Looking over Canada's Western Max lottery (7/50), a combinations calculator shows the chance of getting 7/7 with 50 numbers is 1 in 99,884,400. However, that's if you only had 1 selection per ticket. Since there are 3 selections printed per ticket, their website accurately shows the chance is 3 times better (99,884,400/3 = 33,294,800). Here's the Western Max official website:
http://www.wclc.com/games/western-max.htm
Along with the main draw with a jackpot of 2 million dollars, there are also 14 additional draws of 1 million dollars each. Their website shows that winning 1 of the 14 additional draws has an equal chance of 1 in 33,294,800 just as the main draw. This gives us 15 total possible jackpots.
As an example, let's say your ticket has these 3 selections:
1, 21, 23, 24, 35, 36, 47
7, 18, 19, 21, 32, 33, 43
8, 15, 16, 27, 38, 49, 50
Now, this is where my question comes in: Say you were extremely lucky to have matched 7/7 for not 1 but all 3 of those combinations on your ticket… since we already know that matching 1 of 3 selections on your ticket are 1 in 33,294,800 – my question is what are the odds of matching 2 out of 3 selections and even more incredibly, matching 7/7 on all 3 out of 3 selections on your ticket from the 15 possible jackpots?
Best Answer
However trivial it may seem, the probability of getting 2 is just 1 in 33,294,800^2, or $\frac{1}{1108543707040000}$. Similarly, for getting 3, it's 1 in 33,294,800^3, or $\frac{1}{36908741017155392000000}$.
Why? Each probability is $\frac{1}{33294800}$, and each set of numbers (in your example) doesn't overlap, so we can safely square and cube the fraction. If a number were to overlap, that would be a different story.