A Selection of Problems from A.A. Markov’s Calculus of Probabilities: Problem 2 – A Lottery Game
This problem is similar to the first, but Markov included an application to a lottery game dating back to 17th-century France. Games of chance were the primary motivation for some of the early writings on probability by Fermat, Pascal, Huygens, and others. Lotteries of this type are found today in many states in the US. It would be interesting to compare the probabilities and payoffs in current games to the ones mentioned here.
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Задача 2ая. Из сосуда, содержащего \(n\) билетов с нумерами \(1, 2, 3, \dots, n\) и никаких других, вынимают одновременно или последовательно m билетов, при чем, в сдучае последовательного вынимания, ни один из вынутых билетов не возвращают обратно в сосуд и новых туда также не покладывают. Требуется определить вероятность, что между нумерами вынутых билетов появятся \(i\) нумеров, указанных заранее, напр. \(1, 2, 3, \dots, i\). |
2nd Problem. From a vessel containing \(n\) tickets with numbers \(1, 2, 3, \dots, n\), and no others, we select, simultaneously or consecutively, \(m\) tickets, so that in the case of consecutive selection, none of the selected tickets is returned to the vessel and a new one is removed.
It is required to find the probability that, among the numbers on the chosen tickets, \(i\) numbers, indicated in advance, appear—for example, \(1, 2, 3, \dots, i\).
Continue to Markov's solution of Problem 2.
Skip to statement of Problem 3.
A Selection of Problems from A.A. Markov’s Calculus of Probabilities: Problem 2 – Solution
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.
[6] In modern notation, this becomes \(\frac{\binom{a}{\alpha} \binom{b}{\beta}}{\binom{a+b}{\alpha+\beta}} = \frac{\binom{n-i}{m-i}}{\binom{n}{m}}\), which is equivalent to the stated answer.
A Selection of Problems from A.A. Markov’s Calculus of Probabilities: Problem 2 – An Application to a Contemporary Lottery
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.
[7] One of the earliest such lotteries was the Génoise Lottery in France in the early 17th century. It was used there and in other countries to replenish government treasuries. See [Maistrov 1974], and the references contained therein, for more information.
[8] The Russian term “математическое ожидание” translates as “mathematical expectation.” It is more common to say “expected value” in modern terminology.