normal distribution, as we know it today. When he talks about continuous random variables for the first time in Chapter V, his only examples are those which have a constant density function (i.e., what we would now call the “uniform” distribution) and one whose density is defined for all \(x\) and decreases as \(|x|\) increases. He then claims that the density is proportional to \(e^{-x^2}\), although there are many other such possibilities. This gives us a “normal distribution” with mean 0 and variance \(\frac12\).  So, his approximation can be written as \[P\left(\left| \frac{S_n - \mu}{\sqrt{2} \sigma} \right| < \tau \right) = \frac{2}{\sqrt{\pi}} \int_0^\tau e^{-z^2}\, dz.\]

[21] This is a Fourier transform. Markov chose not to use the symbol \(i\) to represent \(\sqrt{-1}\), although that notation had been around for at least a century.

[22] He is using the facts that \(e^\frac{ix}{2} - e^{- \frac{ix}{2}} = 2i \sin\left(\frac{x}{2}\right)\) and \(\cos(x) = \mathrm{Re}\,(e^{ix})\).

 

 

Review statement of Problem 8.

 

Returning to the approximate calculation and letting \(n \frac{m+1}{2} - \alpha = \beta = \gamma \sqrt{n \frac{m^2-1}{6}}\), we substitute²³ the exponential function \(e^{-n \frac{m^2-1}{24} \phi^2}\) for \(\left( \frac{\sin \frac{m}{2} \phi}{m \sin \frac{\phi}{2}}\right)^n\) based on the concept shown in Chapter III, and for the upper limit of the integral, we put \(\infty\) in place of \(\pi\).

In this way, we get the approximate formula \[P_{n\frac{m+1}{2} - \beta} \approx \frac{1}{\pi} \int_0^\infty \cos\beta\phi \cdot e^{-n\frac{m^2-1}{24} \phi^2}\, d\phi,\] whose right side is equal to \[\frac{1}{\sqrt{\pi}} \sqrt{\frac{6}{n(m^2-1)}} e^{-\frac{6 \beta^2}{n(m^2-1)}} = \frac{1}{\sqrt{\pi}} \sqrt{\frac{6}{n(m^2-1)}} e^{-\gamma^2}.\]

According to these, the probability of the inequalities \[n \frac{m+1}{2} - \tau \sqrt{n \frac{m^2 - 1}{6}} < X_1 + X_2 + \cdots + X_n < n \frac{m+1}{2}  + \tau \sqrt{n \frac{m^2 - 1}{6}}\] is represented approximately by the sum of all products \(\frac{1}{\sqrt{\pi}} \sqrt{\frac{6}{n(m^2-1)}} e^{-\gamma^2}\) for which \(\gamma\) satisfies \(-\tau < \gamma < \tau\) and turns the expression \(n \frac{m+1}{2} - \gamma \sqrt{n \frac{m^2 - 1}{6}}\) into an integer.

All terms in the sum shown contain a factor \(\sqrt{\frac{6}{n(m^2-1)}}\), which is equal to the difference between each two adjacent values of \(\gamma\) and will be arbitrarily small for sufficiently large \(n\).

Replacing on this basis the sum by the integral, we get for the probability of the inequalities \[n \frac{m+1}{2} - \tau \sqrt{n \frac{m^2 - 1}{6}} < X_1 + X_2 + \cdots + X_n < n \frac{m+1}{2}  + \tau \sqrt{n \frac{m^2 - 1}{6}}\] the previous approximate expression²⁴ \[\frac{2}{\sqrt{\pi}} \int_0^\tau e^{-\gamma^2}\, d\gamma.\]

 

This is the end of the solution to this problem.

 

Continue to Markov's analysis of the binomial distribution.

 

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[23] The Taylor series for \(e^{-n \frac{m^2-1}{24} \phi^2}\) and \(\left( \frac{\sin \frac{m}{2} \phi}{m \sin \frac{\phi}{2}}\right)^n\) both begin with \(1 + \frac{n}{24} (1-m^2) \phi^2\). The third terms of these series differ by \(\frac{n}{2880}(1-m^4)\phi^4\). Thus, for sufficiently small \(\phi\), the two functions are approximately equal.

[24] In essence, this example shows that the sum of independent, identically-distributed discrete random variables can be approximated by a normal distribution with appropriate mean and standard deviation. In the previous chapter, Markov showed that the distribution of the sample mean can also be approximated by an appropriate normal distribution; i.e., the Central Limit Theorem. He also relaxed the requirement that the variables have the same mean and variance.

 

 

Turning to the calculation of the sum \(S\), we now show that we can successfully take advantage of an expansion as a continued fraction²⁶ which follows as a particular case of Gauss’ Formula:

\[\frac{F(\alpha, \beta+1, \gamma+1, x)}{F(\alpha, \beta, \gamma, x)} = \cfrac{1}{1-\cfrac{ax}{1-\cfrac{bx}{1-\cfrac{cx}{1-\cfrac{dx}{1-\ddots}}}}}\] where \(F(\alpha, \beta, \gamma, x)\) and \(F(\alpha, \beta + 1, \gamma + 1, x)\) denote the “hypergeometric” series:²⁷

\[ 1 + \frac{\alpha\cdot \beta}{1\cdot \gamma} x + \frac{\alpha(\alpha+1)\beta(\beta+1)}{1\cdot 2 \gamma (\gamma + 1)} x^2 + \cdots \] and \[ 1 + \frac{\alpha(\beta+1)}{1(\gamma+1)} x + \frac{\alpha(\alpha+1)(\beta+1)(\beta+2)}{1\cdot 2 (\gamma + 1)(\gamma + 2)} x^2 + \cdots \] and the coefficients \(a, b, c,d, \dots\) are determined by the equations
\[\begin{align} a & = \frac{\alpha (\gamma-\beta)}{\gamma(\gamma + 1)},&\ \  b & = \frac{(\beta+1)(\gamma+ 1 - \alpha)}{(\gamma+1)(\gamma+2)}, \\
    c&= \frac{(\alpha+1)(\gamma+ 1 - \beta)}{(\gamma+2)(\gamma+3)}, &\ \  d &= \frac{(\beta+2)(\gamma+ 2 - \alpha)}{(\gamma+3)(\gamma+4)}, \dots\\
\end{align}\]

Referring to the result of Gauss’ Formula, we note that it follows from the following simple connections between different hypergeometric series:
\[F(\alpha, \beta+1, \gamma+1, x) - F(\alpha, \beta, \gamma, x) = ax F(\alpha+1, \beta+1, \gamma+2, x),\]
\[F(\alpha+1, \beta+1, \gamma+2, x) - F(\alpha, \beta+1, \gamma+1, x) = bx F(\alpha+1, \beta+2, \gamma+3, x),\]
\[F(\alpha+1, \beta+2, \gamma+3, x) - F(\alpha+1, \beta+1, \gamma+2, x) = cx F(\alpha+2, \beta+2, \gamma+4, x), \dots\]

To apply Gauss’ Formula to the expansion of \(S\) as a continued fraction, we should let \(\alpha = -n + l + 1\), \(\beta = 0\), \(\gamma = l+1\), \(x = -\frac{p}{q}\), which gives us the equation \[S = \cfrac{1}{1-\cfrac{c_1}{1+\cfrac{d_1}{1-\cfrac{c_2}{1+\cfrac{d_2}{1-\ddots}}}}},\] where, in general, \[c_k = \frac{(n-k-l)(l+k)}{(l+2k-1)(l+2k)}\frac{p}{q},\ \ \ \ d_k = \frac{k(n+k)}{(l+2k)(l+2k+1)}\frac{p}{q}.\]

We have here not an infinite, but a finite continued fraction, the last term of which will be \(\frac{d_{n-l-1}}{1}\) since \(c_{n-l} = 0\).

It is also not difficult to see that each of the numbers \(c_k\) is less than 1 only if \(\frac{n-l-1}{l+2}\frac{p}{q} <1\), as we will assume in the following argument.

Hence, denoting for brevity the continued fraction \[\cfrac{c_k}{1+\cfrac{d_k}{1-\cfrac{c_{k+1}}{1+\cfrac{d_{k+1}}{1-\ddots}}}}\] by the symbol \(\omega_k\), we have \(0<\omega_k< c_k\) and then we can establish a series of inequalities:
\[ S = \frac{1}{1-\omega_1} < \frac{1}{1-c_1}, \ \ S > \cfrac{1}{1-\cfrac{c_1}{1+\cfrac{d_1}{1-c_2}}},\]
\[S <  \cfrac{1}{1-\cfrac{c_1}{1+\cfrac{d_1}{1-\cfrac{c_2}{1+\cfrac{d_2}{1-c_3}}}}}, \cdots\]

 

Continue to the third part of Markov’s analysis of the binomial distribution.

Skip to Suggestions for the Classroom and Conclusion.

 

 

 

It remains to compare the previous inequalities with those which \(P\) satisfies and which we have shown above, and we will have the possibility of forming a series of approximate values of the probability that event \(E\) occurs more than \(l\) times in the \(n\) experiments considered, where for each of these approximate values, we will know whether it is greater than the probability or, conversely, less than it.²⁸

Based on these inequalities, interchanging \(p\) with \(q\) and substituting \(l\) by \(n-l'\), we find a series of approximate values of the probability that event \(E\) occurs less than \(l'\) times in the \(n\) considered experiments, where for each of the given approximate values of this new probability, we will also know whether it is larger than the probability or, conversely, smaller than it.

And with approximate values of the probability that event \(E\) occurs more than \(l\) times and the probability that event \(E\) occurs less than \(l'\) times, where \(l > l'\), it is not difficult to obtain an approximate value of the probability that event \(E\) occurs not more than \(l\) times and not less than \(l'\) times, since the sum of these three probabilities must be 1.

For example, let \(p=\frac{3}{5}\), \(q=\frac{2}{5}\), \(n=6520\) and we will seek the probability that the ratio of the number of occurrences of event \(E\) to the number of experiments will differ from \(\frac{3}{5}\) by less than \(\frac{1}{50}\).

In other words, we will seek the probability that event \(E\) occurs not more than 4042 times, and the complement to the event not more than 2738 times.²⁹

According only to the established remarks, calculation of the desired probability reduces to calculating the probability that event \(E\) occurs more than 4042 times and the probability that the complementary event occurs more than 2738 times.

Returning to the probability that event \(E\) occurs more than 4042 times, in the above formulas and inequalities, we must let \(p=\frac{3}{5}\), \(q= \frac{2}{5}\), \(n=6520\), \(l = 4042\).

Then: \[P_1 = \sqrt{ \frac{3260}{\pi\cdot 4043 \cdot 2477}} \left(\frac{3912}{4043}\right)^{4043} \left(\frac{2608}{2477}\right)^{2477},\] \[H =  e^{\frac{1}{12\cdot 6520} - \frac{1}{12\cdot 4043} - \frac{1}{12\cdot 2477}},\] and by means of logarithmic tables, we find...

 

The rest of this page in the original text is filled with highly-detailed arithmetic involving logarithms with 10-digit mantissas. We shall omit this and jump to Markov's conclusions.

 

\[\log H > -0.00003,\] \[ 0.00004094 < P < 0.00004095.\]

On the other hand, we have:
\[c_1 = \frac{2477}{4044} \cdot  \frac{3}{2} = \frac{7431}{8088},\ \ d_1 = \frac{6521}{4044\cdot 4045} \cdot \frac{3}{2} = \frac{19563}{32715960},\]
\[c_2 = \frac{2476 \cdot 4044}{4045 \cdot 4046}\cdot \frac{3}{2} = \frac{7509708}{8183035},\ \ d_2 = \frac{6522 \cdot 3}{4046 \cdot4047} = \frac{3261}{2729027},\]
\[c_3 = \frac{2475 \cdot 4045}{4047 \cdot 4048} \cdot \frac{3}{2} = \left(1 - \frac{2}{4047} \right) \frac{7425}{8096}.\]
and, by performing simple calculations we get, consecutively,
\[c_3< 0.9167,\ \ \frac{d_2}{1-\omega_3} < \frac{3261}{0.0833\times 2729027} < 0.01435,\]
\[0.918 > c_2 > \omega_2 >\frac{c_2}{1.01435}  > 0.9047,\]
\[0.0074 > \frac{d_1}{0.082} > \frac{d_1}{1-\omega_2} > \frac{d_1}{0.0953} > 0.00626,\]
\[0.912 < \frac{c_2}{1.0074} < \omega_1 < \frac{c_1}{1.00626} < 0.9131,\]
\[11.36 < \frac{1}{0.088} < S < \frac{1}{0.0869} < 11.51;\]
Consequently,
\[SP < \frac{0.4095}{869} < 0.0004713,\] but
\[SP > \frac{0.4094}{880} > 0.000465.\]

Proceeding to the probability that the event complementary to \(E\) occurs more than 2738 times, we must let \(p=\frac{3}{5}\), \(q= \frac{2}{5}\), \(n=6520\), \(l = 2738\), ...

 

This page is similar to the previous pages in featuring detailed arithmetic using logarithms.

 

\[\log H < -0.00003,\] \[0.00004280 < P < 0.00004281,\] \[\dots\] \[SP < 0.0005002,\] \[SP > 0.000491.\]

So, the probability that event \(E\) occurs more than 4042 times in the 6520 considered experiments is included between 0.0004713 and 0.000465, and the probability that in these experiments, event \(E\) occurs less than 3782 times is included between 0.0005002 and 0.000491.

And, therefore, the probability that event \(E\) occurs not less than 3782 times and not more than 4042 times lies between \(1 - 0.000972 = 0.999028\) and \(1 - 0.000956 = 0.999044\).

 

Continue to Suggestions for the Classroom and Conclusion.