Help calculating odds

[Loster]

Hello everyone. My first post.

I am starting a syndicate (volunteer run and no profits taken btw) based on the Virtual World Direct approach (NOT affiliated with them, just using the same system), and I need some help calculating the full and actual odds of that system.

OK, so, as you may know, the VWD/e-Lottery system works by entering 44 tickets per syndicate; five number on every ticket are the same and the sixth is each consecutive remaining number.

This wikipedia article, "Lottery Mathematics," (I can't post links yet!) details how to calculate the odds for a 49-ball, 6-balls drawn lottery, which are:

3 1 : 55.65593
4 1 : 1,031.397
5 1 : 55,490.33
5 + Bonus ball 1 : 2,330,365
6 1 : 13,983,815


Calculating odds based when balls fall in consecutive line

Now, because that sixth number is effectively taken out of the equation with this system (every ball drawn will appear somewhere on one of the fourty four tickets) I have been assuming that to calculate the odds you just take the odds of a 49-ball, 5-balls-drawn lottery and take the number of matched balls minus 1. Example: The odds of matching 3 balls across the 44 tickets is the odds of:

matching 2 balls in a 5-balls-drawn, 49-ball lottery. Using this you get:


3 1 : 14.4
4 1 : 201.57
5 1 : 8667.65
5 + Bonus ball 1 : 381,376.8
6 1 : 1,906,884

(I used an online "Lottery Odds Calculator" for this)


That makes this system look quite favourable - though of course the payout is always split. Under a £2Mill jackpot, each get ~£47,000. BUT, here's where it gets complicated.


The Repeated Payouts thing

Those odds above are the ones you get when one of the matched numbers is in that sixth position - the consecutive number position. AND it's important to realise that when you match 4 numbers in this way, you also get fourty three 3 number matches as well. If you match 5 numbers on one ticket, you also get fourty three 4 number matches too. So you do get the benefit of repeating payouts too. I'm assuming that this phenomenon only affects the payout and not the odds (it's still 14.4-to-1 to get 3, 201.57-to-1 to get four etc etc)


The odds when balls fall in repeating numbers

What I CANNOT for the life of me work out is what are the odds of matching 3, 4, or 5 balls just on the numbers that don't change from ticket to ticket. Are they the same as the normal odds on the 6-ball lottery, so, like this:


3 (x44) 1 : 55.65593
4 (x44) 1 : 1,031.397
5 (x44) 1 : 55,490.33

(amounts in brackets are how many tickets you get these matches on)


Is this correct though? Are the odds different because you're only matching five balls (you're taking the sixth, consecutive, ball out of the equation)? Usually when you have 44 tickets your odds get multiplied by 44 don't they? But here, since the numbers are the same, that multiplier probably doesn't apply does it??


I guess then if I get the odds for numbers matched with the consecutive number and odds for numbers matched on the repeating numbers I can add them together to get the full odds of 3 balls matched, 4 balls matched etc etc. Can I??



ANYWAY, any help whatsoever with this would be appreciated. I have a graphic of the table I drew up of all this but this board won't let me post it yet.... I will in the replies if I can...

Any comments most welcome.

[Loster]

[Spot]

Interesting question; sorry I don't have an answer.

Would it help if we eliminate the odds of those 44 numbers that can never win any prizes by themselves (i.e. there are no combinations among the 44 numbers) ?

[Loster]

Hi Spot,

hmm, yes, that might be a good way to do it. Every time a number of 'five repeating' numbers comes out it surely influences the overall result in a large way compared with when the number if amongst the remaining 44.

Shall have to have a go at that and see what I come up with.

Carl.