Irish Lottery probability I'm currently trying to work out what the probability is of winning the Irish (daily) Lottery by picking two balls and am a bit stuck.
There are 39 balls in total, of which 6 balls are chosen as per The normal lottery. The player selects two balls, and if these are in the 6 chosen balls, they win.
In my head, I have the probability of these two balls winning as:
$$\frac{\binom{6}{2}}{\binom{39}{2}} = 0.02025\dots $$
Is this correct?
If so, am I right in saying that if I then put on 50 bets, that I am guaranteed to win? This doesn't seem right to me.
 A: Yes, your computation are correct. And yes, 50 bets are not enough to ensure a win, because each one have a little bit more that $2\%$ chance of winning, but they are not disjoint (several tickets among your 50 can win at the same time).
To have a bound on the minimal number of tickets required to ensure a win, you can consider the problem as a graph problem :


*

*Consider a complete graph (a clique) with 39 vertices and 741 edges.

*Now suppose that by removing $n$ edges, you obtain a graph $G'$ that is $K_6$ free (without a clique of size 6). It means that by buying each $n$ tickets (one for each such edge), you're sure to have at least a win for any 6 balls chosen.

*Use Turan's theorem to prove that $G'$ has at most 
$\frac{4}{10}.39^2=608 $ edges. Hence you should remove at least $741-608=133$ edges


So to ensure a win, you must at least buy 133 tickets. This is possible if you create a Turan graph $G'$ : Buy all tickets $(i,j)$ such that both $i$ and $j$ belongs to the same subset among the five next one :
$\{1,2,3,4,5,6,7,8\},\{9,10,11,12,13,14,15,16\},\{17,18,19,20,21,22,23,24\},\{25,26,27,28,29,30,31,32\},\{33,34,35,36,37,38,39\}$
A: Sorry if this is horribly organised and unclear.
the amount of pairs of numbers per ticket would be 15, as you can pair the first number with any of the other 5, the second with any of the other 4 (as the pair of the first and second has already been counted) and so on.
so 5+4+3+2+1. you can use:
$$ n(n+1)/2 \\
5*6/2 \\
15 $$
The total number of combinations of possible pairs to make would be by the same maths $$38+37+36+... \\
(38*39)/2=741 $$
So therefore the number of tickets needed to have every combination of numbers would surely be $$ 741/15=49.4 $$
which would be 50 tickets? feel like i'm missing something out
