On Sun, 15 Jun 2014, montgrimpulo wrote:

> this is a placeholder of the task which I am trying to solve in Ada.
> As you may know in German Lotto there are 49 numbers 1..49
> from which you have to select 6 numbers to place a bet.
>
> First Round:
> By defining a subtype valid for positive numbers from 1 .. 49,
> the use of Ada.Discrete_Random, and the definition of an appropriate Generator
> you may get a random number in the range of 1 .. 49.

Call that number R_1.

> Second Round:
> Here does my problem start. Now I have a set of numbers where
> one number - which was randomly selected in the first round -
> is missing. How do I use the random function to select another
> random number out of the rest ?

There are a couple of different solutions.

First Solution (in pseudo-code, not in Ada):

repeat choose a random R_2 between 1 and 49 
until R_2 /= R_1.

Third round: Choose a random R_3 between 1 and 49
until R_3 /= R_1 and R_3 /= R_2.

ect.

Second Solution:

choose a random R_2 between 1 and 48
If R_2=R_1 then R_1 := 49

choose a random R_3 between 1 and 47
If R_3=R_1 then R_3 := 48
elsif R_3=R_2 then R_3 := 49

ect.

Now, for the second solution, you need random numbers from different 
intervals. In principle, you can (2a) instantiate a new generator from 
Ada.Numerics.Discrete_Random for each of the R_i. The alternative is to 
(2n) instantiate a single random generator (I've forgotten the package 
name, but it is not A.N.Discrete_Random), which just generates a Float 
F_Rand between 0 and 1. Then set R_I := Truncate(1 + F_Rand * (50-I)).

> A workaround would be
> to test in each round, if that number has already been selected.
> However, the probability would not be the same as with a reduced
> set as the selection would be always from the full set.

That sounds like my first solution. If you repeat the random choice until 
you R_i is different from all the R_(i-1), ..., R_1, the probability is 
exactly the probability you expect. (Well, assuming the random generator 
is good.)



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