DAC, the same mathematics apply if large numbers of people in a consortium play the same "game" making sure that they have chosen different numbers.
Taking your example above (which is true for one person) and accepting that fourth prize, four numbers, is a suitable target outcome, the chance that you will win at random is one in 1031.4 or, more or less, .0001 ... this means the chance you will not win is .9999 and each time you play those are your odds.
Now if I play all year, the chances that I will not win once are (.9999 x .9999 x .9999 etc ) and this probably brings it down a bit, let's say for the sake of argument to .99 ... in other words, I have a one in a hundred chance of winning this prize once a year.
But so do you and so does Red Dog etc etc. So it is like the birthday example, only with different odds so it depends on whether you're saying, how many times until somebody hits a fourth, or how many will win a fourth during the year?
If you think about it, you'll see what I mean.
Of course, it is unlikely that anyone will hit the top prize without quite a large group participating, but my guess (a very difficult computation is required) is that if 3,000 people bought the same numbers every time, it would take them five years before somebody won the grand prize. For perhaps second prize money of about $100,000 it would likely take the same group less than a year.
That's what Red Dog had in mind in general terms, I think.
The sports lottery works on somewhat the same principle. With a group of people, and assuming you don't care if you lose small amounts but you want somebody in the group to win one large amount, you are far better off to bet the longshots all the time, because they will pay off a few times and hit larger multiples than the more certain favourites with their low odds.


