Question 1195519
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In the 6/49 lotto game, suppose that the minimum prize is 20 pesos, 
which you can win by guessing 3 out of 6 numbers correctly. 
The prize goes up as you guess more numbers correctly. 
What is the probability of not winning prizes of any amount?
~~~~~~~~~~~~~~~~~~~~~


Surely, such problem should start with clear formal description of this
lotto game - - - but this one does not.


Therefore, I will give my own description, as I understand the problem.


<pre>
      Six numbers are drawn from a set of 49 different numbers 1,2,3,...,49.
      If a ticket matches at least three numbers, the ticket wins.
      The prize goes up as you guess more numbers correctly. 
      What is the probability of not winning prizes of any amount?
</pre>

<H3>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Solution</H3>

<pre>
First, let's calculate the number of all issued/(possibly existing) tickets.


The total number of all issued/(possibly existing) tickets is the number 
of combinations of 49 items taken 6 at a time. It is the number of combinations

    {{{C[49]^6}}} = {{{(49*48*47*46*45*44)/(1*2*3*4*5*6)}}} = 13983816.



Favorable (winning) tickets are those that carry at least 3 of 6 different drawn numbers.

In accordance to it, <U>UNWINNING</U> tickets are those that 

    - (a) carry zero of 6 drawn numbers;
    
    - (b) carry only one of 6 drawn numbers;

    - (c) carry only two of 6 drawn numbers.


So, we divide 49 numbers in two categories: 6 selected (drawn) numbers and the rest 49-6 = 43 numbers.


In category (a), we have  {{{C[43]^6}}}  tickets;

In category (b), we have  {{{C[6]^1*C[43]^5}}}  tickets;

In category (c), we have  {{{C[6]^2*C[43]^4}}}  tickets.


    +---------------------------------------------------------------------+
    |    Notice that the order of numbers in tickets does not matter -    |
    |               - therefore we use combinations.                      |
    +---------------------------------------------------------------------+


These three categories are DISJOINT - - - therefore, the total of UNWINNING tickets is the sum

    {{{C[43]^6}}} + {{{C[6]^1*C[43]^5}}} + {{{C[6]^2*C[43]^4}}} = 6096454 + 6*962598 + 15*123410 = 13723192.


Now the last step to find the probability of getting unwinning ticket is this fraction

    P = {{{unwinnig_tickets/total_tickets}}} = {{{13723192/13983816}}} = 0.981362455.


<U>ANSWER</U>.  The probability of getting unwinning ticket is 0.98126 (rounded).
</pre>

Solved.


It is a good problem of an advanced level.