p_{k,i} 
k = 1 
k = 2 
k = 3 
k = 4 
i=0 



(nt)(nt1)(nt2)
n(n1)(n2)




(nt)(nt1)(nt2)(nt3)
n(n1)(n2)(n3)



i=1 



3t(nt)(nt1)
n(n1)(n2)




4t(nt)(nt1)(nt2)
n(n1)(n2)(n3)



i=2 




6t(t1)(nt)(nt1)
n(n1)(n2)(n3)



i=3 




4t(t1)(t2)(nt)
n(n1)(n2)(n3)



i=4 




t(t1)(t2)(t3)
n(n1)(n2)(n3)



Table 2: Summary of Problems 14 
Euler's reasoning is entirely elementary, and each Problem builds on the results of the previous one. If k=1, the player is choosing just one number from among n. Since t of these will be winning numbers, p_{1,1} = t/n and p_{1,0}=1 p_{1,1}.
To calculate the probabilities when k=2, we can think of the player as choosing the two numbers one at a time. Therefore, there are n1 ways to choose the second number, having made the first choice from among n. This accounts for the denominator n(n1) in all three of the probabilities in the second column of Table 2. The case k=2, i=2 arises from choosing one of the t winning numbers first, and one of the remaining t1 winning numbers next. Similary, in the calculation of p_{2,0}, the player first chooses one of the nt losing numbers, and then chooses one of the remaining nt1 losing numbers. The interesting case is i=1. There are two ways this can happen, either by choosing a winning number followed by a losing number, or by choosing a losing number followed by a winning number. There are t good choices and nt bad choices, so the numerator of p_{2,1} has to be 2t(nt).
Elements of the pattern in Table 2 gradually come into focus quickly. Particularly noteworthy is the presence of binomial coefficients. The details of Euler's solutions of Problems 14 make this identification certain, as the coefficient of every entry in each column, except the first and last, arises a sum of two coefficients in the previous column, thereby obeying precisely the same recursive formula as the entries in Pascal's triangle. For example, one can bet on k=4 numbers, and get i=2 of them correct in two different ways: either by matching 2 of the first 3 numbers bet upon, and then failing to match on the fourth one, or by matching only 1 of the first 3, and then succesfully matching the fourth one. Therefore, the binomial coefficient
will always be a factor of
p_{k,i}. Let's introduce one more piece of modern notation, by letting
s_{k,i} = 
p_{k,i}

, so that p_{k,i} = 
æ
ç
ç
è 

ö
÷
÷
ø 
s_{k,i}. 

In Problem 5, Euler asks for the general form of p_{k,i}. He introduces r=nt and writes out the cases k=1, 2, ¼, 6 in this new notation. Observing the pattern, s_{k,i} is seen to be:

t(t1)¼(ti+1)r(r1)¼(r(ki)+1)
n(n1)(n2)¼(nk+1)

, 

where 0
£ i £ k, 1
£ k £ t £ n, and
k £ r=
nt.
Euler does not use this notation, nor does he explicitly give a formula for the general value of s_{k,i}. His notation is as follows:
In Corollary 1 to Problem 5, Euler observes that the s_{k,i} can be given recursively. We may give an explicit formulation as follows. Let
and
from Problem 1; then for
k > 1, and 0
£ i <
k:
s_{k,i} = 
r(ki)+1
nk+1

s_{k1,i}. 

Also,
s_{k,k} = 
tk+1
nk+1

s_{k1,k1}. 

In Euler's notation, these formulas are given as follows:
A^{1}= 
t
n

, A^{0}= 
r
n

; 

B^{2}= 
t1
n1

A^{1}, B^{1}= 
r
n1

A^{1}, B^{0}= 
r1
n1

A^{0}; 

C^{3}= 
t2
n2

B^{2}, C^{2}= 
r
n2

B^{2}, C^{1}= 
r1
n2

B^{1}, C^{0}= 
r2
n2

B^{0}; 

D^{4}= 
t3
n3

C^{3}, D^{3}= 
r
n3

C^{3}, D^{2}= 
r1
n3

C^{2}, D^{1}= 
r2
n3

C^{1}, D^{0}= 
r3
n3

C^{0}; 
