spanish lotto

[spaneurofor]

hi all,

nice forum here.
first time for me.

2 Q :

would anybody be able to calculate the following probabilities

in a 6/49 situation;

I kindly want to know HOW MANY tickets would i need to purchase in the following cases in order to get the 6 numbers out of 49:

a) ANY 4 consecutive numbers + any other 2

b) ONE 4 consecutive numbers + any other 2

many thanks

 

[Nick Koutras]

Case 1, any 4 consecutive and 2 others:
The first consecutive set is 1,2,3,4
the second is 2,3,4,5 ...and the last is
46,47,48,49.
So you have 46 cases of 4 consecutive.
The two other numbers are from
a field of (49-4)=45 and can be any combination of two such numbers= (45c2)=(45*44)/(1*2)=990

So this case has 45*990 =44550 combinations.

Case 2, one 4 consecutive and 2 others=990 as explained for a single case above..