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Euler's Analysis of the Genoese Lottery - Fair Prize Values

Author(s): 
Robert E. Bradley

In Problem 6, Euler turns his attention to finding the fair prize for a wager one écu. In this discussion, it is clear that Euler is considering a more elaborate lottery scheme than Roccolini's where, for example, a gambler playing terno must match all three of his k=3 selections. In the "Reflections" paper, Euler will award a prize Fk,i if i of the players' k numbers match the t numbers drawn, for any 0 < i £ k.

To do this, Euler simply chooses k positive numbers satisfying ak,1 + ak,2 + ¼+ ak,k = 1, and award prizes

Fk,i = ak,i

pk,i
.
Then the expected payoff for a ticket costing one écu is
k
å
i=1 
pk,iFk,i = k
å
i=1 
ak,i = 1.

Absent a notation uniform in k, the discussion of this simple point is surprisingly tedious, and is handled one case at a time, ending at k=5. "It's not likely that we'd need to consider more than 5 numbers," Euler says, "as the prizes would be too exorbitant". Of course, it's precisely these `exorbitant' prizes that make so many contemporary lotteries so very irresistible.

The ak,is are not uniquely determined, unless k=1. So for k > 1, Euler discusses three possible weighting schemes:

1. uniform weights

ak,i= 1

k
2. binomial weights

ak,i =
æ
ç
ç
è
k
i
ö
÷
÷
ø

2k-1
3. modified binomial weights

ak,i =
(k-i+1) æ
ç
ç
è
k
i
ö
÷
÷
ø

Mk
   where   Mk = k
å
i=1 
(k-i+1) æ
ç
ç
è
k
i
ö
÷
÷
ø

Euler motivates methods 2 and 3 as progressively minimizing the impact of large prizes, corresponding to large values of i, on the bank. Curiously, he gives neither formulas nor even explanations for these weights, but simply tabulates the coefficients up to the case k=5. I am grateful to Prof. Stephen Bloch of my department for help in solving the riddle of how the coefficients in method 3 were arrived at.

To illustrate these methods, let us compare in Table 3 the values of a5,i in the three cases.

Method i=1 i=2 i=3 i=4 i=5
1
1

5


1

5


1

5


1

5


1

5

2
5

31


10

31


10

31


5

31


1

31

3
25

106


40

106


30

106


10

106


1

106

Table 3: coefficients for k=5


 

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