Here we go again
GillesD said:
Moses
The answers to your question are always the same: YES it could be possible that more people would win with an higher number of winning numbers than with a lower number of winning numbers.
But the real question should not be "Is it possible that..." but "Is it likely that… And to this question is definitely NO.
Let's take actual data from Canadian Lott 6/49. For 3 regular draws in 2009 (no rollover and about $4M in prizes), the data from Loto Quebec site provides this following information (all data is a rounded average for the 3 draws):
- tickets sold: 7,102,385
- number of winners (5 / 6): 110
- number of winners (4 / 6): 6400
- number of winners (3 / 6): 119,500
So it could be possible that more people bet with 4 of the 6 winners than those with 5 winners but unlikely. Why? With such an high volume of tickets, bought independently by many, many people, each with its own system (random picks, hot numbers, cold numbers, etc.), then the distribution of numbers chosen is most likely uniform and the higher number of winners will be with 3 out 6, then 4 out 6, then …
It is very, very unlikely to happen with 3/6 winners, 4/6 winners and 5/6 winners but if you check the data for Nov. 11th, 2008, you will find 6 winners for the 6/6 prize and 0 (yes zero) for the 5+/6 prize. So here you have it.
I have been asking this question for sometime but no-one ever answered my question in mathematical terms! I am not interested why is never happened as I know the lottery system is corrupt but what I am looking for is how we can calculate or workout the actual breakdown or possibility, can you understand this and can you help?
And anyways Gilles, how long is Canadian lottery going for and this situation just happened in Nov 2008? I think they must got my hint at last or somebody went and told them to do something about it!
Possible = Likely, can you tell me the difference please? The way I see it if there is a possibility then there is certainty to happen besides in your good example you compared 6 from 6 and 5+ from 6 again! We don’t have 5+/6 and it should be 5 from 7 (power ball, Bonus ball) and it just happened to ZERO with no payments to payout, what a clever move and now you can brag about it too!
What if this was happened in different prize category 3 and 4 or 3 and 5? The person get 5 numbers right will receive much lesser payment and that would have cause some war, wouldn’t it?
This is what is all about the organisers pocketing money due to people negligence!
What will you do Gilles D if you had 5 numbers right and get $10 and you next door neighbour get 3 numbers and wins more than you? Don’t say it is possible BUT unlikely plllllllllllllllllllllllease!
Now, this is my view which I am working on it and I know that my approach is correct but welcome other views!
If these six numbers are the lottery numbers then
01 02 03 04 05 06
Matching any three from 6 equals 20 wheels or possibilities
01 02 03
01 02 04
01 02 05
01 02 06
01 03 04
01 03 05
01 03 06
01 04 05
01 04 06
01 05 06
02 03 04
02 03 05
02 03 06
02 04 05
02 04 06
02 05 06
03 04 05
03 04 06
03 05 06
04 05 06
There are big possibilities that some of the wheels above get saturated by the players and some never get used at all and the only people would know the inside information is the Camelot or the lottery organiser and no-one else!
Matching any four from 6 equals 10 wheels or possibilities
01 02 03 04
01 02 03 05
01 02 03 06
01 03 04 05
01 03 04 06
01 04 05 06
02 03 04 05
02 03 04 06
02 03 05 06
02 04 05 06
Possibility; more players cover 10 wheels than 20 wheels above
Matching any five from 6 equals 6 wheels or possibilities
01 02 03 04 05
01 02 03 04 06
01 02 03 05 06
01 02 04 05 06
01 03 04 05 06
02 03 04 05 06
Possibility; more players cover 6 wheels than 10,20 wheels above
Matching all six from 6 equal 1 wheels and no other possibilities
01 02 03 04 05 06
Possibility; more players cover 1 wheel than 6,10,20 wheels above
Moses