Small math help?

[tomtom]

Well, got a flu, so a small math problem looks like a big one....
A 6 numbers combination covers 20 triples. Winning 7 numbers make 35 triples. What are chances to have at least one triple in these 6 numbers, which is either3/6 or 2+b? Actually, what are chances for either 3 or 2+b?

 

[Teufellj]

RE: TOMTOM

Hello,Tomtom,

In my opinion, your chances are "50/50" because of the randyness of random numbers. Algebraic configurations work out great on paper or a calculator but the real world of "ping-pong" type plays and electronic sequencing only works at best, a few times to get to the proposed set(s) of numbers indicated.
I too, have the flu (stomach) and it ain't no picnic, is it?

Teufellj...

"If At First You Don't Succeed, Get A Hammer And Grey Tape!"

 

[tomtom]

Re: RE: TOMTOM

In my opinion, your chances are "50/50" because of the randyness of random numbers.

It seems it's a bit difficult question, but it is far from the 50% for sure...you may expect to have one 3/6 among 55 - 60 tickets without counting on luck that much, and the 2 +B odds are a little bit higher…however, doing some " under the flu" math I'm getting something like 3.8 %...which means one among 26 - 27 tickets should have either 3/6 or 2+B...

 

[iago31416]

Assuming you are drawing from 49 numbers, for 3/6, your odds are 1/57... for 2+/6, your odds are 1/81..... since winning one category precludes you from winning the other, your odds of (3/6 or 2+/6) is simply (1/57 + 1/81) = around 3.0%.

This means you would expect to have one of (3/6 or 2+/6) out of every 33-34 tickets, provided no triple appears in more than 1 ticket.